Environmental Physics: Sustainable Energy and Climate Change

Environmental Physics

Environmental Physics Sustainable Energy and Climate Change

Third Edition

EGBERT BOEKER and RIENK VAN GRONDELLE VU University Amsterdam

A John Wiley & Sons, Ltd., Publication

This edition first published 2011 © 2011 John Wiley & Sons, Ltd Registered office John Wiley & Sons Ltd, The Atrium, Southern Gate, Chichester, West Sussex, PO19 8SQ, United Kingdom For details of our global editorial offices, for customer services and for information about how to apply for permission to reuse the copyright material in this book please see our website at www.wiley.com. The right of the author to be identified as the author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, except as permitted by the UK Copyright, Designs and Patents Act 1988, without the prior permission of the publisher. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic books. Designations used by companies to distinguish their products are often claimed as trademarks. All brand names and product names used in this book are trade names, service marks, trademarks or registered trademarks of their respective owners. The publisher is not associated with any product or vendor mentioned in this book. This publication is designed to provide accurate and authoritative information in regard to the subject matter covered. It is sold on the understanding that the publisher is not engaged in rendering professional services. If professional advice or other expert assistance is required, the services of a competent professional should be sought. The publisher and the author make no representations or warranties with respect to the accuracy or completeness of the contents of this work and specifically disclaim all warranties, including without limitation any implied warranties of fitness for a particular purpose. This work is sold with the understanding that the publisher is not engaged in rendering professional services. The advice and strategies contained herein may not be suitable for every situation. In view of ongoing research, equipment modifications, changes in governmental regulations, and the constant flow of information relating to the use of experimental reagents, equipment, and devices, the reader is urged to review and evaluate the information provided in the package insert or instructions for each chemical, piece of equipment, reagent, or device for, among other things, any changes in the instructions or indication of usage and for added warnings and precautions. The fact that an organization or Website is referred to in this work as a citation and/or a potential source of further information does not mean that the author or the publisher endorses the information the organization or Website may provide or recommendations it may make. Further, readers should be aware that Internet Websites listed in this work may have changed or disappeared between when this work was written and when it is read. No warranty may be created or extended by any promotional statements for this work. Neither the publisher nor the author shall be liable for any damages arising herefrom. Library of Congress Cataloging-in-Publication Data Boeker, Egbert. Environmental physics : sustainable energy and climate change / Egbert Boeker and Rienk van Grondelle. – 3rd ed. p. cm. Includes bibliographical references and index. ISBN 978-0-470-66675-3 (cloth) – ISBN 978-0-470-66676-0 (pbk.) 1. Environmental sciences. 2. Physics. 3. Atmospheric physics. I. Grondelle, Rienk van. II. Title. GE105.B64 2011 628–dc22 2011011525 A catalogue record for this book is available from the British Library. Print ISBN: Cloth 978-0-470-66675-3, Paper 978-0-470-66676-0 ePDF ISBN: 978-1-119-97418-5 oBook ISBN: 978-1-119-97417-8 ePub ISBN: 978-1-119-97519-9 Mobi ISBN: 978-1-119-97520-5 Set in 10/12pt Times by Aptara Inc., New Delhi, India.

Contents

Preface Acknowledgements 1

2

3

xiii xv

Introduction 1.1 A Sustainable Energy Supply 1.2 The Greenhouse Effect and Climate Change 1.3 Light Absorption in Nature as a Source of Energy 1.4 The Contribution of Science: Understanding, Modelling and Monitoring Exercises References

1

Light and Matter 2.1 The Solar Spectrum 2.1.1 Radiation from a Black Body 2.1.2 Emission Spectrum of the Sun 2.2 Interaction of Light with Matter 2.2.1 Electric Dipole Moments of Transitions 2.2.2 Einstein Coefficients 2.2.3 Absorption of a Beam of Light: Lambert-Beer’s Law 2.3 Ultraviolet Light and Biomolecules 2.3.1 Spectroscopy of Biomolecules 2.3.2 Damage to Life from Solar UV 2.3.3 The Ozone Filter as Protection Exercises References

7

Climate and Climate Change 3.1 The Vertical Structure of the Atmosphere 3.2 The Radiation Balance and the Greenhouse Effect 3.2.1 Simple Changes in the Radiation Balance 3.2.2 Radiation Transfer 3.2.3 A Simple Analytical Model 3.2.4 Radiative Forcing and Global Warming 3.2.5 The Greenhouse Gases

1 3 4 5 6 6

7 7 9 12 12 14 16 19 20 21 22 28 28

31 32 36 39 41 44 45 48

vi

4

Contents

3.3 Dynamics in the Climate System 3.3.1 Horizontal Motion of Air 3.3.2 Vertical Motion of Ocean Waters 3.3.3 Horizontal Motion of Ocean Waters 3.4 Natural Climate Variability 3.5 Modelling Human-Induced Climate Change 3.5.1 The Carbon Cycle 3.5.2 Structure of Climate Modelling 3.5.3 Modelling the Atmosphere 3.5.4 A Hierarchy of Models 3.6 Analyses of IPCC, the Intergovernmental Panel on Climate Change 3.7 Forecasts of Climate Change Exercises References

51 53 58 59 59 62 63 66 67 70

Heat Engines 4.1 Heat Transfer and Storage 4.1.1 Conduction 4.1.2 Convection 4.1.3 Radiation 4.1.4 Phase Change 4.1.5 The Solar Collector 4.1.6 The Heat Diffusion Equation 4.1.7 Heat Storage 4.2 Principles of Thermodynamics 4.2.1 First and Second Laws 4.2.2 Heat and Work; Carnot Efficiency 4.2.3 Efficiency of a ‘Real’ Heat Engine 4.2.4 Second Law Efficiency 4.2.5 Loss of Exergy in Combustion 4.3 Idealized Cycles 4.3.1 Carnot Cycle 4.3.2 Stirling Engine 4.3.3 Steam Engine 4.3.4 Internal Combustion 4.3.5 Refrigeration 4.4 Electricity as Energy Carrier 4.4.1 Varying Grid Load 4.4.2 Co-Generation of Heat and Electricity 4.4.3 Storage of Electric Energy 4.4.4 Transmission of Electric Power 4.5 Pollution from Heat Engines 4.5.1 Nitrogen Oxides NOx 4.5.2 SO2 4.5.3 CO and CO2

77

70 70 74 76

78 79 82 82 83 84 87 90 91 91 95 97 98 101 103 103 104 105 107 110 113 114 115 117 123 125 125 126 126

Contents

5

vii

4.5.4 Aerosols 4.5.5 Volatile Organic Compounds VOC 4.5.6 Thermal Pollution 4.5.7 Regulations 4.6 The Private Car 4.6.1 Power Needs 4.6.2 Automobile Fuels 4.6.3 Three-Way Catalytic Converter 4.6.4 Electric Car 4.6.5 Hybrid Car 4.7 Economics of Energy Conversion 4.7.1 Capital Costs 4.7.2 Learning Curve Exercises References

127 128 129 129 129 130 131 132 133 134 134 134 138 138 142

Renewable Energy 5.1 Electricity from the Sun 5.1.1 Varying Solar Input 5.1.2 Electricity from Solar Heat: Concentrating Solar Power CSP 5.1.3 Direct Conversion of Light into Electricity: Photovoltaics PV 5.2 Energy from the Wind 5.2.1 Betz Limit 5.2.2 Aerodynamics 5.2.3 Wind Farms 5.2.4 Vertical Wind Profile 5.2.5 Wind Statistics 5.2.6 State of the Art and Outlook 5.3 Energy from the Water 5.3.1 Power from Dams 5.3.2 Power from Flowing Rivers 5.3.3 Power from Waves 5.3.4 Power from the Tides 5.4 Bio Energy 5.4.1 Thermodynamics of Bio Energy 5.4.2 Stability 5.4.3 Solar Efficiency 5.4.4 Energy from Biomass 5.5 Physics of Photosynthesis 5.5.1 Basics of Photosynthesis 5.5.2 Light-Harvesting Antennas 5.5.3 Energy Transfer Mechanism 5.5.4 Charge Separation 5.5.5 Flexibility and Disorder 5.5.6 Photoprotection 5.5.7 Research Directions

145 146 146 150 152 159 160 162 165 165 167 168 169 169 170 170 174 175 175 180 180 182 183 184 185 187 190 193 193 195

viii

6

7

Contents

5.6 Organic Photocells: the Gr¨atzel Cell 5.6.1 The Principle 5.6.2 Efficiency 5.6.3 New Developments and the Future 5.6.4 Applications 5.7 Bio Solar Energy 5.7.1 Comparison of Biology and Technology 5.7.2 Legacy Biochemistry 5.7.3 Artificial Photosynthesis 5.7.4 Solar Fuels with Photosynthetic Microorganisms: Two Research Questions 5.7.5 Conclusion Exercises References

196 196 199 202 203 203 204 207 209

Nuclear Power 6.1 Nuclear Fission 6.1.1 Principles 6.1.2 Four Factor Formula 6.1.3 Reactor Equations 6.1.4 Stationary Reactor 6.1.5 Time Dependence of a Reactor 6.1.6 Reactor Safety 6.1.7 Nuclear Explosives 6.2 Nuclear Fusion 6.3 Radiation and Health 6.3.1 Definitions 6.3.2 Norms on Exposure to Radiation 6.3.3 Normal Use of Nuclear Power 6.3.4 Radiation from Nuclear Accidents 6.3.5 Health Aspects of Fusion 6.4 Managing the Fuel Cycle 6.4.1 Uranium Mines 6.4.2 Enrichment 6.4.3 Fuel Burnup 6.4.4 Reprocessing 6.4.5 Waste Management 6.4.6 Nonproliferation 6.5 Fourth Generation Nuclear Reactors Exercises References

221

Dispersion of Pollutants 7.1 Diffusion 7.1.1 Diffusion Equation 7.1.2 Point Source in Three Dimensions in Uniform Wind 7.1.3 Effect of Boundaries

261

213 213 215 217 222 222 226 229 231 233 234 237 238 244 244 245 247 247 247 248 249 249 252 252 253 256 257 258 259 262 262 267 269

Contents

8

ix

7.2 Dispersion in Rivers 7.2.1 One-Dimensional Approximation 7.2.2 Influence of Turbulence 7.2.3 Example: A Calamity Model for the Rhine River 7.2.4 Continuous Point Emission 7.2.5 Two Numerical Examples 7.2.6 Improvements 7.2.7 Conclusion 7.3 Dispersion in Groundwater 7.3.1 Basic Definitions 7.3.2 Darcy’s Equations 7.3.3 Stationary Applications 7.3.4 Dupuit Approximation 7.3.5 Simple Flow in a Confined Aquifer 7.3.6 Time Dependence in a Confined Aquifer 7.3.7 Adsorption and Desorption of Pollutants 7.4 Mathematics of Fluid Dynamics 7.4.1 Stress Tensor 7.4.2 Equations of Motion 7.4.3 Newtonian Fluids 7.4.4 Navier-Stokes Equation 7.4.5 Reynolds Number 7.4.6 Turbulence 7.5 Gaussian Plumes in the Air 7.5.1 Statistical Analysis 7.5.2 Continuous Point Source 7.5.3 Gaussian Plume from a High Chimney 7.5.4 Empirical Determination of the Dispersion Coefficients 7.5.5 Semi-Empirical Determination of the Dispersion Parameters 7.5.6 Building a Chimney 7.6 Turbulent Jets and Plumes 7.6.1 Dimensional Analysis 7.6.2 Simple Jet 7.6.3 Simple Plume Exercises References

270 271 275 277 278 280 281 282 282 283 286 290 295 298 301 302 304 304 308 309 310 311 313 317 319 321 322 323 324 325 326 328 329 331 333 334

Monitoring with Light 8.1 Overview of Spectroscopy 8.1.1 Population of Energy Levels and Intensity of Absorption Lines 8.1.2 Transition Dipole Moment: Selection Rules 8.1.3 Linewidths 8.2 Atomic Spectra 8.2.1 One-Electron Atoms 8.2.2 Many-Electron Atoms

337 337 341 341 342 345 345 346

x

9

Contents

8.3 Molecular Spectra 8.3.1 Rotational Transitions 8.3.2 Vibrational Transitions 8.3.3 Electronic Transitions 8.4 Scattering 8.4.1 Raman Scattering 8.4.2 Resonance Raman Scattering 8.4.3 Rayleigh Scattering 8.4.4 Mie Scattering 8.4.5 Scattering in the Atmosphere 8.5 Remote Sensing by Satellites 8.5.1 ENVISAT Satellite 8.5.2 SCIAMACHY’s Operation 8.5.3 Analysis 8.5.4 Ozone Results 8.6 Remote Sensing by Lidar 8.6.1 Lidar Equation and DIAL 8.6.2 Range-Resolved Cloud and Aerosol Optical Properties Exercises References

347 347 349 353 359 359 360 361 362 362 362 362 362 364 368 368 369 371 376 377

The Context of Society 9.1 Using Energy Resources 9.1.1 Energy Consumption 9.1.2 Energy Consumption and Resources 9.1.3 Energy Efficiency 9.1.4 Comparing Energy Resources 9.1.5 Energy Options 9.1.6 Conclusion 9.2 Fresh Water 9.3 Risks 9.3.1 Small Concentrations of Harmful Chemicals 9.3.2 Acceptable Risks 9.3.3 Small Probability for a Large Harm 9.3.4 Dealing with Uncertainties 9.4 International Efforts 9.4.1 Protection of the Ozone Layer 9.4.2 Protection of Climate 9.5 Global Environmental Management 9.5.1 Self-Organized Criticality 9.5.2 Conclusion 9.6 Science and Society 9.6.1 Nature of Science 9.6.2 Control of Science 9.6.3 Aims of Science 9.6.4 A New Social Contract between Science and Society

379 380 380 382 383 384 387 388 389 389 390 392 393 394 396 396 396 398 398 401 401 401 402 402 404

Contents

Exercises and social questions Social questions References Appendix A: Physical and Numerical Constants Appendix B: Vector Algebra Appendix C: Gauss, Delta and Error Functions Appendix D: Experiments in a Student’s Lab Appendix E: Web Sites Appendix F: Omitted Parts of the Second Edition Index

xi

405 405 406

409 411 419 423 425 427 429

Preface

This third edition of the textbook ‘Environmental Physics’ has been thoroughly revised to give more focus on sustainable energy and climate change. As fossil fuels and nuclear power will be with us for many years to come, the physical and environmental aspects of these ways of energy conversion are given ample attention as well. The textbook is suitable for second year students in physics and related subjects as physical chemistry and geophysics. It assumes a basic knowledge in physics and mathematics, but all equations are derived from first principles and explained in a physical way. Therefore the book may serve as an introduction to physics in the context of societal problems like energy supply, pollution, climate change and finite resources of fossil fuels and uranium. Even where parts of the text will be familiar from other courses it is advisable to read those parts in order to grasp the way the material is presented here. At some places we included parts that may be too ‘heavy’ for most second year students, but will not easily be found in the literature. They will be suitable for the final undergraduate year or for a Master course. They are indicated by ↓ at the beginning and with ↑ at the end. Usually we will indicate them in the introductory lines of each section as well. As in any text, later chapters refer to concepts and properties discussed in earlier chapters. But the book is written such that the reader may read any chapter on its own. Therefore the teacher may select a few chapters as a module or for background reading. A distinguishing feature of the text is the discussion of spectroscopy and spectroscopic methods, again from basic concepts, as a crucial means to quantitatively analyze and monitor the condition of the environment, the factors determining climate change and all aspects of energy conversion. The emphasis in the book is on physics, on the concepts and principles that help in understanding the ways to produce energy efficiently or to mitigate climate change. Extra attention is given to photosynthesis, not only because of its importance in the field of renewable energy, but also because a comprehensive physics approach is lacking in the literature. With regard to international treaties and conventions, we discuss the most important one for the subject matter of this book: the climate convention (Chapter 9) and more briefly the non-proliferation treaty (Chapter 6). The actual political situation is heavily influenced by the internal situation in major actors like the United States, Russia and China. That may change rapidly. As the need for a guaranteed energy supply is a constant in international relations, this book should give the reader enough background to judge the policy which his or her country is putting forward.

xiv

Preface

The structure of the book follows its emphasis. After an introduction, solar radiation is discussed as the input for most of the renewable energies (Chapter 2), next the factors influencing climate change (Chapter 3), then energy from fossil fuels with an excursion to the private car (Chapter 4) and renewable energy (Chapter 5). Because of its importance and the ongoing discussion of safety we have devoted a separate chapter to nuclear power (Chapter 6). Most if not all ways of energy conversion produce pollution as an inevitable side effect; therefore we discuss transport of pollutants in Chapter 7, albeit less detailed than in the second edition of this book. Monitoring is the subject of Chapter 8 and, finally, in Chapter 9 the social aspects are discussed. In the Appendices some helpful information is compiled. The changed emphasis of this third edition results in many new sections and paragraphs and the omission or condensing of several sections of the earlier editions. For the interested reader these parts of the second edition may be downloaded free of charge from the website http://www.few.vu.nl/environmentalphysics. From this website also a description of environmental experiments for a student’s lab may be downloaded, described briefly in Appendix D. The site also contains a few computer codes that will illuminate some points made in the text. Finally, any mistakes or omissions we discover will be put on the site as well. Almost all equations in this book will be derived from first principles. After a, perhaps lengthy, derivation we often analyze the resulting formula by looking at the units on both sides of the equality sign (=). This provides more physical insight than comparing dimensions length, mass and time only. This same procedure was followed in the authors’ book ‘Environmental Science, Physical Principles and Applications’ which is meant for a general science readership; there it worked very well. For clarity, units are given between square brackets [ ]; also dimensions are given between these brackets. In practice no confusion will arise. Finally we mention some practical points. References are given between square brackets [ ] and printed at the end of each chapter. Also exercises are put there and for teachers the publisher has a teachers’ manual available with the worked out solutions. In the final chapter ‘social questions’ ask for a considered opinion of the student. The student of course is entitled to have his or her own opinion, so there is no ‘right’ answer. But the arguments should be physically sound. In the teacher’s manual some arguments on a certain statement will be given, without the pretention to present the ultimate truth.

Amsterdam, 20 January 2011

Acknowledgements

The authors acknowledge help and suggestion as follows: Chapter 2: Dr Annemarie Huijser for her contribution on melanins. Chapter 3: Dr Schuurmans for his suggestions on the first edition of the book and for suggesting table 3.5. The authors are indebted to Dr Aad van Ulden for his suggestions; dr Rob van Dorland has provided a simple model for radiative transfer, discussed in Sect. 3.2.2 which is grateful acknowledged. IPCC staff kindly provided us with high resolution versions of Plates 1 and 2. Chapter 4: Dr Andriesse for pointing out the real heat engine of Sect. 4.2.3 Chapter 5: Dr Jo Hermans for suggesting an alternative explanation for the Lift force in Sect. 5.2.2. We are grateful to Drs Barzda, Cisek, Roszak, Nield, Nelson, Schulten, Magnuson, Cogdell, T. Moore, Dekker, Boekema, Novoderezhkin and Hambourger for providing us with high resolution pictures of their work, which we used in our colour plates. Chapter 6: Dr Hugo van Dam for valuable suggestions to improve the text of the second edition; Dr Adri Bos for suggestions on sect. 6.3.2; Dr Frank Barnaby for suggestions on sects. 6.1.7 and 6.4.6; Dr Wim Hogervorst for his helpful contribution to Sect. 6.4.2; Dr Joern Harry for helping us with a description of the Japan Fukushima calamity, which was added in proof. Chapter 8: Dr Ilse Aben for introducing us to the secrets of Sciamachy and correcting our draft; Dr Apituley and Dr Donovan for providing a text and illustrations for backscatter lidar of sect. 8.6.2 Chapter 9: Dr John Grin for his valuable social insights Prof Karel Wakker is gratefully acknowledged for his careful reading of chapters 1 to 4 and his many suggestions. The authors are pleased with the way in which Dr Bart van Oort and Ademir Arapovic patiently have drawn and redrawn the many figures in this book and acknowledge the help of Joris Snellenburg with the colour plates.

1 Introduction Physical science is a fascinating subject. Mechanics, quantum physics, electrodynamics, to name a few, present a coherent picture of physical reality. The present book aims at inspiring students with the enthusiasm the authors experienced while working in the field. The work of the physical scientist always takes place in a social context. Ultimately, the scientist has to contribute to society, sometimes by increasing our knowledge of fundamental processes, more often by employing his or her skills in industry, a hospital, a consultancy firm or teaching. In the majority of cases the scientist contributes by tackling societal problems with a physics aspect or by educating students in understanding the strengths and limitations of the physics approach. The text Environmental Physics focuses on two problems where physical scientists can contribute to make them manageable. The first is the need for a safe and clean supply of energy now and in the future, the second is the way to deal with the forecasted climate change. The major part of the text deals particularly with the physics aspects of these two problems. A brief discussion of the social context is given below with a section on the contribution of science. Science can point out the physical consequences of political choices, or of not making choices, but the decisions themselves should be taken through the political process. A more comprehensive discussion of the societal context is given in the last chapter of this book.

1.1

A Sustainable Energy Supply

The concept ‘sustainable development’ became well known by the work of the World Commission on Environment and Development, acting by order of the General Assembly of the United Nations. In 1987 it defined sustainable development as ([1], p. 8): Meeting the needs of the present generations without compromising the ability of future generations to meet their own needs. Environmental Physics: Sustainable Energy and Climate Change, Third Edition. Egbert Boeker and Rienk van Grondelle. © 2011 John Wiley & Sons, Ltd. Published 2011 by John Wiley & Sons, Ltd.

2

Environmental Physics

This is not a physics definition, as the meaning of ‘needs’ is rather vague. Does it imply an expensive car, a motorboat and a private plane for everybody? The definition leaves this open. In the political arena the precise meaning of ‘needs’ is still to be decided. Still, the concept forces one to take into account the needs of future generations and rejects squandering our resources. Indeed, sustainable development, the World Commission emphasized, implies that we should be careful with natural resources and protect the natural environment. Since 1987 many governments have put the goal of a ‘sustainable society’ in their policy statements. Besides protection of the environment and a safe energy supply it then comprises objectives like good governance, social coherence, a reasonable standard of living. In this book we focus on a sustainable energy supply and adapt the 1987 definition as follows: A sustainable energy supply will meet the energy needs of the present generations without compromising the ability of future generations to meet their own energy needs. The environmental consequences of energy conversion should be such that present and future generations are able to cope.

Like with the previous definition the precise meaning of this statement is the subject of political debate. From a physics point of view one may make the following comments: 1. An energy supply based on fossil fuels is not sustainable. The resources of coal, oil and gas are limited, as will be illustrated in Chapter 9. So in time other energy sources will be required. In the meantime the environmental consequences of fossil fuel combustion should be controlled. 2. Renewable energy sources like solar energy, wind energy or bio fuels may be sustainable. Their ultimate source, solar irradiation, is inexhaustible on a human time scale. To be sure of the sustainability of renewable energy sources, one has to perform a lifecycle analysis: analyse the use of energy and materials of the equipment and their environmental consequences from cradle to grave. This book will provide building blocks for such an analysis. 3. It is under debate whether nuclear fission power is sustainable. The resources of 235 U, the main nuclear fuel, are large, but limited. Also, during the fission process many radioactive materials are produced. Proponents of nuclear power argue that most of these ‘waste’ materials may be used again as fuel and the remainder may be stored; in practice, it is claimed, nuclear fission power would be ‘virtually sustainable’. Power from nuclear fusion may be sustainable, but its commercial exploitation is still far off. Governments all over the world are stimulating renewable energies. Not only because of their sustainability. Another strong reason is the security of energy supply. This requires diversification of energy sources. Fossil fuels, especially oil and gas, are unevenly distributed over the world. Industrial countries do not want to be too much dependent on the willingness of other countries to supply them with oil and gas. One may put forward that solar irradiation is unevenly distributed as well, but even at moderate latitudes the irradiation is substantial and the wind blows everywhere. Apart from these considerations, the combustion of fossil fuels produces CO2 , which has climatic consequences, to be discussed in the next section.

Introduction

aS

R

S 1 [m2]

3

Earth

Figure 1.1 Solar radiation is entering the atmosphere from the left, S [Wm−2 ]. A fraction a, called the albedo, is reflected back.

1.2

The Greenhouse Effect and Climate Change

In the simplest calculation the temperature of the earth is determined by the solar radiation coming in and the infrared (IR) radiation leaving the earth, or energy in = energy out

(1.1)

The amount of radiation entering the atmosphere per [m2 ] perpendicular to the radiation is called S, the total solar irradiance or solar constant in units [Wm−2 ] = [J s−1 m−2 ]. Looking at the earth from outer space it appears that a fraction a, called the albedo, is reflected back. As illustrated in Figure 1.1, an amount (1 − a)S penetrates down to the surface. With earth radius R the left of Eq. (1.1) reads (1 − a)Sπ R 2 . In order to make an estimate of the right-hand side of Eq. (1.1) we approximate the earth as a black body with temperature T . A black body is a hypothetical body, which absorbs all incoming radiation, acquires a certain temperature T and emits its radiation according to simple laws, to be discussed in Chapter 2. At present the student should accept that according to Stefan–Boltzmann’s law a black body produces outgoing radiation with intensity σ T 4 [Wm−2 ]. The total outgoing radiation from the earth then becomes σ T 4 × 4π R 2 . Substitution in Eq. (1.1) gives: (1 − a)S × πR 2 = σ T 4 × 4πR 2

(1.2)

or S (1.3) = σT4 4 Numerical values of σ, R and S are given in Appendix A. For albedo a one finds from experiments a = 0.30. Substitution gives T = 255 [K], which is way below the true average earth surface temperature of 15 [◦ C] = 288 [K]. The difference of 33 [◦ C] is due to the greenhouse effect, for which the earth’s atmosphere is responsible. As will be shown later, the emission spectrum of the sun peaks at a wavelength of 0.5 [μm], whereas the earth’s emission spectrum peaks at 10 [μm], the far IR. Several gases in our atmosphere, the so-called greenhouse gases, absorb strongly in the IR. In that way a large part of the solar radiation reaches the surface, but the emitted IR radiation has difficulty in escaping. The same effect happens in a greenhouse, hence the name. It will be discussed later how human activities contribute to the greenhouse effect by increasing the concentration of greenhouse gases like CO2 , tropospheric O3 , N2 O, CH4 and many HFCs. The increase of their concentrations necessarily leads to an increase in the surface temperature of the earth and consequently to climate change. It will be a challenge (1 − a)

4

Environmental Physics

to science, technology and policy to find ways to slow down the increase in greenhouse gas concentration.

Example 1.1: Geothermal Heat Flow As an illustration of the physics of this section, consider the heat flow F from the interior of the earth reaching the earth’s surface. This flow originates from decay of radioactive elements in the earth interior and the friction between the tectonic plates caused by their ‘tidal motion’. The flow is estimated1 as F ≈ 42 × 1012 [W]. What is the flow per [m2 ]? Assume the solar influx would not be present. Calculate the equilibrium surface temperature of the earth arising from the internal heat flow only. Answer The heat flow per [m2 ] is found with F/(4π R 2 ), where R is the radius of the earth. With the data from Appendix A we find that this becomes 0.082 [Wm−2 ]. If there is no solar input the internal flow is the only source. In a steady-state situation the inflow from the interior equals the outflow at temperature T : energy in = energy out. Following Stefan–Boltzmann’s law we have F/(4π R 2 ) = σ T 4 , which gives T = 35 [K]. So, without the influx of solar energy the temperature of the earth’s surface would be down to a value about 35 [K]. Therefore, as we know well, the present, pleasant temperature on earth is due to the sun.

1.3

Light Absorption in Nature as a Source of Energy

The sun also provides energy for plants, algae and so-called cyanobacteria by photosynthesis, a mechanism, which will be discussed extensively in Sections 5.4 to 5.7. Figure 1.2 shows the solar absorption by the important pigments of several photosynthetic organisms. All the light energy absorbed by the pigments can be used for photosynthesis. For comparison the solar spectrum as it enters the top of the atmosphere is shown. At the earth’s surface it is more structured (as we will find in Figure 2.2), but the message remains that the pigments shown absorb in important parts of the spectrum. Chlorophyll-a, the major pigment of higher plants, algae and cyanobacteria, absorbs red and blue light. In combination with carotenoids such as β-carotene they provide plants with their typical green colour. The major photosynthetic organisms of the world oceans, the cyanobacteria (often called blue-green algae), absorb sunlight by a specialized protein which is called phycobilisome and contains as major pigments phycoerythrin (absorbing around 580 [nm]), phycocyanin (absorbing around 620 [nm]) and allophycocyanin (absorbing around 650 [nm]). As is shown in the picture bacteriochlorophyll-a and bacteriochlorophyll-b are the major pigments of two classes of photosynthetic bacteria absorbing in the near IR part of the solar spectrum. 1

In this book data without a reference are taken from refs. [2] and [3], in this case from [2], p. 92.

Introduction

5

Phycocyanin

Pigment absorption / [a.u.]

1.0 Phycoerythrin

0.8

Allophycocyanin Chlorophyll-a

Solar spectrum

Bacteriochlorophyll-a

0.6

0.4 and 0.2 Chlorophyll carotenoids Bacteriochlorophyll-b

0 400

600

800

Visible Blue

1000

1200

Wavelength / [nm] Red

Figure 1.2 Light absorption by important pigments of a variety of photosynthetic organisms in arbitrary units (a.u.) as a function of wavelength. For comparison the solar spectrum is sketched, adapted to the same scale.

1.4

The Contribution of Science: Understanding, Modelling and Monitoring

The first question for a physical scientist is always, ‘Why are things as they are?’ For example, why is the surface temperature on earth 33 [K] higher than calculated by means of Eq. (1.3). Understanding starts in this case by indicating the role of the earth’s atmosphere and the greenhouse effect. Then one has to proceed and understand the 33 [K] quantitatively. In Chapter 3 we show that an analysis of the composition of the atmosphere and the radiative properties of the atmosphere and the surface allow us to give a quantitative explanation. A second question would be how the greenhouse effect is influenced by human activities. Here one needs a model to project these activities into the future, calculate emission of gases and other factors influencing climate, and finally calculate the future greenhouse effect. One has to check how the greenhouse gases behave over time and therefore one has to monitor their concentrations continuously. Comparisons of predictions with reality will increase our understanding of climate and climate change. One will find this continuous loop of understanding, modelling, monitoring and increased understanding in many parts of science, if not all. In order to increase the production of bio fuels from crops or algae, for example, one needs to increase the efficiency of photosynthesis. Again the first step is to understand the process (Sections 5.4 and 5.5). The next step is to construct a model which reproduces the real situation with parameters that may be changed by human interference. The third step is to make a pilot with parameters

6

Environmental Physics

which are supposed to increase the bio-yield of the crop, monitor all external influences and check the results against the model. This, in fact, is the way in which technology works. A realistic model needs to take into account many variables in a consistent and coherent way. That is the subject of computer science, a field which is outside the scope of this book. The present text does provide the student with the building blocks of modelling in the realm of climate and sustainable energy and makes it possible to judge a model as to its applicability in certain situations. Because of the importance of monitoring as part of the scientific process, Chapter 8 is devoted to this subject.

Exercises 1.1 Check the units to the left and right of Eq. (1.3) using Appendix A. Check the calculation which produces a temperature of 255 [K] for the earth. 1.2 Repeat the calculation following Eq. (1.3) for the planets venus and mars. (a) Calculate the solar ‘constants’ SV and S M for these planets, using that the distance from venus to the sun averages 0.72 times the distance from earth to the sun. For mars the corresponding number becomes 1.52. (b) Ref. [3], Sect 14 gives the albedo of venus as 0.65 and that of mars as 0.15. Find their radiation temperatures, that is, the temperatures following from Eq. (1.3). Compare with the surface temperatures of 730 [K] and 218 [K] and explain. 1.3 Calculate the temperature of the solar surface, using the total solar irradiance S, the solar radius Rs = 6.96 × 108 [m] and the sun–earth distance Rse = 1.50 × 1011 [m]. Use the assumption that in this regard the sun behaves as a black body.

References [1] Brundtland, G. (Chair) (1987) Our Common Future, World Commission on Environment and Development, Oxford University Press, Oxford. [2] Kaye, G.W.C. and Laby, T.H. (1995) Tables of Physical and Chemical Constants, 16th edn, Longman, Harlow, England. [3] Haynes, W.M. (1996) Handbook of Chemistry and Physics, 77th edn, CRC Press.

2 Light and Matter On a microscopic level the interaction of solar radiation with matter is described as the interaction of photons of a certain frequency υ or wavelength λ with atoms or molecules. Below, first the emission of a hot black body with temperature T is presented, both for the earth’s temperature and the sun’s. Then the measured solar spectrum is discussed in a general way. In order to describe the interaction between photons and atoms or molecules one needs a few quantum mechanical formulas. For light absorption in a medium one needs to sum over all atoms or molecules the light will encounter. That leads to the concept of optical density, which is necessary to understand the absorption of the earth’s IR radiation and the greenhouse effect. In a solar cell, to be described in Chapter 5, the interaction of a single photon with the cell may result in an (albeit very small) electric current. A plant is a much more complicated system than a solar cell; it requires the cooperation of some 10 photons to complete one full turnover of the photocell in which the solar energy is stored as chemical energy. As an appetizer for Chapter 5 some elements of the photosynthetic process are discussed below. Finally, solar light, especially in the ultraviolet (UV) region not only is beneficial, but may also damage biomolecules, such as those present in the human skin. It is explained how the ozone filter in the outer atmosphere is protecting humans against too much shortwavelength radiation.

2.1 2.1.1

The Solar Spectrum Radiation from a Black Body

The term ‘black body’ originates from experiments with a black cavity held at a certain temperature. It appears to emit a smooth spectrum with the intensity Iλ as a function of wavelength λ given by the shape of the curves in Figure 2.1. The position of the peak shifts to larger wavelengths for lower temperatures. Environmental Physics: Sustainable Energy and Climate Change, Third Edition. Egbert Boeker and Rienk van Grondelle. © 2011 John Wiley & Sons, Ltd. Published 2011 by John Wiley & Sons, Ltd.

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Environmental Physics

Figure 2.1 Emission spectra of a black body with intensity Iλ [Wm–2 μm–1 ] as a function of wavelength λ, calculated with Eq. (2.4). The left spectrum corresponds with Planck’s curve (2.2) for a temperature T = 5800 [K] (the radiation temperature of the solar surface, which gives the general behaviour of the spectrum); the right spectrum corresponds to T = 288 [K] (the temperature of the earth’s surface). Note that the vertical scales differ by many orders of magnitude. Also note that the left curve peaks at approximately 0.5 [μm] and the right spectrum at about 10 [μm].

In the beginning of the twentieth century attempts were made to calculate the spectral shape of black-body radiation. In these calculations it was assumed that at each wavelength all energies could be emitted, so there was no relation between wavelength and energy. However, the resulting spectral composition of the black- body radiation did not correspond to experiments. In a historic contribution to quantum mechanics Planck solved the problem by making the assumption that at any wavelength λ the energy [J] of the emitted electromagnetic field is restricted to integral multiples of the energy E = hν = h

c [J] λ

(2.1)

Here υ is the frequency of the electromagnetic field and λ is the corresponding wavelength while c is the speed of light. The symbol h denotes Planck’s constant, given in Appendix A. In other words, the electromagnetic field consists of photons with energies E = hv. Thus, Planck arrived at the so-called Planck energy distribution for the radiation IE dE emitted by the black body with energies between E and E + dE: I E dE =

1 1 2π ν 3 2π E 3 dE = dE [Wm−2 ] c2 ehν/kT − 1 h 3 c2 e E/kT − 1

(2.2)

Here T is the absolute temperature of the black body [K] and k is Boltzmann’s constant [J K−1 ]. The radiation IE is also called the radiation flux, or briefly the flux, as it refers to the radiation energy passing a square metre per second in a certain direction.

Light and Matter

9

The energy density W(E)dE [J m−3 ] of the radiation field at energies between E and E + dE is found by realizing that radiation in a certain direction can pass a [m2 ] in two directions and may have two independent polarizations. This results in W(E) = 4 × IE /c. Below we will need the expression of energy density as a function of ω = 2π υ.1 Using E = ω gives:   1 ω3 (2.3) dω [J m−3 ] W (ω)dω = 2 3 π c eω/kT − 1 which is the energy density for frequencies between ω and ω + dω. In Eq. (2.2) the flux is given as a function of the energy E. We are also interested in the flux as a function of wavelength λ. We may use Eq. (2.1) and find (Exercise 2.1): 1 2π hc2 (2.4) dλ [Wm−2 ] λ5 ehc/λkT − 1 Of course, Iλ dλ again is the radiation energy with wavelengths between λ and λ + dλ passing a square metre per second [Wm−2 ]. In Figure 2.1 the intensity Iλ was displayed with λ and dλ in [μm], hence the dimension Iλ [Wm−2 μm−1 ]. For large frequencies υ or small wavelengths λ the (−1) in the denominator may be ignored and one finds the peak value λmax of the flux Iλ as (Exercise 2.2) Iλ dλ =

λmax T = 2898 [μm K]

(2.5)

This approximate, but useful, relation is called Wien’s displacement law. It shows how the maximum of the emitted radiation distribution depends on the temperature of the black body. For T = 5800 [K] the maximum of the emitted radiation is at λ ≈ 0.5 [μm], which agrees with the left curve in Figure 2.1. By integrating Eq. (2.2) over all energies one finds another well-known equation, Stefan–Boltzmann’s law (Exercise 2.3). This law expresses how the total flux I [Wm−2 ] emitted by a black body depends on its temperature T: I (T ) = σ T 4 [Wm−2 ]

(2.6)

Here, σ is the Stefan–Boltzmann constant, which is independent of the material. 2.1.2

Emission Spectrum of the Sun

In Figure 2.2 the emission spectrum of the sun is shown as it enters the top of the atmosphere; the lower and more structured curve represents the solar spectrum as it is observed at sea level. The difference is due to the absorption of the solar light in the earth’s atmosphere. The gases mainly responsible for this absorption are indicated: CO2 , water vapour H2 O, and O2 and O3 . The dashed curve in Figure 2.2 indicates the shape of the black-body emission spectrum for T = 5900 [K], which fits the curve equally as well as T = 5800 [K]. This implies that for simple considerations, as in Chapter 3, the solar spectrum may be approximated by that of a black body. Note that the left curve in Figure 2.1 gives the emission from the surface of the black body (the sun), whereas Figure 2.2 shows what is arriving at the earth. This explains the difference in scales of both figures (Exercise 2.4). 1

As ω = 2πυ it follows W(ω)dω = 2π W(ω)dυ = W(υ)dυ. In consulting handbooks be aware of the variables used.

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Environmental Physics

Figure 2.2 Spectral distribution of incident solar radiation outside the atmosphere and at sea level. Major absorption bands of the important atmospheric gases are indicated. The shape of the black-body curve at 5900 [K] is shown for comparison. Note that Figure 2.1 refers to the emission at the solar surface while Figure 2.2 refers to the intensity on the earth. (Reproduced by permission of McGraw-Hill, copyright 1965, from [1], Figure 16.1.)

At temperatures such as we experience on earth, the black-body radiation has peak intensity far in the infrared, which is way out of the visible region, see Figure 2.1 right curve. Bodies at temperatures of for example 100 [◦ C] = 373 [K] still do not emit appreciably in the visible part of the spectrum. We cannot see, but ‘feel’ the IR radiation when we put our hands near a hot-water bottle.

Example 2.1 (a) Write the black-body spectrum (2.2) with T = 5800 [K] as function of electronvolt [eV], the energy unit of atomic and molecular physics. (b) Also plot the number of photons as a function of energy in [eV]. Check the validity of Stefan–Boltzmann’s relation (2.6) from the curves.

Light and Matter

11

Answer (a) Equation (2.2) is written in [J]. Following Appendix A one can write 1 [eV] = a [J] with a = 1.602 × 10−19 . All energies have to be replaced by aE, where E is the value in [eV]. This gives for the intensity IE IE =

2πa 3 E 3 1 × a E  /kT [m−2 s−1 ] h 3 c2 e −1

(2.7)

I

On the left of Figure 2.3 the black-body intensity is plotted as a function of E [eV]. The product of peak times width approximately gives 3.8 × 1026 [eV s−1 m−2 ]; 1 [eV] = a [J], which gives 6.1 × 107 [Wm−2 ], close to the Stephan-Boltzmann value of 6.4 × 107 [Wm−2 ], which indicates that the derivation of Eq. (2.7) is correct. Numerical integration will show that this relation holds exactly.

Figure 2.3 Left: Black-body spectrum for T = 5800 [K] as function of energy E [eV]. Right: the number of photons per [eV] passing 1 [m2 ] per second. Note that the maximum of the curve on the right is shifted somewhat to the left with respect to the curve on the left-hand side. The visible region is indicated on both curves.

(b) Let NPE be the number of photons per [eV], passing 1 [m2 ] in 1 [s]. Then NPE × dE will be the number of photons with energies [eV] between E and E + dE passing 1 [m2 ] in 1 [s]. Multiplied by the energy E this should give IE dE . Therefore IE 2πa 3 E 2 1 = × a E  /kT (2.8) [m−2 s−1 eV−1 ] E h 3 c2 e −1 The result is given on the right in Figure 2.3. The maximum is shifted to the low energy part and is outside of the visible region. NP E  =

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Environmental Physics

From a physics point of view it is interesting to notice that most transitions in atoms and molecules occur at energies of less than 1 [eV]. Solar photons therefore have energy enough to trigger these transitions. On the other hand, ionizing potentials (the energy required to eject an electron from the least bound state) of virtually all atoms and molecules are larger or much larger than 5 [eV], which literally saves our skins ([2], Section 4.1.2 or [3], 10–214/235).

2.2

Interaction of Light with Matter

The interaction of solar radiation with matter is described on a microscopic scale. Below, the formulas for a light-induced transition between two quantum states are derived, leading to the so-called transition dipole moments. Next, the Einstein coefficients for absorption and emission are introduced and finally the absorption of a beam of light in a medium like the atmosphere is discussed. Note: this section applies some basic quantum mechanics. Students who feel this is still somewhat too difficult are advised to read it anyway and try to understand Eqs. (2.17), (2.21) and (2.22). Then continue with Section 2.2.2

↓

2.2.1

Electric Dipole Moments of Transitions

Consider an atomic or molecular system described by a Hamiltonian H 0 . It has discrete energy levels Ek and wave functions k0 which are the solutions of the time-independent Schr¨odinger equation of the system H0 k0 = E k k0

(2.9)

Without perturbations the system can remain in a single state k0 . At t = 0 light with a single frequency ω starts interacting with the system, which is described as a perturbation H 1 (t). At time t the system will be found in a state (t), which is the solution of the time-dependent Schr¨odinger equation i

d = (H0 + H1 (t))(t) dt

(2.10)

The time-dependent wave function (t) can be expressed as (t) =



ck (t)k0 e−i Ek t/

(2.11)

k

Inserting (2.11) in (2.10) gives a set of equations for the time derivatives of the coefficients ck (t). One will get rid of some summations by closing with the bra k0 , which gives     i  dck (t) cn (t)eiωkn t k0  H1 n0 =− dt  n

(2.12)

Here ωkn = (Ek – En )/. Remember that at t = 0 the system is in the stationary state 10 . This means that at t = 0 one has c1 (0) = 1 and ck (0) = 0 for all k = 1. Thus for small times t, the right hand side of Eq. (2.12) reduces to a single term. Integration for an arbitrary

Light and Matter

13

k = 1 gives ck (t) = −

i 

t



eiωk1 t k |H1 | 1dt 

(2.13)

0

The probability of finding the system in the excited state k, having absorbed an energy quantum ωk1 from the light beam, is given by |ck (t)|2 . Eq. (2.13) shows that |ck (t)|2 is directly related to the squared matrix element |k| H1 |1|2 . To proceed we will derive the perturbation H 1 (t). We assume that the perturbation is caused by the interaction of a plane electromagnetic wave with an atomic or molecular system at the origin of the coordinate system. For electrically neutral systems the leading term of the perturbation is the interaction between the electrons of the system and the electric field E(t) = E0 cos ωt. To first order this may be described by H1 (t) = −μ · E(t) where μ=−



qi ri

(2.14)

(2.15)

i

is the electric dipole operator of the system and the sum extends over all electrons with charge qi and position ri . One finds the probability Pk (t) = |ck (t)|2 that the system is in state k at time t by substituting Eq. (2.14) into Eq. (2.13). One assumes that the frequency ω of the light is close to the resonance frequency ωk1 of the transition, so one may ignore terms (ωkl + ω)2 in the denominator and find

|E0 |2 cos2 θ sin2 12 (ωk1 − ω)t 1 2 (2.16) Pk (t) = 2 |k |μ|1|  (ωk1 − ω)2 As it represents a probability Pk (t) is dimensionless (Exercise 2.5). In (2.16) θ is the angle between the vectors μ and E0 . It is clear that the squared matrix element |μk1 |2 = |k |μ|1|2

(2.17)

determines the probability Pk (t) that the system is in state k at time t. The matrix element μk1 is called the transition dipole moment. Note that the probability of absorption (2.16) depends on the orientation of the vector μk1 with respect to the electric field E0 by the factor cos2 (θ ). For isotropic solutions or gases one has to average over all directions. For ordered systems the factor cos2 (θ ) leads to polarization effects. In a few steps the transition probability [s−1 ] of the transition 1 → k will be derived. The transition probability [s−1 ] is the increase dPk of the probability Pk (t) that the system is in state k during a time dt, divided by dt which gives dPk /dt. One therefore has to differentiate Eq. (2.16) and reorganize the resulting sine and cosine functions. This gives |μk1 |2 dPk sin((ωk1 − ω)t) |E0 |2 cos2 θ = 2 dt  2(ωk1 − ω)

(2.18)

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Environmental Physics

One may show (Exercise 2.6) that after a long time t the fraction at the right of Eq. (2.18) peaks around ω = ωk1 . In such a case it is convenient to use the delta function, defined in Appendix C to describe the limit of Eq. (2.18) for t → ∞ (see Eq. (C6). This gives |μk1 |2 π dPk |E0 |2 cos2 θ × δ(ωk1 − ω) = dt 2 2

(2.19)

It is useful to check the dimensions of Eq. (2.19) (Exercise 2.7). One then should keep in mind the definition of the δ– function δ(ωk1 − ω)dω = 1 from which it follows that in the present case its dimension should be [ω–1 ] = [s]. For a plane electromagnetic wave the energy density of the field may be written as U = ε0 × E0 /2 [J m−3 ]. Finally, the average of cos2 (θ ) in three dimensions gives 1/3. Taking all factors together one finds dPk π |μk1 |2 U δ(ωk1 − ω) = dt 3ε0 2

(2.20)

In this equation ωk1 is defined by the energy difference between states k and 1. Therefore Ek − E1 = ω. In practice the transition can also take place in a band of frequencies around ω = ωk1 . Therefore one has to replace Uδ(ωk1 – ω) in Eq. (2.20) by an energy-dependent W(ω) [J m−3 s], which we already met in Eq. (2.3) for a black body. This finally results in π dPk |μk1 |2 W (ω) = dt 3ε0 2

(2.21)

As an extracheck one may compare the integral over ω in Eqs. (2.20) and (2.21). It is  apparent that Uδ(ωk1 − ω)dω = U (ωk1 )[Jm−3 ] then is replaced by W (ω)dω, which is the integral of the energy density integrated over the band around ω = ωk1 . Eq. (2.21) represents the rate of transition 1 → k. The last equality defines the Einstein coefficient for absorption and stimulated emission as Bk1 =

↑

π |μk1 |2 3ε0 2

(2.22)

The dimensions of Eqs. (2.21) and (2.22) are discussed in Exercise 2.8. 2.2.2

Einstein Coefficients

Consider a simple system with two energy levels. A lower level with energy E1 populated by N 1 atoms or molecules and a higher level with energy E2 = E1 + ω and populated by N 2 atoms/molecules. This is indicated schematically in Figure 2.4. First note, that in deriving Eq. (2.21) it was not stated whether state k0 was the state of higher or of lower energy. Therefore the transition rate (2.21) applies both to absorption from an electromagnetic field, which gives B12 W(ω) and to stimulated emission in a field, given by B21 W(ω). Spontaneous emission from state 2 to state 1, which may occur ‘in the dark’ is given by a transition rate A21 for which we will derive an expression now. The rate of population of level 1 in the presence of a light field W(ω) may be written as dN1 = −B12 W (ω)N1 + B21 W (ω)N2 + A21 N2 dt

(2.23)

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15

E2, N2

hω

A21

B12W(ω)

B21W(ω)

Spontaneous emission

Absorption

Stimulated emission

E1, N1

Figure 2.4 The three basic types of radiative processes. Indicated are the Einstein coefficients and the transition rates for two levels with energies E1 and E2 , which are occupied by N1 and N2 atoms/molecules, respectively.

Assuming a steady state, where the population of the levels does not change, we have dN1 /dt = 0 and it follows that W (ω) =

A21 B12 N1 /N2 − B21

(2.24)

In the steady state with temperature T the ratio N 1 /N 2 is found from the Boltzmann distribution for N 2 /N1 , N2 = e−ω/kT N1

(2.25)

This fundamental relation expresses that the temperature motion of the atoms and molecules represented by the energy kT[J] will populate the higher energy level with energy E2 = E1 + ω with a relative probability e–ω/kT . For a higher energy difference ω the population of the upper level will decrease and for a higher temperature T it will increase. Finally, in the steady state the energy density (2.24) of the radiation field will be determined by the Planck distribution for a black body (2.3). When one substitutes (2.25) into (2.24) and put that equal to (2.3) one finds   A21 1 ω3 [J m−3 s] (2.26) = W (ω) = 2 3 π c e ω/kT − 1 B12 e ω/kT − B21 The last equality should hold for all temperatures T, which implies B12 = B21

(2.27)

A21 ω3 = 2 3 B21 π c

(2.28)

and

or A21 =

ω3 |μ21 |2 3π c3 ε0

(2.29)

Consequently the same squared matrix element |μ21 |2 determines the rates and properties of all radiative processes.

16

2.2.3

Environmental Physics

Absorption of a Beam of Light: Lambert-Beer’s Law

The microscopic quantities |μ21 |2 , B12 , B21 and A21 are connected to the macroscopic phenomenon of absorption as described by Lambert-Beer’s law, which will be derived now. When a beam of light passes through a sample of material, the light beam is usually absorbed, but sometimes, as in a laser, amplified. Absorption dominates when most of the atoms/molecules are in their ground states (N 1 N 2 ) and when the beam intensity is weak. In that case the rate equation for transitions between states 1 and 2 is given by Eq. (2.23) without the term for stimulated emission as dN1 = −B12 W N1 + A21 N2 dt

(2.30)

Here B12 N 1 W is the rate at which light is removed from the incident beam. The last term of Eq. (2.30) describes emission from the excited state. That light is radiated into all directions. Also, N 2 N 1 so the contribution of the last term of Eq. (2.30) may be ignored. In Figure 2.5 a light beam is entering a slice of medium with thickness dz along the z-axis. Its energy density will depend on ω and z. Due to the absorption the density will decrease as well, making it a function of the time. This leads to W = W(ω, z, t) [J m−3 s]. The beam energy in the frequency interval ω to ω + dω is given by W(ω, z, t) dω a dz where a [m2 ] is the surface area of the slice. The energy decrease of the beam in the slice must be equal to the absorption by the medium. This results in −

a dz ∂W a dω dz = B12 N1 F(ω) dω W (ω, z)ω ∂t V

(2.31)

The left-hand side represents the energy decrease of the beam. The right-hand side represents the absorption, which in Eq. (2.30) was indicated by B12 N 1 W. Expression (2.30) had to be modified as follows. The population N 1 was replaced by the fraction within the slice, i.e. N 1 a dz/V where V is the total volume of the sample. Furthermore, Eq. (2.30) refers to the rate of absorption [s−1 ]. In order to find the absorbed energy per second [J s−1 ], this rate has to be multiplied by the photon energy ω, which explains the far right of Eq. (2.31). Finally, the factor F(ω)dω was inserted because the atom/molecule may absorb over a certain region around ω = ωk1 . This was discussed in deriving Eq. (2.21). The incoming light may comprise many frequencies ω. We are interested in the band around W d ω a dz area a

⎛ ∂I ⎞ dz⎟ dω ⎜I + ⎝ ∂z ⎠

I dω

z

dz

z+dz

Figure 2.5 A beam of light along the z-axis, drawn horizontally, is absorbed by a thin slice of a sample perpendicular to the beam.

Light and Matter

ω = ωk1 . Therefore, the weight function F(ω) was added with the property  F(ω) dω = 1

17

(2.32)

band

From Eq. (2.31) one finds ∂W ω = −N1 F(ω)B12 W [J m−3 ] ∂t V

(2.33)

In practice, one is interested in the extinction of a beam with intensity I(ω) dω [Wm−2 ] between ω and ω + dω. In this case we use the fact that the energy loss of the beam in the slice of Figure 2.5 over a time dt (the left-hand side of Eq. (2.31)) is equal to the decrease of the intensity of the beam over the slice. This gives −

∂I ∂W dω a dz = (I (z) − I (z + dz)) a dω = − dz a dω ∂t ∂z

(2.34)

which leads to ∂ W/∂t = ∂ I/∂z. The intensity of the beam and its energy density are related by W = I/(cn), where c is the speed of light in vacuum and n the refractive index. On the left of Eq. (2.33) we substitute ∂ W/∂t = ∂ I/∂z. On the right we substitute W = I/(cn). This leads to a simple differential equation for I(z) B12 N1 ωF(ω) ∂I =− I ∂z V nc

(2.35)

On the right-hand side all quantities except I are independent of z, which gives LambertBeer’s Law I (z) = I (0)e−kz

(2.36)

where the absorption coefficient k is given by k=

B12 N1 ωF(ω) V nc

(2.37)

 We mention in passing that the exponent kz in Eq. (2.36) may be replaced by k(z) dz, as is proven in Eq. (8.45). The quantity B12 in Eq. (2.37) was given by Eqs. (2.22) and (2.27). It is independent of the frequency ω of the field. One therefore may extract F(ω) from Eq. (2.37) and use (2.32) and will find  k(ω) V cn dω (2.38) B12 = N1  ω band

where the integral runs over a band around the transition frequency ω12 . Eq. (2.36) shows that one can measure k(ω) over the band and find the Einstein coefficients from (2.38). Using (2.22) and taking n = 1 one then will find the transition dipole moment |μk1 |2 as  k(ω) 3ε0 c 2 |μk1 | = dω (2.39) π (N1 /V ) ω band

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Environmental Physics

In the following few equations we will replace the official MKS units on the right-hand side by more convenient units. Correction factors will be introduced on the right-hand side in order to keep the correct units [μk1 ] = [C m] on the left. Often one works with a concentration C [mol/L = mol/dm3 ]. Using Avogadro’s number NA one finds N1 /V [m−3 ] = N1 /(VN A )[mol/m3 ] = N1 /(VN A 103 )[mol/L] = C[mol/L] This leads to C=

N1 [mol/L] V N A 103

(2.40)

The factor N 1 /V in Eq. (2.39) may be extracted from Eq. (2.40), which leads to  k(ω) 3ε0 c |μk1 |2 = dω (2.41) 3 πC N A 10 ω band

The function k(ω) was defined in Eq. (2.36). That equation may also be written as I (l 0 ) = I (0) × 10−OD

(2.42)

where l0 [m] is the path length in the medium and OD is the optical density. In atmospheric applications this exponent usually is called optical depth. By comparing Eqs. (2.36) and (2.42) one will find k=

l0

OD log e

(2.43)

10

This gives 3ε0 c |μk1 | = πC N A 103 l 0 10 log e



2

band

OD dω ω

(2.44)

where the dipole matrix element μk1 is found with its MKS unit [C m]. It is interesting that the macroscopic measurement of the optical density OD directly gives the microscopic property μk1 .

Example 2.1 Extinction in Plants For applications in, e.g. plants one expresses the path length as l [cm] and writes OD = εlC

(2.45)

(a) Find the dimension of the molar extinction coefficient ε. (b) Reduce Eq. (2.44) to an integral which contains ε. Answer (a) OD is dimensionless (see (2.42). so [ε][cm][mol/L] = dimensionless, which gives [ε] = [L mol−1 cm−1 ]

Light and Matter

(b) Substitution of (2.45) in (2.44) gives   ε ε 3ε0 c 2 −61 |μk1 | = dω = 1.023 × 10 dω, 10 π N A 10 × log e band ω band ω

19

(2.46)

where we used l0 = 10–2 l and used the numerical values in Appendix A.

Example 2.3

Optical Density in Leaves

A leaf containing chlorophyll has in its absorption maximum at 680 [nm] an extinction coefficient ε =105 [dm3 mol−1 cm−1 ]. The chlorophyll concentration in the leaf is about 10−3 [mol dm−3 ] and its path length for light passing through the leaf about 0.02 [cm]. Calculate OD and the reduction of intensity. Answer OD = 105 × 0.02 ×10−3 = 2. So through a leaf with a thickness of 0.02 [cm] the intensity of the (red) 680 [nm] light is reduced by a factor of 10−2 = 100 (from Eq. (2.42). As a consequence, no red light is detected below the outer array of leaves on a tree. Finally, we have a second look at Figure 2.5. The decrease of energy over the slice was given in Eq. (2.34) as a(I(z) − I(z + dz))dω [W]. This must be proportional to the number of particles in the slice, which is (N/V) adz and also proportional to the intensity of the beam I(z)dω. The proportionality constant σ 0 has the dimension [m2 ], as follows from a(I (z) − I (z + dz))dω[W] = (N/V )adz × I (z)dω × σ 0

(2.47)

On the right the product (N/V) adz × σ 0 may be interpreted as the total area with which the incoming beam interacts. Then I(z)dω may be interpreted as the number of photons impinging. Their product, the right-hand side of Eq. (2.47) would represent the decrease of intensity of the beam, which corresponds with the left hand side. Consequently σ 0 may be interpreted as the cross section of the particles in the slice. From Eq. (2.47) it follows that ∂ I/∂z = −(N /V )σ 0 I and I (z) = e−(N/V ) σ

0

z

= e−kz

(2.48)

where (2.36) was used. Consequently k = (N/V)σ 0 . The cross section usually is given as σ [cm2 ], from which σ 0 [m2 ] = 10–4 σ [cm2 ]. Using Eqs. (2.43), (2.45) and (2.40) one finds ε = 10−3 σ N A 10 log e

2.3

(2.49)

Ultraviolet Light and Biomolecules

In this section we discuss the absorption of light in the ultraviolet (UV) region by biological molecules such as proteins and nucleic acids. Since both the amount and the spectral

20

Environmental Physics

composition of UV light incident on the earth’s surface is totally determined by the ozone present in the atmosphere, we shall summarize processes that lead to the building up and destruction of the ozone layer and show that small variations in ozone concentration may lead to dramatic biological effects. These effects are largely related to the highly non-linear properties of Lambert–Beer’s law (Eqs. (2.36) and (2.37)) where the concentration C = N 1 /V appears in the exponent. 2.3.1

Spectroscopy of Biomolecules

Figure 2.6 shows the solar emission spectrum below 340 [nm] at the earth’s surface in combination with the absorption spectrum of two important biomolecules: DNA, which is the carrier of the genetic code and α-crystallin, which is the major protein of the mammalian eye lens. The absorption of light by these biological molecules is essentially zero in the region 320 [nm] < λ< 400 [nm], which is called the near UV or UV-A; the absorption is intense in the region 200 [nm] < λ < 290 [nm], which is called the far UV or UV-C and only overlaps with the solar spectrum in the wavelength region 290 [nm] < λ < 320 [nm], the mid-UV or UV-B. From Figure 2.6 it appears that the interaction between sunlight and the two biomolecules will take place essentially in the UV-B region. The absorption of DNA is due to the DNA bases, guanine, thymine, cytosine and adenine, and it peaks about 260 [nm], with a maximum extinction of ε = 104 [L mol−1 cm−1 ] (expressed per mole of base). Note that here the units of Eq. (2.45) are used because they are convenient for biological applications.

OD /[a.u.]

0.8

0.75

0.6 0.5 0.4

UV-C

UV-B

UV-A 0.25

0.2

Relative sunlight intensity

1.0

0.0 240

260

280

300

320

340

Wavelength /[nm] Figure 2.6 Absorption spectra of DNA (dotted) and α-crystallin (dashed) in the wavelength region 240–340 [nm]. The solid line indicates the solar spectrum at the earth’s surface in the same region normalized to one at 340 [nm]. The left axis refers to the optical density in arbitrary units and the right axis to the relative solar intensity.

Light and Matter

21

The absorption of proteins like α-crystallin in the wavelength region 250 [nm] < λ < 300 [nm] is mainly due to the aromatic amino acids tryptophan and tyrosine, and it peaks at about 280 [nm]. For a protein with a number nTRP of tryptophan amino acids and a number nTYR of tyrosine amino acids the extinction at 280 [nm] is given by ε280 = 5600 n TRP + 1100 n TYR [L mol−1 cm−1 ]

(2.50)

For a typical biological cell about 10% of the total absorption of solar UV due to proteins would be due to the absorption of nucleic acids. Often proteins carry additional cofactors which allow proteins to carry out their specific tasks. For instance, in haemoglobin the haem group (which belongs to the class of porphyrin molecules), is attached to the protein in a haem–protein complex which is active in oxygen binding of blood. In these cases the absorption extends into the near-UV, visible and sometimes near infrared region of the spectrum. Although in several cases the absorption of light in this wavelength range is crucial for the biological function (photosynthesis, vision), we shall not further discuss it here.

2.3.2

Damage to Life from Solar UV

The interaction of solar UV with biological cells may lead to severe damage of the genetic material or even cell death. In complex, multicellular organisms, exposure to solar UV may lead to damage of crucial parts of the organism, and in the case of humans, to skin cancer. As Figure 2.6 showed, the UV-B wavelength range will be of prime importance in these processes. In particular, the light-absorbing parts of the DNA, called the chromophores, form the prime target in the process that leads to photo-damage. Figure 2.7 shows the action spectrum E(λ), which is the damage done by a unit of irradiation of a certain wavelength in producing erythyma (sunburn) in human skin. It also shows the UV tail I(λ) of the solar spectrum. The damage done by solar radiation at wavelength λ is therefore given by the product E(λ)I(λ), which is called the solar erythemal effectiveness. Note that Figure 2.7 is drawn with a vertical logarithmic scale. It shows that the efficiency E(λ) of producing erythyma increases four to five orders of magnitude between 350 and 280 [nm], precisely in the region where the solar UV spectrum collapses. The action spectrum E(λ) must be largely ascribed to the direct absorption of solar UV by DNA. Although for λ > 320 [nm] the action spectrum tends to be higher than the DNA absorption in Figure 2.6, implying that there are more absorbers, the primary target is probably still DNA. Similarly, studies on cellular cultures where cell survival was measured have identified the chromophores of DNA as the primary target molecules. The major product of photo-absorption in DNA is the pyrimidine dimer, a covalent structure involving two thymine molecules or a cytosine–thymine pair. Its presence in DNA has major destructive effects on the reading and translation of the genetic code. The damage D done to a biological system by solar UV can be calculated from the action spectrum E(λ) by multiplying with the incident solar radiation at the earth’s surface I(λ)dλ

22

Environmental Physics

with wavelengths between λ and λ + dλ and integrating ∞ D=

E(λ)I (λ)dλ

(2.51)

0

Since E(λ) is a function with a large negative slope and I(λ) is a function with a large positive slope, even a minor change in I(λ) may lead to large changes in the damage D. 10

Relative units

10 10 10 10 10 10

0

-1

Solar spectrum

Erythema action spectrum

-2

-3

Solar erythemal effectiveness

-4

-5

-6

260

280

300

320

340

360

380

Wavelength /[nm] Figure 2.7 Erythema (sunburn) action spectrum, plotted together with the solar spectrum and the resultant ‘solar erythemal effectiveness’ E(λ)I(λ) of Eq. (2.51). Note the vertical logarithmic coordinate. (Reproduced by permission of Plenum Press, copyright 1978, from [2], Fig 8.4 p. 119.)

2.3.3

The Ozone Filter as Protection

Ozone (O3 ) forms a thin layer in the stratosphere, with a maximum concentration between 20 and 26 [km] above the earth’s surface (Figure 3.2). The atmospheric ozone absorbs essentially all the radiation below a wavelength of 295 [nm], due to a strong optical transition at about 255 [nm], which extends into the mid-UV region. Figure 2.8 shows the spectrum of ozone between wavelengths of 240 and 300 [nm]. At the maximum of 255 [nm] the absorption cross section is about 10−17 [cm2 ]; it is about half maximum at 275 [nm] and has decreased to about 10% of its maximum value by 290 [nm]. That ozone forms a very thin shield indeed, is probably best demonstrated by the fact that the amount of O3 in the atmosphere corresponds to a layer of 0.3 [cm] at standard temperature and pressure (Exercise 2.9). Ozone is constantly formed in the upper layer of the atmosphere through the combination of molecular oxygen (O2 ) and atomic oxygen (O). The latter is formed through the

Absorption cross section /[10-17 cm2]

Light and Matter

23

1. 2

x 100 0. 8

DNA 0. 4

0 240

260

280

300

320

340

360

Wavelength /[nm] Figure 2.8 Absorption spectrum of ozone in the wavelength region 240 [nm] to 300 [nm] (solid line), reproduced by permission of the Optical Society of America, copyright 1953, from [3], Figure 1, p. 871. On the right the spectrum is given on a 100-fold enlarged scale from 300 [nm] to 350 [nm], kindly made available by Dr Kylling who used data going back to [4]. For comparison, the absorption cross section of DNA shown in Figure 2.6 is given, redrawn to the same scale as the ozone spectrum on the left.

photo-dissociation of O2 in the 100 [km] region by light of wavelengths shorter than 175 [nm]. Sunlight excites the electronic transition between the triplet ground state of O2 (3 u− ) and a triplet excited state (3 g− ). Once excited, the O2 molecule may dissociate into two oxygen atoms: one in the ground state of O (3 P), and one in an excited state of O (1 D). The latter has an energy of 2 [eV] above the ground state. As a consequence of these processes, sunlight with λ < 175 [nm] is totally extinguished above the stratosphere. Once formed, atomic oxygen attaches to O2 to form O3 . The efficiency of O3 formation by UV light is sensitive to a large number of factors, amongst which are the availability of O2 , changes in stratospheric temperature, chemicals and dust from volcanic eruptions. Most of the ozone is produced above the equator, where the amount of incident solar UV light is maximal. Ozone formed at these latitudes then diffuses to the poles, where it is ‘accumulated’. The effective thickness of the ozone layer may increase from ≤0.3 [cm] at the equator to ≥0.4 [cm] above the pole at the end of winter. The ozone concentration displays significant daily and seasonal fluctuations and tends to be highest in late winter and early spring. Ozone is not only permanently being formed, but also constantly broken down in the stratosphere and only a very small fraction of the formed ozone escapes down to the troposphere. There are basically two pathways for the destruction of ozone and the

24

Environmental Physics

reformation of O2 : O + O3 → 2O2 O3 + O3 → 3O2

(2.52) (2.53)

These reactions are the net result of a complex set of reactions catalysed by various gases and radicals ([5]). We mention explicitly atomic chlorine Cl, nitric oxide NO and hydroxyl radicals OH. The Cl radical, for example, strips off an O atom from O3 forming ClO. It then loses an O atom to a free O atom and returns to the radical Cl state. The two O atoms recombine to O2 . The net effect is Eq. (2.52). Note that this scheme requires free O atoms to be present at an altitude of 25 [km] or more. How are the free radicals NO, Cl and OH produced? The OH radical is a product of the breakdown of H2 O vapour, for instance produced in the exhaust of supersonic aeroplanes. Although part of the Cl radical may be formed from HCl, released by volcanoes, the major input of Cl into the stratosphere originates from chlorofluorocarbons (CFCs), which are used as foam-blowing agents, refrigerants and propellants. CFCs are extremely stable in the troposphere. However, a small fraction may escape into the stratosphere, eventually reaching the upper stratosphere where they may be decomposed under the influence of UV light, thereby producing Cl and ClO. N2 O is partly of anthropogenic origin and is released from soils and water where it has been formed as a fertilizer waste product. Like the CFCs, the N2 O released at the earth’s surface may eventually be photo-decomposed and NO is formed. These radicals, together with the OH radical are responsible for 90% of the removal of stratospheric ozone. The crux of the ozone problem is demonstrated by Figures 2.6 and 2.8. Any small variation in the ozone concentration will lead to changes in both the amount of UV light at a particular wavelength and the transmission of shorter wavelength radiation. This is of course a direct consequence of Lambert–Beer’s law (2.42). A certain percentage change in concentration of ozone yields the same percentage change in optical density OD. A decrease of one OD unit at a particular wavelength implies a 10-fold increase in the amount of light of that wavelength reaching the earth’s surface. Since the action spectrum for damage to living cells or tissue is an exponentially increasing function with decreasing λ (see Figure 2.7) the amount of damage may dramatically increase, even with a relatively small decrease in the amount of ozone. Figure 2.9 shows experimentally that the amount and spectral distribution of UV light is in fact a strong function of the ozone concentration. In this figure the irradiance at ground level was measured for three different ozone concentrations. The atmospheric pressure was converted to a fictitious pressure of 1 [atm]. From the logarithmic scale it is obvious that at the lower wavelengths the increase in irradiance going from C to A is considerable. Predictions from models for atmospheric ozone production and breakdown indicate a stratospheric ozone depletion ranging from 5–20% due to CFC and N2 O production. From results as presented in Figures 2.6 and 2.8 one may predict that a 10% ozone depletion would result in a 45% increase in effective UV-B radiation. These alarming numbers illustrate the necessity to monitor the structure of the ozone layer accurately and to quantify the effects of increased UV on living organisms. In Chapter 9 we will discuss briefly that international agreements to ‘save the ozone layer’ seem successful.

Light and Matter

25

Figure 2.9 UV intensity at the earth’s surface for various effective thicknesses of the ozone layer. The three cases are A = 0.273 [cm], B = 0.319 [cm] and C = 0.388 [cm]. All measurements were in the clear sky and converted to a fictitious pressure of 1 [atm] = 105 [Pa.]. (Reproduced from T. Ito, Frontiers of Photobiology, 1, Figure 1, p 515. Copyright Elsevier 1993.)

2.3.3.1

Melanin Pigments in the Human Skin2

Melanins are a broad class of macromolecules found throughout nature. In humans they are responsible for a wide variety of colorations of skin, hair and eyes, and they are also present in the inner ear and in neurons of the substantia nigra in the human brain. Epidermal pigments can be divided into two classes: the black to brown eumelanin and the yellow to reddish pheomelanin. Darker skins contain more eumelanin and have a higher ratio of eumelanin to pheomelanin, as compared to lighter skins [7]. The chemical composition of the two pigments is completely different. Eumelanin is a complex polymeric structure based on 5,6-dihydroxyindole (DHI) and 5,6-dihydroxyindole-2-carboxylic acid (DHICA), while pheomelanin is a highly heterogeneous polymer based on benzothiazine and benzothiazole derivatives. The structures of these compounds are shown in Figure 2.10. Both eumelanin and pheomelanin show an intriguing absorption, increasing monotonically towards the UV, as shown in Figure 2.11. This featureless absorption is very uncommon amongst organic chromophores, which generally show distinct absorption bands 2 This subsection is based on a gratefully acknowledged contribution by Annemarie Huijser (Lund University, Department of Chemical Physics, Sweden), which was slightly edited by the authors.

26

Environmental Physics

Figure 2.10

Building blocks of eumelanin (A) and pheomelanin (B).

(see Figure 5.28). The peculiar absorption of eumelanin and pheomelanin are generally attributed to a combination of a large variety of different absorption bands due to the large heterogeneity in composition. It might allow for protection of DNA in the skin against UV-induced damage, although the true function of epidermal pigments is so far not at all understood. There are indications that eumelanin may be photo-protective, while pheomelanin is thought to become toxic upon UV-exposure due to formation of reactive species

Figure 2.11 The broad-band absorption spectra of eumelanin and pheomelanin. (Reprinted from Biophysical Journal, 90, 3, Tran et al, 743–752, copyright 2006 with permission from Elsevier.)

Light and Matter

27

Figure 2.12 UV induced reaction mechanisms in physiologically relevant states of DHICA, the building block of eumelanin: the carboxylate anion (A) and the neutral state (B). (Reproduced by permission of Wiley-VCH.)

(such as radicals). Knowledge of UV-induced reactions is, however, largely lacking. A difference in UV response between eumelanin and pheomelanin, if it exists, might explain the higher susceptibility to skin cancer of people with a fair complexion. Spectroscopic studies to unravel the UV-induced reaction mechanisms in eumelanin and pheomelanin have been highly complicated by the very complex structures of the pigments and the unknown properties of even the small constituents. This has driven researchers to investigate smaller building blocks with a well-defined structure. As an example, the UV-induced reaction mechanisms in the eumelanin building block DHICA are discussed. At physiologically relevant conditions DHICA can occur in two forms: the carboxylate anion and the neutral state. Recent time-resolved fluorescence experiments taking into account isotope effects have shown that these two states respond completely different upon UV exposure, as visualized in Figure 2.12. After excitation the carboxylate anion decays back to its ground state by fluorescence and via non-radiative decay channels (Figure 2.12 (A)). The neutral state, on the other hand, quickly dissipates part of the absorbed UV energy via sub-[ps] excited state intramolecular proton transfer (ESIPT) from the carboxylic acid group towards the nitrogen, which gets a positive charge (Figure 2.12(B)). The formed state subsequently mainly decays back to the ground state non-radiatively and shows an only weak fluorescence and short lifetime [9]. In terms of photo-protection, the neutral state (B) would be preferable above the anionic state (A). In (B) a significant part of the excitation energy is already dissipated through ESIPT on the sub-[ps] time scale and the short lifetime of the formed state (240 [ps], vs 2.1 [ns] for the anionic state (A)) reduces the probability of forming reactive species and causing mutagenic reactions. Further research suggests that, although monomers by themselves are not likely to exist in the full pigment, monomer-like processes still occur to some extent,

28

Environmental Physics

in addition to other (e.g. intermonomer) processes. This suggests that this conclusion not only holds for the DHICA building block, but also for the complete eumelanin pigment.

Exercises 2.1 Check the dimensions of Eqs. (2.2) and (2.3). Derive (2.3) from (2.2) using (2.1). 2.2 Find Wien’s law (2.4) from (2.3) using the approximation of small wavelengths. 2.3 Find Stephan–Boltzmann’s law (2.5) by integration of (2.2) using E = hν. Look for the nasty integral in a book or use mathematics software to find it. Compare the value of σ you find with the value given in Appendix A. 2.4 Deduce from Figure 2.1 the peak value of solar black-body radiation to be expected on earth. Use for the radius of the sun Rsun = 6.96 × 108 [m] and for the distance from the centre of the sun to the centre of the earth Rse = 1.49 × 1011 [m]. Check with Figure 2.2. 2.5 Show that the probability Pk (t) in Eq. (2.16) is dimensionless. 2.6 In the transition from (2.18) to (2.19) we used π sin ((ωk1 − ω)t) → δ(ωk1 − ω) 2(ωk1 − ω) 2 for long times t. (a) Plot the left hand side for a few long times and (b) by change of variables reduce the integral of the left-hand side to a well-known form. 2.7 Check the dimensions of Eq. (2.19) 2.8 (a) Check the dimensions of Eq. (2.21) and (b) deduce the dimensions of the Einstein coefficient (2.22). 2.9 Consider the ozone filter at a wavelength of 255 [nm]. Calculate the optical density OD for a path length of 0.3 [cm] using Figure 2.8 and Eq. (2.49). Use Eq. (2.42) to calculate the reduction in intensity over the path. Verify that the remaining intensity is negligible. Perform the same calculation for λ = 290 [nm] and compare.

References [1] Valley, S.l. (ed.) (1965) Handbook of Geophysics and Space Environments, McGraw Hill, New York. [2] Parrish, J.A., Anderson, R.R., Urbach, F. and Pitts, D. (1978) UV-A: Biological Effects of Ultraviolet Radiation with Emphasis on Human responses to Long wave Ultraviolet, Plenum Press, New York. [3] Inn, E.C.Y. and Tanaka, Y.J. (1953) Ozone absorption coefficient in UV and visible. Journal of the Opical Society of America, 43, 870–973. [4] Molina, L.T. and Molina, M.J. (1986) Absolute aborption cross sections of ozone in the 185–350-nm wavelength range. Journal of Geophysical Research, 91, 14501–14508, Table 1 [5] Jagger, J. (1985) Solar UV Actions on Living Cells, Praeger Special Studies, Praeger Publishing Division of Greenwood Press Inc., Westport, Conn., USA. A general text on the interaction of light with biomolecules in relation to the ozone problem.

Light and Matter

29

[6] Ito, T. (1993) UV-B observation network in the Japan Meteorological Agency. Frontiers Photobiology, 1, 515–518. [7] Simon, J.D., Peles, D., Wakamatsu, K. and Ito, S. (2009) Current challenges in understanding melanogenesis: bridging chemistry, biological control, morphology, and function. Pigment Cell and Melanoma Research, 22, 563–579. [8] Tran, M.L., Powell, B.J. and Meredith, P. (2006) Chemical and structural disorder in eumelanins: A possible explanation for broadband absorbance. Biophysical Journal, 90, 743–752. [9] Huijser, A., Pezzella, A., Hannestad, J.K. et al. (2010) UV-dissipation mechanisms in the eumelanin building block DHICA. ChemPhysChem, 11, 2424–2431.

3 Climate and Climate Change Climate at a certain location is defined as the average weather conditions over a period of some years. It not only comprises the temperature, but also the temperature variations during a year, the rainfall, the cloudiness, the humidity and the like. The climate system described by physical models consists of many components, depicted in colour Plate 1 at the end of this book and in Figure 3.1 in shades of grey. The main drivers of climate are the solar input (top left in the figure), the terrestrial radiation (centre-left) and the atmospheric gases and aerosols (centre). Climate change is caused by changes in these drivers, which for a large part are located in the atmosphere. Therefore in Section 3.1 the vertical structure of the atmosphere is discussed, together with the vertical motion of air. The radiation balance and the greenhouse effect follow in Section 3.2. The uneven distribution of solar input over the globe, the rotation of the earth and the geography of land and oceans cause a horizontal transport of energy by air and water. Their equations of motion are input for climate modelling and given in Section 3.3. The currently observed climate change is the combined effect of natural variability and human-induced climate change. The natural variability can be studied by analysing the climate change before the beginning of the industrial revolution, around 1750. That will be the subject of Section 3.4. The remainder of this chapter deals with modelling (Section 3.5) and the work of the Intergovernmental Panel on Climate Change IPCC (Sections 3.6 and 3.7; its social context is discussed in Section 9.4.2). IPCC is not, by itself, a research organization. It consists of groups of scientists, organized in working groups, who summarize and interpret the literature on climate and climate change. They refer to papers appearing in established, peer-reviewed journals, compare the papers and draw conclusions as to the state-of-the art knowledge in their field. IPCC has a role in political decision-making as it publishes the consequence of policy or the lack of policy concerning the climate. In practice IPCC working groups also have a steering role in research policy, as they point out gaps in knowledge and understanding of climate phenomena. Some of the IPCC graphs which we use in this book are produced by the IPCC Environmental Physics: Sustainable Energy and Climate Change, Third Edition. Egbert Boeker and Rienk van Grondelle. © 2011 John Wiley & Sons, Ltd. Published 2011 by John Wiley & Sons, Ltd.

32

Environmental Physics

Figure 3.1 The components of the climate system, which define climate and climate change. (Reproduced with permission of Cambridge University Press, copyright 2007, from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the IPCC (Chapter 3, Ref. [1], FAQ 1.2, Fig. 1).) See also Plate 1.

working groups, others are taken from the literature. For simplicity we only refer to pages in the IPCC reports and leave it to the reader to go back to the original papers, if necessary. This chapter should provide the student with enough insights to appreciate the science basis of climate modelling and the consequences of an unrestrained combustion of fossil fuels and other climate-change drivers. Note: in order to appreciate this chapter the student should be familiar with Appendix B, up to Eq. (B11).

3.1

The Vertical Structure of the Atmosphere

The atmosphere extends horizontally with a scale in the order of the circumference of the earth, which is 40 000 [km]. The vertical scale, shown in Figure 3.2, is much smaller, at most 100 [km]. The changes of the weather are taking place in the so-called troposphere. Looking at the graph from bottom to top, it shows that the fall or rise of temperature with altitude defines in order the troposphere, stratosphere, mesosphere and thermosphere. The regions separating these ‘spheres’ are called the tropopause, stratopause and mesopause. Their altitude varies with latitude.

Climate and Climate Change

33

60

Mesosphere Stratopause

40 Stratosphere 20

Tropopause

Summer pole Noctilucent cloud Temp e

ratur e

Mother-of-pearl cloud

Cloud vapor aerosol

Troposphere 0 -80

Air density / [kg m-3]

Mesopause

Ionosphere

80

Turbosphere Uniform N2,O 2,Ar proportions

Altitude above sea level / [km]

Winter pole

-60

-40

-20

0

O3

}

10-6

0.001

10-5

0.01

10-4

0.1

10-3

1

10-2

10

10-1

100

1

Atmospheric pressure / [hPa]

Thermosphere 100

1000

20

Temperature / [ °C] Figure 3.2 The vertical structure of the atmosphere. The left vertical scale denotes the altitude which on the right vertical scale corresponds with respectively the air density and the atmospheric pressure. Horizontally one finds the temperature with a significant dependence on the seasons around 80 [km] of altitude. The names of the atmospheric regions are indicated. (By permission of Oxford University Press.)

One notices the increase in temperature above 80 [km]. This is mainly due to the photodissociation of molecular oxygen O2 into atomic O. These atoms strongly absorb solar radiation of wavelengths between 100 and 200 [nm]. Alternatively, dissociation into O+ ions and electrons may occur, giving rise to the ionosphere which reflects radio waves. Similarly, absorption of UV sunlight by O2 between 20 and 40 [km] of altitude produces stratospheric ozone O3 , which strongly absorbs solar radiation with wavelengths of 200 to 300 [nm] and causes a general increase in temperature in these regions (cf. Chapter 2). The troposphere is the scene of strong vertical motion of air caused by the heating of the earth’s surface. Its temperature profile can be understood by the adiabatic expansion of a rising parcel of air. ‘Adiabatic’ means that no heat is exchanged with the local environment. To a very good approximation a parcel of air is in hydrostatic equilibrium with the air above and below. In Figure 3.3 a cylindrical parcel is shown with cross section A. Let the pressure at height z be p(z) and at height z + dz be p + dp. In vertical equilibrium the force upwards pA must be equal to the force downwards (p + dp) A + gρAdz. This leads to the equations p A = ( p + d p) A + gρ A dz

(3.1)

d p = −gρ dz

(3.2)

or

34

Environmental Physics

p+dp

A

z+dz

z

A p

ρgA dz Figure 3.3 Derivation of the hydrostatic equation (3.2). An air parcel of area A is in hydrostatic equilibrium with the ambient air.

which is called the hydrostatic equation. In the lower troposphere one may take the gravity acceleration g as constant; the density ρ has a strong exponential decrease with increasing altitude (as can be seen in Figure 3.2) and it may not be taken constant. However, one may derive a relation between ρ and p as the thermodynamic behaviour of the air in the atmosphere can be approximated rather well by the equation of state of an ideal gas: pV = n RT

(3.3)

Here, n is the number of moles in the volume V, where 1 [mole] equals M [g] and M is the molecular weight of the air as a mixture of gases (Exercise 3.1). Also, R is the universal gas constant, given in Appendix A. The mass m of the air is given by m = n M × 10−3 [kg]

(3.4)

The specific gas constant for air is defined as R =

103 R M

(3.5)

Its numerical value is given in Appendix A. By substituting n from (3.4) and R from (3.5) into Eq. (3.3) one obtains p=

m  R T = ρ R T V

(3.6)

With the hydrostatic equation (3.2) it follows that ∂p g p = −gρ = −  p = − ∂z RT He

(3.7)

In most of the atmosphere (Figure 3.2) the temperature T varies between 200 and 300 [K]. Taking some average value T av for T one may define an effective scale height H e = R T av /g [m] as was done on the right-hand side of Eq. (3.7). Then Eq. (3.7) gives an exponential decrease of pressure with height as p = p0 e−z/He

(3.8)

Climate and Climate Change

35

For T = 250 [K] one finds H e ≈ 7.3 [km] and a decrease of a factor of 10 in pressure for an increase of altitude with 16.8 [km]. This indeed agrees with the logarithmic character of the scales on the left and right in Figure 3.2 and correctly gives their relationship. When a parcel of dry air moves up without exchange of heat with its surroundings, the relationship between temperature and altitude follows the so-called dry adiabat. Similarly, a parcel of wet, saturated air will condense part of its water vapour and follow the saturated adiabat. The resulting formation of clouds is an important element of the climatic situation and is accounted for by sophisticated models. These three subjects will be discussed below. 3.1.1.1

The Dry Adiabat

Consider 1 [kg] of air. The first law of thermodynamics δ Q = cV dT + p dV

(3.9)

expresses that some of the added heat δQ is used to increase the temperature by dT, while the remaining heat energy is used to perform work p dV. For constant volume the only use for the heat is to increase the temperature, therefore cV is the specific heat for constant volume [J kg−1 K−1 ]. Consider a parcel of air, which may deform and expand, but retains its mass of 1 [kg]. Its volume then is related to its density ρ by V = 1/ρ. Eq. (3.9) then may be rewritten with the help of Eq. (3.6) as     1 p 1 1 = cV dT + d − d p = cV dT + R  dT − d p δ Q = cV dT + p d ρ ρ ρ ρ 1 (3.10) = c p dT − d p ρ Here, cp = cV + R is the specific heat at constant pressure, which follows from taking dp = 0. For dry air without exchange of heat with its surroundings δQ = 0 and the relation between temperature and pressure follows from the right-hand side of Eq. (3.10) as cp dT = dp/ρ. This may be rewritten as dT = d p/(ρc p ) = R  T d p/( pc p ) or R T dT = dp pc p

(3.11)

where Eq. (3.6) was used. Eq. (3.11) may be used to derive the famous Poisson equation for adiabatic change of an ideal gas pV κ = constant

(3.12)

in which κ = cp /cV (Exercise 3.2). The dry adiabat is given as the change of temperature with altitude z −gp ∂T ∂p R T g ∂T × = × = =− ∂z ∂p ∂z pc p RT cp

(3.13)

36

Environmental Physics

The quantity  d = g/cp is called the dry adiabatic lapse rate and represents the slope of the curve T(z). With the data of Appendix A one finds  d ≈ 0.01 [K m−1 ]. The minus sign in Eq. (3.13) illustrates that rising air will expand and consequently will cool. For wet, but not saturated, air with a mass fraction ω of water vapour Eq. (3.13) may be used with the understanding that c p = (1 − ω)c p,air + ωc p,water vapour 3.1.1.2

(3.14)

The Saturated Adiabat

If the parcel of humid air rises and cools, the water vapour it can contain decreases and part of the water vapour will condense. The mass fraction ω of water vapour in the parcel will increase an amount dω < 0. Here (–dω) is the mass fraction that has condensed. The evaporation heat per unit of mass is given as Hv [J kg−1 ]. Eqs. (3.9) and (3.10) give δ Q = (Hv )(−dω) = c p dT −

1 dp ρ

(3.15)

A derivation similar to the one from Eqs. (3.11) to Eq. (3.13) yields dT =

Hv 1 dp − dω ρc p cp

Hv ∂ω Hv ∂ω g ∂T × × =− − = −d − = −s ∂z cp cp ∂z cp ∂z

(3.16) (3.17)

The saturated adiabatic lapse  s rate is smaller than the dry rate  d , as ∂ω/∂z < 0. This is clear because the liberated condensation heat compensates somewhat for the cooling by expansion. This effect is largest for warm air, as it can contain a lot of water vapour, leading to a factor of 2 or even 3 between both lapse rates. 3.1.1.3

Clouds

The principle of cloud formation may be illustrated by the formation of a cumulus cloud, as depicted in Figure 3.4. At ground level a parcel of air is heated and starts rising. It contains water vapour, and its temperature decreases according to Eqs. (3.13) and (3.14). In the example of the graph the parcel is somewhat warmer than its surroundings and keeps rising following its adiabat (this situation is called dry, unstable). At level A condensation starts and the parcel follows the saturated adiabat with a somewhat smaller slope; in the graph the parcel now is considerably warmer than the ambient air, it keeps rising and condensing until all water vapour has been condensed at level B (this situation is called wet, unstable). It then follows again a dry adiabat, where from level C the parcel appears to be cooler than the ambient air. It therefore stops rising. (This situation is called dry, stable). In between levels A and B a high cloud, a cumulus, has been formed.

3.2

The Radiation Balance and the Greenhouse Effect

The two principal factors determining the earth’s surface temperature are the solar flux in and the terrestrial flux out. The solar input is proportional to πR2 , which is distributed

Climate and Climate Change

37

C B Γs Altitude z

cloud

Γd

A

Temperature T Figure 3.4 Formation of a cumulus cloud. A rising parcel of air first follows a dry adiabat, starts condensation at level A following a saturated adiabat. At level B all vapour has left and a dry adiabat is followed again. The ambient temperature is drawn; the ‘path’ of the parcel of air is dashed.

over the total surface of 4πR 2 and results in S/4 as average input, in the same way as in Section 1.2. In an equilibrium situation the solar influx and the terrestrial outflux are in balance. This radiation balance of the earth, in its simplest version is given in Figure 3.5. The atmosphere, in the middle region of the figure, absorbs solar radiation and scatters or reflects part back into outer space. The remainder reaches the earth’s surface, where part is reflected and part is absorbed. The 30% reflection and absorption in the figure comprises reflection by the earth’s surface and the atmosphere, including clouds, and represents the albedo of the earth as viewed from outer space, given in Section 1.2. The surface loses energy by emission of IR, most of which is radiated back, that is, downwards, by the atmosphere. The relevant data are given on the right in Figure 3.5. The net effect of the radiation is warming of the surface by 31 units and cooling of the atmosphere by the same amount. This is compensated by two effects. The first is the evaporation of water, which cools the surface and after condensation heats the atmosphere (26 units). The second is the convection of heated air upwards following the dry adiabat discussed before (5 units). The student may check that for outer space, for the atmosphere and for the surface, the incoming energy equals the outgoing, as it should in an equilibrium situation. On the right in Figure 3.5 one observes that most of the IR emission of the earth is absorbed by the atmosphere. Only 7% of the emitted 114% reaches outer space; this may be interpreted as a transmission of t = 7/114 = 0.06 of the atmosphere for IR radiation. The absorption of IR radiation by the atmosphere is strongly dependent on its wavelength. This is illustrated in Figure 3.6, which shows the upward IR flux at the tropopause.

38

Environmental Physics Solar radiation

Thermal radiation

30%

63%

7%

a Sc tte ga

or Abs

rin

Kinetic energy 0.8 [MJ m -2 ]

nd

n ptio

ec

l ref

23%

1%

Deep space

Emission

100% = S/4

1%

tio

Static energy 2620 [MJ m -2]

n

Internal energy 1880 [MJ m -2]

Atmosphere

Potential energy 740 [MJ m -2] 98%

47% Absorption

26%

5%

Emission

Latent heat 64 [MJ

m-2]

107% + 15 [°C] 114%

Evaporation Convection Abs Emission

Earth

Figure 3.5 Radiation balance for the earth and the atmosphere. The solar input is taken as 100% and its destiny is displayed. The bottom line depicts the earth surface, the upper part outer space and the atmosphere occupies the region in between. For the atmosphere estimates of the energy content are displayed. (Reproduced with permission of the Dept of Geography, ETH Zurich, copyright 1990 from [3], p. 28.) ¨

Figure 3.6 The intensity of the radiation (2.4) is displayed as a function of μ = 1/λ. The solid line shows the upward Infrared flux Iμ at the tropopause [Wm–1 ]. It is fitted with three dashed lines, corresponding to black-body radiation at temperatures of 294 [K], 244 [K] and 194 [K]. Note that the wavelength λ is shown (nonlinearly) on top of the figure. (Reproduced with permission of Cambridge University Press, copyright 2007, from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the IPCC (Chapter 3, Ref. [1], FAQ 1.2, Fig. 1).)

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39

It is fitted with three black-body curves of which the curve with T = 294 [K] gives the best results. This is a little larger than the T = 288 [K] which is adopted in this book as the temperature of the earth, but the curve will be very similar. It appears that the atmosphere is relatively transparent for wavelengths in a ‘window’ between 8 [μm] and 12.5 [μm], but opaque at wavelengths around 15 [μm]. In general, greenhouse gases are gases in the atmosphere that absorb IR radiation. The most important are CO2 and H2 O, but we will see later that others also contribute. The transmission t = 0.06 of the atmosphere refers to the radiation passing the window. The rest of the IR emission is absorbed, re-emitted, absorbed and so on, in a process of radiative transfer to be discussed in Section 3.2.2. At this point we just notice that the thermal radiation upwards (63 units) and downwards (98 units) in Figure 3.5 is different. This can be explained by the fact, shown in Figure 3.2, that the lower atmosphere is warmer than the upper atmosphere. The lower atmosphere emits more black-body radiation than the upper atmosphere because of the Stephan–Boltzmann law σ T 4 (see Eq. (2.6)). Both layers emit equally up and down. Outside the window shown in Figure 3.6 most upward radiation from the lower atmosphere is absorbed, like most of the downward radiation from the upper atmosphere. One is left with warm radiation downwards and cooler radiation upwards. In the following we first discuss the effect of simple changes in the radiation balance (Section 3.2.1), next we discuss radiation transfer in the atmosphere (Section 3.2.2) with an analytical model in Section 3.2.3. The results are used to describe global warming (Section 3.2.4) and the properties and effects of the greenhouse gases (Section 3.2.5) 3.2.1

Simple Changes in the Radiation Balance

From Figure 3.2 it follows that most of the mass of the atmosphere is concentrated in the troposphere. Most changes occur in the troposphere; the higher layers of the stratosphere are much more stable and adapt quickly to changes in the radiation balance. This is the reason why below the radiation balance is studied at the top of the troposphere. At the top of the troposphere Figure 3.5 may be summarized as S (3.18) = σ Ta4 + tσ Ts4 4 The left-hand side summarizes the incoming solar radiation; the right-hand side adds the contribution from the higher troposphere with the terrestrial IR radiation transmitted through the window. The albedo of the earth a = 0.30 is composed of contributions from all parts of the earth, the troposphere and the clouds. It represents some average of the individual albedos shown in Table 3.1. We cannot proceed without a model, but for lack of it we just take the difference Ts – Ta = constant. We then may make a few qualitative observations of the behaviour of the equilibrium surface temperature Ts under extreme conditions (Exercise 3.5). (1 − a)

3.2.1.1

The White Earth

Assume that the earth’s surface is completely covered with snow, both on land and on the oceans (ice and snow). The resulting albedo is high, Table 3.1 suggests a value a = 0.50. One then finds a surface temperature Ts = 268 [K], which is well below the freezing point

40

Environmental Physics Table 3.1 Mean values of albedos for various surfaces. Surfaces

%

Clouds

Horizontal water (low solar angle) Fresh snow Sand desert Green Meadow Deciduous Forest Coniferous Forest Crops Dark soil Dry earth

5 85 30 15 15 10 10 10 20

Cumuliform Stratus Altostratus Cirrostratus

% 70–90 60–85 40–60 40–50

With permission of John Wiley, copyright 1977, reproduced from [5], p. 30.

of sea water, which is 271.1 [K]. This means that the earth would remain covered with snow and ice; in other words a white earth is a stable solution of the energy balance equations. It is, however, so remote from the present situation that the transition to a white earth is hardly thinkable.

3.2.1.2

The Nuclear Winter/Extinction of Dinosaurs

The climatic consequences of nuclear war were seriously discussed in the 1980s. It was argued that the explosion of a few hundred nuclear warheads would lead to enormous fires all over the world. That would bring great quantities of small particles of dust and smoke into the atmosphere, essentially cutting much of the earth’s surface off from sunlight. Even a slightly larger value a = 0.35 would lead to Ts = 284, a cooling of 4 [◦ C]. It is to be hoped that mankind will be spared a nuclear winter. A similar phenomenon may have occurred some 66 million years ago with the extinction of the dinosaurs. Many specialists believe that an asteroid hit the earth, which would have raised a global dust cloud with the same cooling effect, possibly resulting in lack of food.

3.2.1.3

The Cool Sun

If the sun behaves like other stars of its type, some billions of years ago the solar irradiance S must have been considerably smaller than nowadays. Two billion years ago for example, it would have been 85% of its present value and 4 billion years ago some 72% of its present value [5]. In our approximation this would lead to temperatures of 278 and 268 [K], respectively. Still life existed even then, requiring temperatures well above the freezing point of water (273 [K]). It is therefore believed that due to an elevated concentration of greenhouse gases in the atmosphere, the smaller solar irradiation was largely compensated. The increase of the solar irradiance with time must to a significant extent have been compensated by the fixation of greenhouse gases like CO2 .

Climate and Climate Change

Solar

Terrestial

41

Terrestial

Window

WINDOW

Visible

k

(9 [m])

(9100 [m])

(9 [m])

Figure 3.7 Absorption of radiation in the atmosphere. A value of 100% indicates that the radiation is not able to pass through the entire atmosphere. For solar radiation this is defined from top to surface, while for IR radiation this is defined from the earth surface to the top of the atmosphere. The left-hand figure refers to a broad range of wavelengths and is drawn horizontally on a logarithmic scale. The right-hand figure is restricted to the most important greenhouse gases H2 O and CO2 . Underneath the averaged values of the absorption coefficient k for part of the spectrum is indicated (cf. Eq. (3.19)). (With permission of Oxford University Press, copyright 2010, reproduced from [2], Figure 8.5, p. 290.)

3.2.2

Radiation Transfer1

In Figure 3.6 it was shown that the atmospheric gases absorb IR radiation at many wavelengths outside a ‘window’ between 8 and 12.5 [μm]. In Section 3.2.5 we will find that the gases which contribute most significantly to this absorption are CO2 and H2 O. This is shown explicitly in Figure 3.7. The left picture refers to the complete solar and terrestrial spectrum and therefore is drawn horizontally on a logarithmic scale. On the vertical scale 100% absorption at a wavelength means that radiation at that wavelength cannot cross the atmosphere directly. As to the IR window around 10 [μm], it appears that CO2 and H2 O hardly absorb there, although there is some absorption by O3 . The absorption lines shown appear in bands, much broader than the narrow lines of individual molecules that one may find in the literature [7]. One reason is that there are often several absorption lines close together. Each of them is broadened, amongst other things due to collisions between molecules in the air which shorten the lifetimes of their excited states (Section 8.1.3). These facts together result in the appearance of absorption bands. The right-hand side of Figure 3.7 shows a very schematic representation of the absorption of CO2 and H2 O. Underneath the averaged values of the absorption coefficient k for the part of the spectrum is indicated. These values are to be considered as approximations connected with the simplified analytical model to be discussed below.

1 The authors are indebted to Dr Rob van Dorland for his assistance with this subsection. This subsection and the next one are based on Van Dorland’s thesis [6].

42

Environmental Physics

We are interested in the upward flux of long-wave radiation at the top of the troposphere and its change by the addition of greenhouse gases. For global temperature changes the effects of changes of atmospheric greenhouse gas content on the net flux at the top of troposphere are dominant (Section 3.2.4). Since the changes in upward flux are more important than the changes in the downward component, we consider here the upward flux and its change by the addition of greenhouse gases. We will derive a general formula for the upward flux at an altitude z in the atmosphere. Figure 3.8 shows that there are two contributions, one from the surface and one from the underlying layers of the atmosphere. To be more specific, we consider a certain spectral range λ of the black-body spectrum (2.4). Its emission will be called B [Wm−2 ]; from the surface of the earth it will be denoted by an index s which gives Bs and from an arbitrary altitude z in the atmosphere it will be indicated by B(z ). Its dependence on the wavelength will not be indicated explicitly. The contribution from the surface follows from Lambert–Beer’s law (2.36), which has already been derived as Fs+ = Bs e−kz

(3.19)

In order to find the contribution from the underlying layers we consider 1 [m2 ] in a strip between z and z + z , as indicated in Figure 3.8. Such a strip of gas does not emit in the same way as a solid surface. Its emission, for example, will be proportional to its thickness. Its emission is found by imagining the strip in a cavity which is in thermodynamic equilibrium. In this case black-body radiation is absorbed and the same amount of radiation is emitted again. Even without a cavity local thermodynamic equilibrium would require that the black-body emission equals the absorption of a hypothetical incoming black-body beam. We assume that this is the case. From Eq. (2.35) or (2.36) it follows that over the strip the absorption of a beam with intensity I will be given by I = –kIz . If a black-body spectrum B(z ) entered this strip the absorption would be B = –kBz . For local thermodynamic equilibrium the emission + (z  ) from should be equal to kBz . Therefore, for any incident beam the emission Fatm the strip should have this behaviour: + (z  ) = k B(z  )z  Fatm

Bse-kz

ΔF +atm

2

1 [m ] Bs

(3.20)

z

z’+Δz’ z’

surface

Figure 3.8 The upward radiation flux at height z in the atmosphere has a contribution from the surface and from underlying layers. The black-body contribution from the strip between z + and z + z is indicated by F atm .

Climate and Climate Change

43

The upward flux from the atmosphere at location z is given by the integral, where the absorption is taken into account by Eq. (2.36) with the proper distance z – zt

+ Fatm

z =



k B(z  ) e−k(z−z ) dz 

(3.21)

0

For the downward flux a similar relation can be derived. Equations (3.19) and (3.21) describe the radiation transfer in a vertical direction. Note that the absorption coefficient k defined in Eq. (2.37) is dependent on the absolute concentration of the gas, which is N 1 /V. In deriving Lambert–Beer’s Law (2.36) k was assumed to be independent of the coordinate. In order to proceed with simple formula we stick with this assumption knowing that precise results have to be computed numerically. Then the decrease in absolute concentration with altitude has to be taken into account. From the derivation of Lambert–Beer’s Law (2.36) it also follows that the absorption coefficient is dependent on the transition probability B12 and therefore is a function of the wavelength of the radiation. In an extremely simplified way, values of k are given in Figure 3.7), where for the atmospheric window a value kw = (9100 [m])–1 is given and outside the window koutw = (9 [m])–1 . This means that at a distance of 9 [m] the intensity of an IR beam outside the window decreases its intensity by a factor 1/e ≈ 0.37. If the human eye was only sensitive to λ ≈ 15 [μm] it would be difficult to survive. In fact at this wavelength the CO2 absorption is very strong, leading to k ≈ [cm−1 ] in this wavelength region resulting in a visible depth of only a few [cm]. In order to proceed one needs to know the function B(z). We distinguish between the window region 9–12.5 [μm] and the rest of the spectrum. At the surface with a temperature of 288 [K] one finds Bsw = 110 [Wm−2 ] and outside the window Bsout = 280 [Wm−2 ] (Exercise 3.6). For the change in B(z) with altitude we make two different assumptions. The first is a linear decrease in the atmospheric temperature with altitude up to the tropopause, which is suggested by Figure 3.2. Then it is simple to integrate Eqs. (3.19) and (3.21) numerically and find the upward going fluxes at the top of the troposphere. This will be done in Example 3.1; the second assumption is a linear decrease of B(z) with z; this will be done in Section 3.2.3.

Example 3.1

Radiation at the Top of the Troposphere

(a) Put the temperature at the top of the troposphere (13000 [m]) as 247 [K] and write down the linearly decreasing function T(z). (b) Calculate the surface contribution (3.19) at the top of the troposphere inside the window and outside the window. Compare with the data in Figure 3.5. (c) Similarly compute the up going atmospheric radiation on top of the atmosphere inside and outside the window and compare with Figure 3.5.

44

Environmental Physics

Answer: (a) Write T(z) = a + bz; substitute 288 [K] for z = 0 and 247 [K] for z = 13 000. It follows that T(z) = 288 – 0.00315z. (b) Inside the window Bsw = 110.45 [Wm−2 ], to be multiplied by e–13000/9100 which gives 26.5 [Wm−2 ]. Outside the window one has Bsout = 280 [Wm−2 ], to be multiplied by e–13000/9 , which is negligible. The result 26.5 [Wm−2 ] has to be compared with 0.07 × S/4 = 23.9 [Wm−2 ] of Figure 3.5, which is a reasonable agreement. (c) The Planck function (2.4) is dependent on temperature. Strictly speaking the fraction of the spectrum inside and outside the window will be a function of temperature, and therefore also of altitude. We ignore this for the rather small range of temperatures and write inside the window Binside (z) = (110.45/390.5) × σ T(z)4 ; integration of (3.21) up to 13 000 [m] is straightforward and gives 58.8 [Wm−2 ]. Outside the window, similarly, Boutside (z) = (280/390.5) × σ T(z)4 , which results in 151.4 [Wm−2 ]. Together this gives 210.2 [Wm−2 ] to be compared with 0.63 × S/4 = 215.2 [Wm−2 ], again a reasonable agreement. Note the following: (1) The temperature of 247 [K] at the tropopause was chosen so as to give reasonable results for this simple model. A more realistic value would be 220 [K]. (2) Because of the large value of k outside the window only the highest strip in the atmosphere contributes to Eq. (3.21). The fact that the decrease of k with altitude was not taken into account overestimates the absorption in the top layers, which here was corrected by using a high value for the temperature at the tropopause. (3) The contribution of the atmospheric emission at the top of the troposphere is significant both inside and outside the window. (4) A realistic calculation should divide the spectrum into many bands.

The addition of more gases to the atmosphere, which absorb in the IR, the so-called greenhouse gases, will increase the value of the absorption coefficient k at certain wavelengths. Without a temperature change of the earth or of the atmosphere the total upward going flux at the tropopause will become smaller. This effect will be largest in the window, where the absorption is still small. This statement may be checked numerically for the model in Example 3.1 (Exercise 3.8).

3.2.3

A Simple Analytical Model

Van Dorland [6] introduced an analytical model to study the radiation balance. The decrease in black-body radiation with height is assumed to be linear B(z) = Bs − z

(3.22)

where  = −∂ B/∂z > 0. This is a drastic assumption and has to be qualified, but for our purpose it is sufficient to mention that the variable z in Eq. (3.22) increases with height [6].

Climate and Climate Change

Next substitute Eq. (3.22) into Eq. (3.21) and find after some algebra    + (1 − e−kz ) − z Fatm = Bs + k

45

(3.23)

Add the surface contribution (3.19) and find for the total upward flux  (3.24) (1 − e−kz ) k The effect of the increase of the concentration of greenhouse gases and hence of k is shown by differentiating (3.24) with respect to k. This gives F + (z) = Bs − z +

   dF + (3.25) = − 2 1 − (1 + kz)e−kz dk k One may verify that the expression between brackets always is positive, hence the derivative (3.25) always is negative and the upward flux decreases with increasing k. In an atmospheric window one will have k  1 and may approximate (3.25) by dF + /dk = –z2 /2, which would increase with z. Outside the window(s), in an opaque region one will have k > 1 and (3.25) may be approximated by dF + /dk = –/k2 , which is very small for large k. Increase of the absorption k gives hardly any decrease in flux, a phenomenon which is known as saturation. 3.2.4

Radiative Forcing and Global Warming

At this point we understand that one may calculate the decrease of upward going flux at the top of the troposphere by the addition of greenhouse gases. To be complete one also has to compute the change in downward flux originating from changes in the stratosphere. All taken together there will be a decrease I of the upward going flux at the top of the troposphere. The climate system will react by increasing the temperature of the earth’s surface and changing the temperature of the atmosphere. A convenient way to quantify the climatic effect of the human addition of greenhouse gases is given by the concept of radiative forcing. The idea can be illustrated by Figure 3.9, where a simple atmosphere is sketched. The net radiation flux at the top of the atmosphere will vanish under equilibrium conditions. Assume a sudden increase in the concentration of greenhouse gases. This would lead to a net reduction in the outgoing long-wavelength radiation at the top of the atmosphere

Flux out - flux in = -ΔI Top atmosphere In Out Earth’s surface

Figure 3.9 A sudden decrease of outward flux I at the top of the atmosphere will lead to an increase in the surface temperature.

46

Environmental Physics

by I. The incoming energy from the sun remains the same. The energy balance at the top of the atmosphere should be restored by an increase T of the earth’s surface temperature. This effect is called radiative forcing: the radiation imbalance at the top of the troposphere enforces a rise in temperature([1] p. 134). The required flux increase I to compensate for the decrease by greenhouse gas absorption will be connected with the increase of the surface temperature T s by the relation I =

∂I Ts ∂ Ts

(3.26)

The intensity I = tσ Ts4 is the earth’s direct outgoing radiation measured at the top of the atmosphere. We ignore changes in the radiation transfer by the atmospheric layers. It then follows with Eq. (1.5) that 4I 4 S ∂I = 4tσ Ts3 = = (1 − a) ∂ Ts Ts Ts 4

(3.27)

Here Eq. (1.5) was used. Substitution of the correct values and a = 0.30 gives ∂ I /∂ Ts = 3.3 [Wm−2 K−1 ]. Eq. (3.26) is often written the other way round as ([8] p. 656) Ts = GI

(3.28)

where G ≈ 0.3 is called the gain factor, which operates on the cause I to produce the effect T s . The concept of radiative forcing dates back to the beginning of climate modelling, but is still widely used to compare models, or to compare the contribution of the individual greenhouse gases to climate change. In fact all changes in the components of the climate system depicted in Figure 3.1, such as changes in land use, may be computed in terms of a forcing. In Section 3.6 we will show the comprehensive summary of all forcings, taken into account by IPCC. This leads to a total, averaged global forcing (since 1750) in 2005 as I = 1.6 [Wm−2 ] ([1], p. 204). With Eq. (3.28) this would give T s = 0.5 [◦ C]. This calculation is too simple for two reasons, which we will briefly discuss below. Firstly, a temperature increase at the surface will lead to effects which will reinforce or mitigate the warming effect. Secondly, the oceans have an enormous heat capacity, which will slow down global warming. 3.2.4.1 Effects Reinforcing Global Warming r Melting of ice and snow will lower the albedo. r More water vapour in the air will lead to smaller transmission t and a higher value of k. r The increase in the cloud cover from higher evaporation will have the same effect. r Several processes will cause (further) increase in CO2 concentration: a higher sea water temperature gives less CO2 absorption in sea water; higher polar temperatures will cause a smaller ocean circulation and decreasing absorption; a faster decay of organic materials will give more CO2 and CH4 (from rotting). r Thawing of permafrost, covering some 20% of the earth’s land surface, may liberate large quantities of CH4 and CO2 trapped in the frozen soil ([1], pp. 77,110).

Climate and Climate Change

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r More CO2 leads to an increased growth of plants, which may lower the albedo, according to Table 3.1. 3.2.4.2 Effects Mitigating Global Warming r An increase in sea water temperature would lead to an increase in the growth of algae. They would use CO2 for photosynthesis and reduce the CO2 concentration in the water, absorb more from the atmosphere and reduce the atmospheric CO2 concentration. r Most CO2 is due to burning of fossil fuels, which liberates aerosols, small particles that will increase the earth’s albedo a by backscattering incoming solar radiation. r With a higher surface temperature more sea water will evaporate. At higher altitudes it will condense, heating the upper layers more and cooling the surface. r More CO2 leads to an increased growth of plants which will bind CO2 . Taking everything together, one can replace Eq. (3.28) by Ts = G f I

(3.29)

where the precise value of Gf [K/(Wm−2 ] has to be found by detailed modelling. The quantity Gf [K/(Wm−2 ] is called the climate sensitivity parameter (λ in [1], p. 133 or S in [1], p. 825), defined as the increase in temperature [K] for a radiative forcing of 1 [Wm−2 ]. Its value may be estimated from the data IPCC [1] gives for a doubling of the CO2 concentration, viz. T s ≈ 3.0 [◦ C] ([1], p. 799) and I ≈ 3.7 [W/m2 ] ([1], p. 140). This gives Gf ≈ 0.8 [K/(Wm−2 )]. With the forcing I = 1.6 [Wm−2 ] mentioned before one can find the temperature rise, since 1750 as 1.3 [◦ C]. 3.2.4.3

Time Delay by Ocean Warming

Equations (3.26) to (3.29) were derived assuming a sudden increase in greenhouse gas concentration with an immediate response of the surface temperature by T s . We now still keep the sudden increase in greenhouse gas concentrations, leading to the same extra downward flux I [Wm−2 ]. That flux is used to heat the surface which, we assume, will be ocean only. The upper layer of the ocean is well mixed. The mixing with the lower layers is much less. Therefore in a first approximation the incoming heat is used to warm the top mixed layer of the ocean. Let the heat capacity of a column with a cross section of 1 [m2 ] in the mixed layer be cs [J m−2 K−1 ]. The temperature increase T s of the water will be a function of time T s (t) Between times t and t + dt the incoming extra heat from above will be Idt [J m−2 ]. The temperature rise of the mixed layer resulting from this extra incoming heat at time t will be d(T s (t)) [K] for which energy cs d(T s ) [J m−2 ] is needed. The temperature rise T s (t) induces up going radiation T s /Gf [Wm−2 ] at time t (see Eq. (3.7)). Between times t and t + dt that becomes (T s /Gf )dt [J m−2 ]. This leads to an equation with dimensions [J m−2 ] on the left- and right-hand sides: I dt =

Ts dt + cs d(Ts ) [J m−2 ] Gf

(3.30)

This equation expresses that the downward flux on the left-hand side is used for the upward flux (first term on the right) and the warming of the top ocean (last term on the

48

Environmental Physics

right). Dividing (3.30) by dt leads to I =

d(Ts ) Ts + cs Gf dt

(3.31)

This differential equation looks somewhat cumbersome, but if the student replaces T s (t) by f (t) it will be recognized as an inhomogeneous differential equation of first order. The solution becomes Ts (t) = G f I (1 − e−t/τ )

(3.32)

where τ = cs Gf . For long times t → ∞ the exponential vanishes and one will get back Eq. (3.29). For a water column with height h = 120 [m] one finds τ = 12.2 [yr] (Exercise 3.10). Note that in deriving Eq. (3.32) a sudden forcing was assumed. In order to find the present temperature rise one has to specify the rise of forcing with time and to calculate the temperature effect by applying Eq. (3.32). With a simple model Exercise 3.11 yields a temperature increase from 1850 to 2005 of 0.87 [◦ C], somewhat higher than the observed 0.76 [◦ C], from [1], p. 37. The difference may be due to the simplicity of our model or to the fact that the observed value is composed of natural and human-induced changes, or both. From Eq. (3.10) it is clear that even with a radiative forcing that is kept constant the temperature will keep increasing for some time. It is also clear that the parameter τ will control this process. The above estimate for τ only took into account the top layer of the oceans. In reality that layer will mix with lower layers by the deep ocean circulation, where water goes down at the poles, moves along the deep ocean and slowly moves up again with a time constant of about 300 years. This implies that even with constant forcing the temperature will keep increasing until the complete ocean has warmed up, which may not be for some 600 years. 3.2.5

The Greenhouse Gases

In Table 3.2 the concentrations of the most important greenhouse gases and their contribution to the greenhouse effect are given numerically. One finds a total warming of 33 [◦ C] Table 3.2 Greenhouse effect of the most important atmospheric gases. The warming effect is given as estimated in 1984 [9].

Trace gas CO2 N2 O CH4 HFCs PFCs SF6 O3 troposphere H2 O vapour

Warming Increase/ effect [%/yr] Conc/ppmv (1984)/[◦ C] Lifetime/(yr) (1998–2005) 379 0.32 1.77 ≈10−5 ≈10−4

5000

Other data are from [1], p 141, 212, 213.

7.2 1.4 0.8 0.6 2.4 20.6

50–200 114 12 1–270 1000 3200

0.49 0.22 0.09 ≈10 ≈3 3.8

GWP (20 yr)

GWP (100 yr)

1 289 72 >1000 >1000 16 300

1 298 25 >1000 >1000 22 800

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produced by these gases [1]. Their concentrations rise because of human activities with a rate also given in Table 3.2. The increase in these concentrations is important for the climate, as they contribute to the human-induced greenhouse effect. For a reliable calculation of the future effects of adding greenhouse gases to the atmosphere, one would need to take into account the absorption spectrum of each gas. Also significant is the time an emitted gas molecule stays in the air and thus contributes to the effect. Estimates of these lifetimes are given in the table. Finally, it is helpful to policy makers to have a simple number, which comprises the combined effect of all gases. For this purpose the concept of global warming potential (GWP) of a certain gas is introduced. As the increase of the CO2 concentration is most widely discussed, it is taken as the standard. The GWP of a certain gas then compares the warming effect of adding 1 [kg] of the gas to the warming effect of the addition of 1 [kg] of CO2 . Because of the different lifetimes the GWP will depend on the time horizon one considers. In the last column of Table 3.2 GWPs are given for 20 years and 100 years after emission. When one has decided on a time horizon, one may convert the emission of a certain greenhouse gases to the emission of CO2 by multiplying by the GWP. This is called the equivalent CO2 emission. The reason why CO2 is taken as the standard for emissions becomes clear from Table 3.2. The top six rows refer to gases which are directly emitted into the atmosphere by human activities. Of these, CO2 gives the largest contribution and its increase with time is well documented. For the period since 1970 it has been monitored accurately, as is shown in Figure 3.10. In summer photosynthesis binds more of the emitted CO2 than in winter, which explains the periodicity in the graph.

Figure 3.10 CO2 concentrations in the atmosphere, measured on Hawaii, far from industrial centres. (Reproduced with permission of Cambridge University Press, copyright 2007, from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the IPCC (Chapter 3, Ref. [1], FAQ 1.2, Fig. 1).)

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Figure 3.11 Increase of concentrations of the most important greenhouse gases over time. (Reproduced with permission of Cambridge University Press, copyright 2007, from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the IPCC (Chapter 3, Ref. [1], FAQ 1.2, Fig. 1).)

In Figure 3.11 earlier concentrations of the top three greenhouse gases CO2 , CH4 and N2 O are shown, as found from air bubbles in ice cores. It appears that the great increase in their concentrations started around 1750 with the beginning of the industrial revolution. Therefore, this year is taken as the reference point in calculating human influences. The three greenhouse gases shown in Figure 3.11 were present in nature even before 1750. Their concentrations are increasing due to human activities as summarized in Table 3.3. The most widely discussed greenhouse gas, CO2 , enters the air by combustion or burning of fossil fuels or biomass. Burning of pure coal would be represented by the chemical reaction C + O2 → CO2

(3.33)

but in practice fossil fuels and biomass consist of C, CH4 , Cx Hy , (CH2 O)n and many other trace elements. They all originate from photosynthetic processes. Table 3.2 shows in rows 4, 5 and 6 gases which originate only from human activities. The HFCs and PFCs comprise a set of chemicals, many of which were shown to contribute to the ozone hole, discussed in Section 2.4.1. Those are banned according to an international agreement, the Montreal protocol; others, however, are not banned and will remain as refrigerants or by-products of Al production and other industrial processes. Although their concentrations are still low, many of them show an increasing concentration and they have a large GWP. The origin of the chemical SF6 is indicated in Table 3.3.

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Table 3.3 Human sources of principal greenhouse gases ([1], 138–145). CO2 CH4 N2 O SF6

Combustion of fossil fuels, gas flaring, cement production (using CaO from CaCO3 ), biomass burning Wetlands (rotting), rice agriculture, biomass burning, ruminant animals Microbes in fertilized agricultural lands Electrical insulation fluid, inert tracer for studying transport processes

In the lower part of Table 3.2 one finds tropospheric ozone. This gas is a result of emissions of nitrogen oxides, carbon monoxide and organic compounds. Its greenhouse effect is taken into account in the calculations. Finally, in the bottom row one finds water vapour, which has the biggest greenhouse effect of all. In model calculations one starts from the major greenhouse gases, computes the climate effect, finds the change in water vapour concentration and cloud formation and takes that into account in a feedback loop.

3.3

Dynamics in the Climate System

In the previous sections the radiation balance was averaged over the globe. Figure 3.12 shows that this approximation is too simple to explain the temperature differences over the earth. The curve shows the absorbed solar radiation as a function of latitude. It peaks at the equator, where the sun every day reaches a high elevation and it is small at the poles, where the sun not only disappears part of the year, but the snow reflects much of the solar light as well. The absorbed solar radiation therefore will be small. The emitted long wave radiation σ T 4 , dashed in the graph, is indeed smaller at the poles than at the equator due to the lower temperature, but the net energy absorbed at the poles is negative, illustrated in the lower curve.

N(φ) / [Wm-2]

300

Absorbed solar Emitted longwave

200 100 0

Net radiation absorbed

-100

-90

-60

-30

0

30

Nnet 60

90

Latitude φ Figure 3.12 Absorbed solar radiation and emitted long wave radiation as a function of latitude. The graph is averaged over latitude circles and over the year. The net radiation absorbed is negative at the poles, implying energy transport from the equator to the poles. (This article was published in Global Physical Climatology, Hartmann, pg 37, copyright 1994, Elsevier.)

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Environmental Physics

Figure 3.12 shows that at high latitudes the emitted energy is larger than the absorbed energy. Consequently there must be an enormous energy transport from the equator to the poles.

Example 3.2 Energy Transport to the Poles Estimate the energy flux, which passes the latitude φ = 35◦ to the north. Answer Assume that all net absorbed energy north of the equator goes to the north. From Figure 3.12 we estimate that at φ = 0◦ the absorbed energy equals 75 [Wm−2 ] and it vanishes at φ = 35◦ We approximate the curve by a straight line and use radians instead of degrees. This gives Nnet (φ) = 75 − 123 φ [Wm−2 ] which has to be integrated over a strip with width R dφ, as shown in Figure 3.13. The resulting flux F is found as 0.61 (75 − 123 φ)2π R 2 cos φdφ F=

(3.34)

0

which gives F = 5.65 × 10 [W] or, accounting for the circumference of the earth at φ = 35◦ one finds 1.7 × 108 [Wm−1 ] = 170 [MWm−1 ] crossing the latitude circle: a medium-sized power station per metre. 15

N

R cosφ

R dφ

φ M

Figure 3.13 latitude φ.

R

The net absorption of Figure 3.12 is integrated over a strip with width Rdφ at

For realistic analyses the energy transport has to be modelled correctly and has to take into account the geography of the earth. For the oceans the situation is sketched in Figure 3.14.

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Figure 3.14 Global ocean circulation. Dark arrows indicate warm surface currents and grey arrows represent deeper return currents. A circle with a cross indicates downwelling and circles with a dot indicate upwelling of deep waters. (Reproduced from Netherlands Tijdschrift Natuurkunde, Dijkstra, 63, 79–83, 1997.)

One notices warm surface currents, indicated by dark arrows and colder return currents, indicated in grey. The surface currents are concentrated in space. The Gulf Stream in the North Atlantic, for example, has a width of 60 [km], is 500 [m] deep and moves with a velocity of 1 [m s−1 ]. This corresponds with a mass transport of 3 × 1010 [kg s−1 ]. The so-called Kuroshio near Japan moves a similar mass. They make up a considerable part of the energy transport, calculated in Example 3.2 (Exercise 3.12). The oceans transport energy as sensible heat, in the air much is transported as latent heat. At warm latitudes water evaporates and moves upwards in the atmosphere; next the humid air is transported and condenses at another place, liberating the evaporation heat as condensation heat. The orders of magnitudes are illustrated in Exercise 3.10. From these examples it follows that computations of climate, as for weather, require the equations for horizontal and vertical motions of air and water. We discuss them briefly, starting with the vertical motion of air.

3.3.1

Horizontal Motion of Air

Horizontally, the velocity of a parcel of air changes frequently in direction and magnitude, so, contrary to the vertical case of Section 3.1, the full equations of motion should be solved. Consider a volume dτ with mass ρdτ and velocity u. Newton’s equations of motion may be written as du ρdτ = Fpress + Fviscous + FCoriolis + Fgravity dt

(3.35)

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On the left-hand side one finds the traditional ‘acceleration times mass’ and on the righthand side the forces which operate on the parcel of air. These forces will be discussed successively below. 3.3.1.1

Pressure Gradient Forces

The first term on the right of Eq. (3.35) is the pressure gradient force Fpress . Its magnitude and direction are found from the left-hand side of Figure 3.15. Consider a rectangular block with its sides parallel to the coordinate axes and look in the x-direction. The force to the right equals p(x, y, z)dydz and the force to the left   ∂p dx dydz (3.36) p(x + dx, y, z)dydz = p(x, y, z) + ∂x The total force in the positive x-direction therefore becomes (−∂ p/∂ x)dτ

(3.37)

for the block dτ = dxdydz. In the y- and z-directions one finds the same expression with x replaced by y or z. This result is summarized by Fpress = −∇ p dτ

(3.38)

where the vector ∇ p has x–, y–, z– components ∂ p/∂ x, ∂ p/∂ y, ∂ p/dz. A short but comprehensive discussion of vector algebra and vector differentiation may be found in Appendix B. 3.3.1.2

Viscous Forces

On the right-hand side of Figure 3.15 one observes the same block as on the left-hand side, but in this case the velocities are indicated. Consider the x– component ux (x, y, z) of the velocity u and assume for the sake of the argument that the velocity of air is to the right. We follow Newton’s assumption that the viscous force at the bottom of the block will be proportional to −∂u x /∂z. The minus sign is understood by assuming that component ux increases with z, so ∂u x /∂z > 0. Then, the drag of the surrounding air on the block should be to the left, hence the minus sign. The student may check that with other assumptions the same equation follows. The proportionality constant μ is called the dynamic p(x+dx,y,z)dydz ux(z+dz) dz

dz

dy p(x,y,z)dydz

Figure 3.15

dy (xyz)

dx

ux(z) (xyz)

dx

Derivation of the pressure gradient force (left) and the viscous force (right).

Climate and Climate Change

viscosity [Pa s]. The force on the bottom therefore becomes  ∂u x  −μ dxdy ∂z x yz Similarly, the force on the top has the other sign  ∂u x  dxdy +μ ∂z x yz + dz The net force in the x– direction therefore becomes   ∂u x  ∂u x  ∂ 2u x dxdy + μ dxdy = +μ 2 dτ Fx = − μ   ∂z x yz ∂z x yz + dz ∂z

55

(3.39)

(3.40)

(3.41)

where, as before, the volume element dτ = dxdydz. The force Fy in the y– direction can be found by replacing ux by uy in Eq. (3.41). In the atmosphere the friction force essentially works in the horizontal (x– and y–) directions as the vertical (z–) component of the velocity can be neglected. Strictly speaking, the vertical is defined as the line connecting the centre of the earth with the location on the surface one is considering; the horizontal plane then is perpendicular to this vertical. The volume element dτ has a mass ρdτ so the viscous force per unit of mass in the x– direction becomes ∂ 2u x μ ∂ 2u x Fx = v =+ ρdτ ρ ∂z 2 ∂z 2

(3.42)

which defines the dynamic viscosity v. Note that for a complete treatment the friction along the surface should be added to Eq. (3.35). 3.3.1.3

Coriolis Force

The Coriolis force, the third contribution in Eq. (3.35) is a fictitious force, to be added to the physical (natural) forces in order to correct for the acceleration of the coordinate system. In the present case, that accelerated system is fixed at a location on the surface of the earth. When in an inertial coordinate system a particle moves in a straight line, that line will seem curved, when measured from a rotating frame. The fictitious extra force should account for the resulting trajectory of the particle. Of these forces only the Coriolis force is relevant for our discussions. The force is written as FCoriolis = −2 × uρdτ

(3.43)

The vector  represents the earth’s rotation, as shown on the left in Figure 3.16. As the air velocity u is essentially horizontal, only the vertical component of  (perpendicular to the local horizontal plane) will determine the acceleration of the air parcel. That vertical component depends on the latitude β; its magnitude is  sin β. Because of the factor of 2 in Eq. (3.43) one calls f = 2 sin β the Coriolis parameter.

(3.44)

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Environmental Physics Ω

Ω β

k β

uG

∇p

β

phigh plow

Figure 3.16 Coriolis force and geostrophic flow. On the left the vector representing the earth rotation  is projected on the local vertical k to find the horizontal component of the Coriolis force. On the right the direction uG of the geostrophic flow is derived for the Northern Hemisphere.

3.3.1.4

Gravity

The last term on the right-hand side of Eq. (3.35) is the gravity force. For a parcel of air with mass ρdτ it may be written as Fgravity = g ρdτ

(3.45)

where g is the acceleration due to gravity. It may be shown that in the absence of vertical velocity the hydrostatic equation (3.2) follows from (3.45) with (3.35) and (3.38) (Exercise 3.13). It may be remarked that vertical motion does occur, for example in rising hot air or in strong downpour from a cumulonimbus cloud. These must be taken into account for detailed weather calculations ([2], p. 210), but in most cases the vertical component of the air velocity is much smaller than the horizontal one and may be ignored.

Example 3.3 Geostrophic Flow Derive an expression for the velocity of air at high altitude (500 [m], say), assume a constant horizontal air velocity u, give arguments why certain terms in Eq. (3.35) may be ignored and check whether the air velocity indeed can be constant. Answer The assumption of constant air velocity implies that the left-hand side of Eq. (3.35) vanishes, as well as the viscous forces. The surface is so far away that friction need not be taken into account and gravity is compensated by the hydrostatic decrease of pressure upwards. This leaves us for the horizontal component of Eq. (3.35) with − ∇ pdτ − 2 × uρdτ = 0 (3.46) or −

1 ∇ p = 2 × u ρ

(3.47)

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It is seen from the left of Figure 3.16 or from Eq. (3.44) that at the equator the vector product 2 × u has vertical components only, as both  and u are in the horizontal plane. Therefore Eq. (3.47) only makes sense at middle latitudes. In that region the horizontal component of 2 × u which has magnitude f × u (see Eq. (3.44) should be equal and opposite to the horizontal component of ∇ p/ρ. The situation is sketched on the right in Figure 3.16 with two isobars plow and phigh . The vector ∇ p/ρ goes from plow to phigh as indicated. The vertical component of  points upwards from the paper. It follows that the velocity u must go to the left, as shown in the figure, for then the vector product 2 × u goes contrary to ∇ p/ρ. Note that in the southern hemisphere the vector  has the opposite direction, so in that case the direction of uG in Figure 3.16 would be reversed. The flow which follows from Eq. (3.47) apparently is directed parallel to the isobars and is called the geostrophic flow. It is constant in direction and magnitude and therefore obeys the assumptions made. Its magnitude is given by uG =

|∇ p| fρ

(3.48)

The measured pressure gradients give values within 10% of the measured wind velocities at high altitudes (Exercise 3.14). Velocities at an altitude of 1000 [m] may be of the order of 25 [m s−1 ] or higher. 3.3.1.5

↓

Coupling of Horizontal and Vertical Properties in the Atmosphere

In the following paragraphs it will be shown that horizontal and vertical changes in the atmosphere are coupled. In modelling, therefore, both horizontal and vertical dimensions have to be considered and one may not restrict oneself to one dimension. We consider the higher parts of the atmosphere where Eq. (3.47) for the geostrophic flow holds: −(∇ p)/ρ = 2 × u. With k the unit vector in the vertical direction this may be rewritten with Eq. (3.44) as k×u=−

1 ∇p ρf

(3.49)

The equation of state (3.6) reads p = ρ R T

(3.50)

and the hydrostatic equation (3.2) may be written as g=−

1 ∂p ρ ∂z

(3.51)

Substitution of p from (3.50) into (3.51) gives g=−

∂T 1 ∂ρ  R T − R ρ ∂z ∂z

Dividing both sides by T and using that R is independent of z gives  R  ∂ρ R ∂ T ∂ ∂ ∂ g   =− − = −R (ln ρ) + (ln T ) + (ln R ) T ρ ∂z T ∂z ∂z ∂z ∂z

(3.52)

(3.53)

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Environmental Physics

or ∂ g = −R  (ln p) (3.54) T ∂z Note that one obtains Eq. (3.49) from Eq. (3.51) if one substitutes k × u for g on the left and ∇ p/ f for ∂/∂z on the right. Therefore, we may deduce from Eq. (3.54) without further derivation that R k×u = − ∇(ln p) T f

(3.55)

This equation is differentiated with respect to z and with the help of Eq. (3.54) we find     ∂ k×u R ∂ g 1 g ∇T (3.56) =− ∇ ln p = ∇ = − ∂z T f ∂z f T f T2

↑

As k points in the vertical direction, the left-hand side of this equation has horizontal components only. Eq. (3.56) therefore expresses that the variation of the temperature T in the horizontal plane is through ∂/∂z coupled to the vertical variation of the velocity u. 3.3.2

Vertical Motion of Ocean Waters

The vertical cross section of the oceans is sketched in Figure 3.17 with the Pacific on the left and the Atlantic on the right. The land masses in the oceans are shaded grey. On top one will note the mixed layer that was introduced in Section 3.2.4. Below this top layer one distinguishes upper waters, deep waters and bottom waters. Note that the vertical scale has been changed at 1000 [m] depth. The vertical structures of the Atlantic and the Pacific appear to be quite different in the Northern, Arctic region. Notice that in the North Atlantic surface water goes down; this was already indicated by the circles with a cross in Figure 3.14. After going down, bottom water is formed, which will eventually go up again

Pacific ocean 0 100

Atlantic ocean 0 100

Mixed layer

Upper waters

Upper waters

1000

1000

Deep waters

Deep waters 4000

4000

Bottom waters

Bottom waters 5500 90°S

Mixed layer

Eq.

90°N

5500 90°S

Eq.

90°N

Figure 3.17 Layered structure of the oceans with the Pacific on the left and the Atlantic on the right. Note the change of scale at a depth of 1000 [m] and the different bottom structure for the Pacific and the Atlantic in the Arctic region. In the South bottom water is formed in both oceans, while in the North it is mainly formed in the Atlantic.

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in the so-called thermohaline circulation. In the South, bottom water is formed in both oceans. The word ‘thermohaline’ indicates the driving force behind the ocean circulation: heat and salinity. The Gulf Stream transports relatively warm water to the North, as discussed earlier. In the Arctic, water evaporates, which cools the remaining waters, and because the evaporated water vapour is pure, the salinity of the remaining water increases. Both effects increase the density to values larger than the water underneath, so the water sinks to the bottom of the ocean, as indicated. The amount of deep bottom water formed this way is about 1010 [kg s−1 ] in the Antarctic Ocean and double that in the North Atlantic ([10], 201). It must be remarked that the wind and the tides also influence the circulation. Therefore the modern term MOC, the Meridional Overturning Circulation, is coming into use. 3.3.3

Horizontal Motion of Ocean Waters

The horizontal circulation in the oceans has already been shown in Figure 3.14. The driving forces are the wind stress, pressure differences and the Coriolis force, while the presence of the land masses puts a boundary to the ocean currents. The wind stress along the surface is especially important where there are dominant winds constantly flowing in the same direction, such as the Easterly trade winds in the tropics. When wind stress is ignored, Eq. (3.35), which was derived for the motion of air in the atmosphere, can be applied to the oceans as well. Except at places of downwelling and upwelling, the approximation that the velocity u is horizontal can be made. A further, more daring, approximation is that viscous forces can be ignored. In that case the discussion on the geostrophic flow can be repeated, leading to Eq. (3.48) which reads uG =

|∇ p| fρ

(3.57)

The pressure variations in the atmosphere are reflected in the ocean surface, leading to a similar value of |∇ p| and the Coriolis parameter f will be the same. The density of sea water, however, is a factor of 1000 larger than that of air, leading to velocities a factor of 1000 smaller. Instead of 30 to 50 [m s−1 ] one would expect 3 to 5 [cm s−1 ]. In practice, most observed surface speeds are between 5 and 50 [cm s−1 ].

3.4

Natural Climate Variability

Climate variations have always happened. That may be illustrated by reconstructions of the evolution of sea surface temperatures during the last 400 000 years, shown in Figure 3.18. The top graph refers to the North Atlantic, the bottom one to the South Atlantic and the middle one to the Equatorial Indian Ocean. Both the North and South exhibit a similar alternation of warmer and colder periods and some fine structure. With the exception of the last few hundred years the variations must have been due to internal variations in the climate system and external drivers of change. The cold periods are called a glacial period or ice age, where large parts of the earth’s surface were covered with snow and ice. Except at the equator there is a succession of these

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Figure 3.18 Trends in sea surface temperatures during the last 400 000 years in the North Atlantic (top), the South Atlantic (bottom) and the Equatorial Indian Ocean (middle). (Reproduced with permission of Cambridge University Press, copyright 2007, from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the IPCC (Chapter 3, Ref. [1], FAQ 1.2, Fig. 1).)

glacial and interglacial periods, which are most clearly to be seen in the South Atlantic curve of Figure 3.18. The Serbian scientist Milankovitch, who worked in the 1920s, attributed this phenomenon to changes in the orientation of the earth’s axis with respect to its orbit around the sun. The situation is sketched in Figure 3.19. The eccentricity of the orbit varies between 0.002 and 0.050 (present value 0.0167), the tilt (the angle between the earth’s axis and the normal on the orbital plane) varies between 22.05◦ and 24.50◦ (present value 23.45◦ ) and the axis itself performs a precession around that normal ([1], 444 - 446). The first two effects are due to gravitational interactions of the earth with the rest of the solar system, while the precession is due to the deviation of the earth from a perfect sphere: it has a ‘belt’ around the equator which experiences a gravitational torque from the sun. The variations

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Tilt Precession

earth axis Sun

Eccentricity

Figure 3.19 Three parameters determine the insolation of the earth: the eccentricity of the orbit (indicated by two extreme positions), the tilt of the axis (changing between 22◦ and 24.5◦ ) and the precession of the axis around the normal to its orbital plane.

are quasi-periodic; the eccentricity with periods of 100 000 years, the tilt with 41 000 years and the precession with 19 000 and 23 000 years. It must be noticed that the precession and the change of tilt do not change the average insolation of the earth. They do lead to a difference in insolation between the hemispheres. This is quickly understood by taking an extreme example, a tilt of almost 90◦ . At the right in the diagram the Southern Hemisphere has full sunlight day and night (a terrible summer) and the Northern Hemisphere has day and night in complete darkness. After half a year the situation is reversed. For a moderate tilt, as indicated in Figure 3.19, consider the equatorial plane of the earth, which passes through the centre of the earth. The orbital plane also contains that centre. The two planes intersect in a straight line through the earth’s centre. In Figure 3.20 the situation Spring equinox (March 20) Summer solstice (June 21)

Perihelion (January 4)

Winter solstice (December 21) Autumnal equinox (September 22)

Figure 3.20 The earth in its elliptical orbit with the sun as one of its focal points. The major axis of the ellipse is indicated as is the earth equatorial plane. Twice a year the sun will be on the intersection between orbital plane and equatorial plane: the vernal equinox and the autumnal equinox. This intersection moves around with the precession. The season dates are for the UK in the year 2013, the perihelion for 2014. Note that the minor axis of the ellipse (not drawn) will be a little to the left of the drawn intersection.

62

Environmental Physics

is sketched. In the summer of the Northern Hemisphere the sun will be above the equatorial plane. In the winter it will be below. Twice in between the sun will pass the equatorial plane of the earth, in which case the intersection of equatorial plane and orbital plane will contain both the earth and the sun. In these two cases day and night will be equally long: the beginning of autumn or spring. Because of the precession the equatorial plane moves around and consequently also the intersect between both planes. So the positions of spring and autumn move along the orbit. If for a highly eccentric orbit midsummer in the Northern Hemisphere occurs at a position close to the sun, the summer will be relatively hot and the winter (with a position far from the sun) relatively cold. For the Southern Hemisphere, the winter will be mild (the planet is close to the sun) and the summer will be mild as well. These examples show that the accumulation of the three effects may affect the hemispheres in different ways. The insolation averaged over the year and the total globe will only vary a little, as a consequence of a change in eccentricity, but this minor effect may be ignored. The onset of ice ages therefore should be attributed to a local decrease of insolation. Furthermore, the coupling between both hemispheres by means of the ocean currents (Figure 3.14) and winds will result in the complex climate variation which is observed. For the onset of the last ice age, 110 000 years ago in the Northern Hemisphere, the decrease in insolation has been estimated as 40 [Wm−2 ] for mid June at 65◦ North with respect to the value of 380 [Wm−2 ] at present ([1], 445). The next ice age is expected 30 000 years from now ([1], 449), most likely too late to compensate for the present temperature rise from the greenhouse effect. The Milankovich variations do not explain climate variations on a shorter time scale. There are at least two other effects. The first is the variation of the solar emission. The sunspot cycle of 11 years has measurable effects on the insolation. Reconstructions from 1700 to 2000 suggest fluctuations of the ‘solar constant’ S between 1365 and 1366.5 [Wm−2 ] ([1], p. 190). The second effect is the eruption of volcanoes. Large eruptions bring huge amounts of dust and sulfate aerosols into the air. This results in an increase of the albedo a of the earth because they prevent part of the light from penetrating the atmosphere. Consequently the temperature will drop. The Pinatubo eruption in 1991 indeed caused a drop of temperature of about 0.3 [◦ C] for a few years ([1], 62, 684). Dust and aerosols slowly precipitated by rain or gravity. Finally, it must be mentioned that the physics of the radiation balance is not sufficient to explain or predict climate change satisfactorily. Figure 3.21 shows the concentration of the greenhouse gases CO2 and CH4 over the last 20 000 years, reconstructed from ice and fern data ([1], 25 448). Their variations before 1750 must be due to biochemical processes, induced by changes in the radiation balance. The strong increase in recent times is most apparent in the figure. Interesting are the vertical bars, which indicate the variability for the past 650 000 years.

3.5

Modelling Human-Induced Climate Change

A perfect model should be able to reproduce the past climate, comprising many more than the graphs shown here, for all possible locations, all possible times and all possible

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Figure 3.21 Concentrations of greenhouse gases CO2 and CH4 during the last 20 000 years. The variations must be due to coupling of the radiation balance with biochemical processes. The increase in recent times is attributed to human-induced emissions. The grey width indicates the range of natural variability during the past 650 000 years. (Reproduced with permission of Cambridge University Press, copyright 2007, from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the IPCC (Chapter 3, Ref. [1], FAQ 1.2, Fig. 1).)

variables. The model may not restrict itself to physics but also calculate the biochemical feedbacks, some of which appeared in Figure 3.21. If the past is reproduced completely, one will have faith in the forecasts as well. Obviously, the scientific community has not reached this phase yet, and it probably never will. Therefore, the best one can do is to give the best possible estimates and point out the uncertainties in the predictions. Policy-makers and society-at-large then may choose a path with the smallest risks and the largest potential benefits. In this section we will give an indication of the elements climate modelling will comprise. First we discuss the carbon cycle as one of the biochemical cycles to take into account. Next, we give an overview of the physical equations and finally we discuss what kind of models are used in practice. 3.5.1

The Carbon Cycle

Carbon in the form of CO2 is one of the most important greenhouse gases. It is therefore important to know how carbon circulates in the ecosystem of the earth. In Figure 3.22 a simplified overview of the global carbon cycle is given, where the numbers in boxes are estimated for the 1990s. They refer to the carbon content [1012 kg] of parts of the ecosystem and the numbers near the arrows refer to the annual fluxes [1012 kg C/yr]. Drawn arrows represent relatively rapid changes and dashed ones are relatively slow. The top of the diagram shows the atmosphere where the carbon content directly influences the greenhouse effect. Far left, the input from the burning of fossil fuels is shown. This number is easiest to estimate as the use of fossil fuels is well-documented in UN statistics.

64

Environmental Physics Fossil CO2 Volcanic outgassing

Atmosphere

6.4

762

CO2 from

changes in landuse

1.6 Land sink Photosynthesis 2.6 120

Weathering Respiration 0.2 119.6

Vegetation + Organic soil carbon 2260

0.8 Rivers

90.6

Gasexchange 92.2

Marine biosphere 3

39 50

918 Surface ocean

11 Dissolved and particulate organic carbon

91.8 101 Deep sea 37200

Sedimentation 0.2

Remineralization

Fossil fuel resources 3450 Sediments, carbonaceous rocks

Figure 3.22 Global Carbon Cycle with data for the 1990s. The numbers in boxes refer to the C content [1012 kg] and the numbers near the arrows to C fluxes [1012 kg/yr]. Drawn arrows indicate fast fluxes and dashed ones slow exchanges. More details are displayed in Plate 2, which is Reproduced with permission of Cambridge University Press, copyright 2007, from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the IPCC (Chapter 3, Ref. [1], FAQ 1.2, Fig. 1).

3.5.1.1

Land-Use Change

The input from changes in land use, also on the left in Figure 3.22, is a little more complicated. Conversion of forests to agriculture releases the biomass present in the trees. Also the carbon and nitrogen contents of the soil decrease. The crops that replace the trees have there own carbon content, but the amount of carbon is less as the volume of crops is smaller than that of trees.

3.5.1.2

Ocean-Atmosphere Interaction

Somewhat to the right of the centre of Figure 3.22 the oceans are shown. From measurements of pressures in the top of the ocean and from studies with radioactive tracers one estimates the exchange between atmosphere and ocean: an absorption of 92.2 × [1012 kg/yr] and an emission of 90.6 × [1012 kg/yr] The difference between these big numbers and the inflow from the rivers 0.8 × [1012 kg/yr] determines the net uptake of carbon by the oceans. This difference, which is essential for estimating the reaction of the oceans to the appearance of additional CO2 in the atmosphere, will have a relatively big uncertainty.

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The extra CO2 gas in the ocean top layers will react with water according to the equilibrium reaction CO2 (gas) + H2 O → H2 CO3 (aq)

(3.58)

Here the addition of (aq) means that the carbonic acid H2 CO3 is dissolved in water. Next the acid dissociates into bicarbonate and carbonate according to the two reactions H2 CO3 (aq) + H2 O → H3 O+ (aq) + HCO− 3 (aq)

(3.59)

2− + HCO− 3 (aq) + H2 O → H3 O (aq) + CO3 (aq)

(3.60)

Both equations are governed by an equilibrium constant. The relative magnitude of the HCO3 − and CO3 2− concentrations strongly depend on the pH of the water. For sea water with pH ≈ 8 one finds that HCO3 − dominates with a relative concentration of 97% (Exercise 3.15). As a consequence the carbon will react with calcium in marine organisms according to the reaction Ca2+ +2HCO− 3 (aq) → CaCO3 +CO2 +H2 O

(3.61)

In this way the carbon is stored in calcium carbonate (or calcite), which is used for the growth of shells and tissues. Part of it slowly sinks down to the bottom of the oceans. In a steady state this downward flux is compensated by an upward flux of carbonrich waters near the equator, along the continental margins and in high latitude waters. The process of transport down and up again is called the ‘marine biological pump’. Its magnitude does not only depend on the carbon content of the water, but for a very large part on the availability of nutrients to sustain marine life such as N, P and micro-nutrients like Fe and Mn. With a doubled CO2 concentration in the atmosphere it is difficult to predict how the availability of nutrients will respond to the resulting change of climate. It may go both ways. If we are lucky more nutrients will result in more photosynthetic micro-organisms in the oceans. They would fix part of the excess of CO2 , thereby mitigating the climate change. If we have bad luck and the nutrients are lacking, the photosynthetic life in the oceans would be strongly reduced. In that case the ‘pump’ would stop functioning and the atmospheric CO2 concentration might increase by another 150 [ppmv]. 3.5.1.3

Atmosphere–Land Interaction

Somewhat to the left of the centre in Figure 3.22 the exchange between atmosphere and vegetation is shown. Again two big numbers determine the exchange: photosynthesis is estimated as 120 × [1012 kg/yr] and respiration at night which includes emission of CH4 due to rotting amounts to 119.6 × [1012 kg/yr]). Besides these two effects there is a ‘land sink’ due to increase of vegetation, which amounts to 2.6 × [1012 kg/yr]. The land uptake is calculated from the carbon budget. This is shown in Table 3.4 for the 1990s and 2000–2005. The input into the atmosphere from fossil fuel emissions can be estimated rather accurately. This also holds for the accumulation in the atmosphere. The uptake into the ocean is modelled from which the net land uptake is inferred. This net uptake is the effect of land use change, which binds less CO2 (which can be estimated) and a sink due to increasing growth of vegetation caused by the higher CO2 concentration in

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Table 3.4 Global mean CO2 budget/[1012 kg C yr−1 ].

Estimated fossil fuel + industrial emissions Measured accumulation in the atmosphere Modelled ocean uptake Inferred net land uptake Partitioning of land uptake Land use change, CO2 increase Residual land sink, CO2 decrease

1990s

2000–2005

6.4 ± 0.4 3.2 ± 0.1 2.2 ± 0.4 1.0 ± 0.6

7.2 ± 0.3 4.1 ± 0.1 2.2 ± 0.5 0.9 ± 0.6

1.6 (0.5–2.7) 2.6 (0.9–4.3)

Not available Not available

With permission of Cambridge University Press, copyright 2007, reproduced from [1], p. 26).

the atmosphere. For the 1990s the residual sink is estimated as around 2.6 × [1012 kg/yr], which explains the number in Figure 3.22; for the 2000s the data were not available. From Table 3.4 one may conclude that half of the extra CO2 emissions are stored in the oceans or by land. It is not certain that this will remain so. As to the oceans, the coin may flip to both sides; as to the land, some models suggest that in the future increased respiration will produce a land source instead of a sink ([1], p. 777). That would strongly reinforce the greenhouse warming and climate change. 3.5.1.4

The Final Steady State

The numbers in Figure 3.22 refer to the present situation. In the future most recoverable fossil fuels will have been mined and combusted, say a quarter, which amounts to 1500 [1012 kg]. Model calculations [13] have compared two time scales, one in which the resources are exhausted in 200 years and another one, where that happens in 400 years. The result of such a calculation is shown in Figure 3.23. In the lower graph two types of emissions are shown, both starting in the year 1900. The drawn curve corresponds to a quick release of carbon, within 200 years, whereas the dashed one corresponds to a slower exhaustion of resources, in about 400 years. The upper graph shows the resulting atmospheric CO2 concentrations as a function of time. In both cases a final equilibrium would be reached some 1500 years after 1900 with a concentration of about 400 [ppmv]. When the pulse is distributed more evenly in time (the dashed line) the peak height is lower. In both cases the climatic consequences will be big. But it stands to reason that in the second case they might be smaller, as the ecological system has more time to adapt. It must be noted, of course, that other greenhouse gases may take over the role of CO2 in the future and consequently the equivalent CO2 concentration would continue to rise. The description of the carbon cycle given here illustrates that climate modelling not only requires a good understanding of physics and computer science, but also an in-depth knowledge of chemistry and biology. 3.5.2

Structure of Climate Modelling

The most advanced climate models are the so-called GCMs, general circulation models or general climate models. They explicitly consider all actors in the climate system, depicted in

600

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Atmospheric CO2 concentration

500 400

Final equilibrium 300

Emissions /[10 12 kg C yr -1]

Concentration /[ppm]

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1900

2300

2700

3100

20 15

CO2 emissions

10 5 0 1900

2300

2700

3100

Year Figure 3.23 Release of all recoverable fossil fuel resources. The lower graph shows a quick release (drawn) and a slower release (dashed). The upper graph indicates the resulting CO2 concentration in the air. Adapted by permission of the European Commission, copyright 1999, from [13], p. 4.

Figure 3.1 and are often called AOGCMs, Atmospheric-Ocean Coupled General Circulation Models, to indicate that the coupling between atmosphere and ocean is taken into account. Their structure is shown in Figure 3.24. They are usually built up from modules, describing separately ocean, land, sea ice and atmosphere. The modules produce fluxes of energy (sensible and latent heat), momentum (winds, sea currents, rising air) and mass (water, carbon). These fluxes are exchanged and taken into account, indicated by the central box. The need for coupling all modules will be apparent from the discussion so far. A higher temperature at high latitudes will cause expansion of the sea water, decreasing its density. If the effect is big enough, the surface waters in the North Atlantic no longer will go down, stopping the thermohaline circulation (MOC). That in its turn would influence the transportation of energy and the carbon cycle and have its impact on the land vegetation. All these effects affect the atmosphere, the box on top in Figure 3.24. 3.5.3

Modelling the Atmosphere

The fields of science that are most important to model the behaviour of the atmosphere are given at the top right in Figure 3.24. Thermodynamics describes air as an ideal gas (3.15) and applies conservation of energy (3.21) in cloud formation and precipitation. Clouds also are responsible for 2/3 of the earth’s albedo ([1], 114) and thereby largely determine the

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Atmosphere Thermodynamics Dynamics Water vapour/clouds Chemistry Radiation

Land Vegetation Soil, lakes, glaciers Drainage

Interconnection of energy, momentum and mass (water, carbon etc.) fluxes

Temperature Thermodynamics

Sea ice

Temperature Salinity Circulation

Oceans Figure 3.24 Structure of general climate models (GCMs). The four players, land, ocean, sea ice and atmosphere are modelled and their interactions taken into account by the central box. (Reproduced from A Climate Modelling Primer, McGuffie, pg 166, fig 5.1, Copyright 2005, John Wiley & Sons Ltd.)

local and global radiation balance. Chemistry is needed to analyse the composition of the atmosphere, the photo reactions that may break down or form greenhouse gases and to model the CO2 uptake of the oceans (see our discussion of the carbon cycle). Dynamics starts from Newton’s equations of motion (3.35) and describes horizontal motion of air while the hydrostatic equation (3.2) describes the vertical motion. In order to appreciate the computational effort of modelling we look at the equations of motion (3.35) in more detail: du (3.62) ρdτ = Fpress + Fviscous + FCoriolis + Fgravity dt The forces on the right-hand side have been discussed earlier (Eqs. (3.36)–(3.45)) and will be a function of longitude, latitude and altitude. The left of Eq. (3.62) describes a parcel of air with mass ρdτ and velocity u. The velocity u is a function of time and position u(x, y, z, t). We write out the x-component of the time derivative du/dt on the left of Eq. (3.62). This leads to ∂u x dx ∂u x dy ∂u x dz ∂u x du x (x, y, z, t) = + + + dt ∂ y dt ∂z dt ∂t ∂ x dt ∂ ∂ ∂ ∂u x (3.63) + uy + uz ux + = ux ∂x ∂y ∂z ∂t ∂u x = (u.∇)u x + ∂t

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The first equality expresses that the parcel of air moves in time. Its position (x, y, z) therefore is also a function of time and has to be differentiated. The second equality is just a way of rewriting dx/dt = ux , dy/dy = uy and dz/dt = uz . The third part of the equality then may be read as a scalar product (see Eq. (B4) in Appendix B) of the real vector u(x, y, z, t) and the symbolic vector ∇(∂/∂ x, ∂/∂ y, ∂/∂z) as is expressed concisely in the last part of Eq. (3.63). The right-hand side of Eq. (3.63) shows clearly that the left-hand side is nonlinear; if, for example, the velocity u doubled, the time derivative would become four times as big. The consequence is that solving Eqs. (3.62) with two slightly different initial conditions (pressure, density, velocity) at time t = 0 may lead to quite different solutions some time later. This effect is studied in chaos theory. In climate modelling one does not need to know the temperature and precipitation at a certain date and location, as in weather forecasting, but rather the change in average temperature and precipitation in a certain month and a certain region as a consequence of human-induced effects like radiative forcing. It is believed that the effects due to nonlinearity of the equations then average out ([1], pp. 105, 117). Equations (3.62) in general cannot be solved analytically. One therefore puts a twodimensional grid over the globe and uses a number of vertical layers. In each little block of the resulting three-dimensional grid all properties are assumed to be the same. The 23 GCMs discussed in [1] typically have 30 or more vertical layers in both atmosphere and ocean and a horizontal resolution of 1◦ by 1◦ ([1], p. 597–599). In this way the models are able to forecast climate as a function of region and time of the year. In fact, global maps are published with forecasts as to temperature, precipitation, ice cover for the different seasons. It must be mentioned that not all details can be found from the models. For processes involving clouds for example, several sets of parameters are in use, which is one of the aspects in which models diverge ([1], p. 602). To be realistic the properties of the surface have to be taken into account. They are summarized in Table 3.5 and need to be known as a function of latitude and longitude. The first two, relief and roughness describe the geography of the surface, the rest its properties.

Table 3.5 Boundary conditions of GCMs. Earth surface properties

Reason properties required for modelling

Relief Roughness Albedo Emissivity Heat capacity Heat conduction Soil humidity Ice and snow cover Salinity

Obstacles for air flow Friction Radiative fluxes Infrared flux Heat currents into and out of the soil Melting and freezing Evaporation and rain absorption Run-off, influence on albedo Mixing run-off water in sea

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A Hierarchy of Models

The point of climate modelling is forecasting the consequences of any mix of greenhouse gas emissions, land-use change and other effects of human interference. That should be done with a variety of realistic parameter sets in order to find the range of properties for the climate to be expected. However, each run of a complete GCM, comprising all aspects of Figure 3.24, will cost a lot of computer time; therefore less complex models have been designed, where aspects of a GCM are summarized in a few effective parameters, which are estimated by comparison with a more complete GCM ([1], 643–645).

3.6

Analyses of IPCC, the Intergovernmental Panel on Climate Change

The science working group of IPCC evaluates the available literature on climate and climate change. As an example Figure 3.25 shows the radiative forcing of climate between 1750 and 2005, due to several origins, which are briefly discussed below ([1], 199–202). At the top the contribution of CO2 amounts to 1.66 [Wm−2 ]. This number is found by averaging the published values, which are based on the well-known past and present concentrations of this greenhouse gas and its well-known absorption of long-range radiation (Section 3.2). In the CO2 case the group judges that there is both sufficient evidence from physics and sufficient consensus amongst the publications to assign a small error bar. The precise value of the error bar follows from expert assessment of the published values and their ranges. This procedure is followed everywhere in the graph. The second row of Figure 3.25 summarizes the effects of three other important greenhouse gases CH4 , N2 O and the group of halocarbons. The physics is understood, but there are some differences between different data sets and various radiative transfer models. For ozone, the third row, the error bar is much higher. For stratospheric ozone the changes prior to 1970 are uncertain, for tropospheric ozone the changes in lightning are uncertain and for both the trends near the tropopause. Stratospheric water vapour from oxidation of CH4 is found in the next row. The error bar arises from the models of oxidation and radiative transfer. The surface albedo, the next row, has a negative and a positive component. Incomplete combustion of fossil fuel will produce finely divided carbon particles, which precipitate on snow and decrease its albedo: a positive forcing effect. Land-use change such as deforestation and desertification increase the albedo: a negative forcing. Aerosols scatter solar radiation back into space, (a higher overall albedo aa ) and produce a negative forcing, either directly or by affecting the cloud albedo. The latter effect follows from GCMs, but there is no direct observational evidence, hence the large error bar. Linear contrails produced by aircraft only have a small effect; changes in solar irradiance are somewhat higher. The resulting net effect of all human activities is estimated as 1.6 [Wm−2 ], the value that was used in Section 3.2.4.

3.7

Forecasts of Climate Change

An increasing concentration of greenhouse gases will result in higher temperatures and changes of climate. Suppose that at some future date the equivalent CO2 concentration,

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Radiative forcing terms

{

CO2

Long-lived greenhouse gasses

N2O CH4

Human activities

Halocarbons Ozone

Stratospheric

Tropospheric

(-0.05)

Stratospheric water vapour Surface albedo

Black carbon on snow

Land use

Direct effect Total aerosol Cloud albedo effect

{

Natural porcesses

Linear contrails

(0.01)

Solar irradiance Total net human activities -2

-1

0

1

2

Radiative forcing /[w m-2] Figure 3.25 Radiative forcing of climate between 1750 and 2005. On the left the contributions to radiative forcing are indicated and on the right they are shown with error bars based on expert assessment. (Reproduced with permission of Cambridge University Press, copyright 2007, from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the IPCC (Chapter 3, Ref. [1], FAQ 1.2, Fig. 1).)

or more precisely the net human-induced radiative forcing, is kept constant at some value Cconstant . The resulting equilibrium temperature rise can be calculated from Eq. (3.9) as Ts = G f I

(3.64)

This equilibrium will be reached after some time because of the ocean delay (3.32). As mentioned earlier the parameter Gf is estimated as about 0.8 [K/(Wm−2 )]. The forcing I

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has a simple relationship with the equivalent CO2 concentration as   [CO2 ] 3.7 × log I = log 2 280

(3.65)

If CO2 has the pre-industrial concentration of 280 [ppmv], the human-induced forcing is defined as I = 0 and when this concentration is doubled it follows from Eq. (3.65) that I = 3.7 [Wm−2 ]. This value is generally accepted and the logarithmic dependence on the concentration is seen as a good approximation. For the 2005 value [CO2 ] = 379 [ppmv], for example, one finds with Eq. (3.65) I = 1.62 [Wm−2 ], in good agreement with Figure 3.25. From Eqs. (3.64) and (3.65) the equilibrium temperature rise can be calculated as   3.7 Cconstant × log . (3.66) Ts = G f × log 2 280 In Figure 3.26 it is assumed that the equivalent CO2 concentration, including all elements of Figure 3.25, will stabilize at one of four values 450 [ppmv], 550 [ppmv], 750 [ppmv] of 1000 [ppmv]. In the lower figure the corresponding emissions are calculated with the so-called simple Hadley model. It is clear that in all cases the emissions must go down as the greenhouse gases have a rather long lifetime in the atmosphere (Table 3.2). The decrease of emissions must be strongest with the lowest equilibrium concentration of 450 [ppmv]. From (3.60) it appears that this concentration would lead to an equilibrium temperature rise of 2 [◦ C].

Figure 3.26 Stabilization of CO2 concentrations at four values (top) and the corresponding emissions in a simple model, including the impact of the carbon cycle (bottom). (Reproduced with permission of Cambridge University Press, copyright 2007, from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the IPCC (Chapter 3, Ref. [1], FAQ 1.2, Fig. 1).)

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The lower graph includes the effects of the carbon cycle on the CO2 concentration. It appears that higher temperatures will reduce the land carbon uptake. The photosynthetic production goes down and the respiration of the vegetation goes up. The difference between the two large numbers of Figure 3.22 then decreases. Figure 3.26 starts from a stabilizing greenhouse gas concentration and calculates the corresponding emissions. One may also work the other way round, start from a realistic economic scenario, then calculate the emissions of all greenhouse gases and all forcings separately, add them up, compute the resulting total forcing and finally find the resulting climate as a function of time and location from a GCM. This in fact has been done, and we will describe one of the scenarios, called A1B and discuss its climatic consequences as computed by the available GCMs. The A1B scenario [15] describes a world of very rapid economic growth, a population peak in the midst of the twenty-first century and a rapid introduction of new and efficient technologies. There is a balance between fossil-fuel intensive and nonfossil energy sources and it is assumed that there is no implementation of internationally agreed emission targets. The model is run from 2000 to 2100 and in some cases even to the year 2300. The emissions of greenhouse gases follow from the model ([1], 803). For the concentrations there is already a range of possible values depending on for example, the precise operation of the carbon cycle. Consequently, there is a range in the radiative forcing, which increases in time. For A1B in 2100 the range is estimated to be between 5.8 and 6.4 [Wm−2 ], see Figure 3.27. The resulting temperature rise is calculated for 19 GCMs, which produce an extra spread, illustrated in the right-hand graph of Figure 3.27. The global temperature rise from 1990 to the year 2100 will be between 1.8 and 4.2 [◦ C] in this scenario, with 3.4 [◦ C] as the most probable value ([1], 763). The sea-level rise in 2100 will be mainly due to thermal expansion (0.13 to 0.32 [m]), while all other effects like melting of glaciers, ice caps, and Greenland and Antarctic ice sheets contribute another 0.08 to 0.16 [m] ([1], 820).

Figure 3.27 Radiative forcing (left) and temperature rise (right) calculated for the A1B scenario for the next three centuries. (Reproduced with permission of Cambridge University Press, copyright 2007, from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the IPCC (Chapter 3, Ref. [1], FAQ 1.2, Fig. 1).)

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A warmer climate allows more water vapour in the air (Exercise 3.18), which will lead to more precipitation. In 2100 the global precipitation is expected to be 4% higher than in the year 2000 ([1], 763). This extra will be distributed unevenly over the globe and will be concentrated in the tropics (monsoons) and at high latitudes. More rain near Greenland will reduce the salinity of the sea water, which in its turn would slow down the thermohaline circulation or MOC, but in this century no complete collapse is predicted. The amount of warm waters that are transported northward from the equator will diminish, however, resulting in a somewhat less warm climate in mid-latitudes in Europe ([1], 775,776). This only changes the distribution of heat over the globe, so elsewhere it will become even warmer. Interesting is the response of the Arctic sea ice to global warming. Melting of more ice during summer will allow open water to receive more solar energy, because of the lower albedo of water as compared to ice. This will lead to open waters deep into September, or even in Autumn ([1], 776).

Exercises 3.1 For a mixture of gases like air: (a) Derive R = xi Ri , where xi = ρ i /ρ is the relative mass of component i, ρ = ρ i and Ri = 103 R /Mi for Mi the molecular weight of component i. (b) Find R for air and compare with the value given in Appendix A. 3.2 From Eqs. (3.11) and (3.3) derive the Poisson equation (3.12) for adiabatic change. 3.3 Consider 1 [m3 ] of dry air, which is heated from 10 [◦ C] to 20 [◦ C]. It picks up evaporated water vapour from the surroundings, by which the humidity becomes 80%. The air stays at ground level. Calculate the increase in sensible heat and the increase (from zero) of sensible heat. Note: for this exercise you need to consult tables like in Chapter 1, refs [2] or [3]. 3.4 According to Figure 2.1 the tails of the solar spectrum and the earth IR spectrum will cross at a certain wavelength. Calculate the wavelength at which the intensity of the solar (black-body) radiation received on earth [Wm−2 ] equals the emitted IR [Wm−2 ]. Use for the radius of the sun Rsun = 6.96 × 108 [m] and the distance from the centre of the sun to the centre of the earth Rse = 1.49 × 1011 [m]. Check with Figure 2.1 that both spectra are well separated. 3.5 Check: (a) that the terms in Eq. (3.18) numerically correspond with Figure 3.5; calculate T s – Ta ; (b) perform the calculations for a = 0.50; a = 0.35; S1 = 0.85 × S; S2 = 0.72 × S. 3.6 Distinguish between the window region 9–12.5 [μm] and the rest of the spectrum. Use the Planck function (2.4) and calculate the contribution of both parts of the spectrum to the black-body emission from the earth surface. For a temperature of 288 [K] find for the window region Bsw = 110.45 [Wm−2 ] and outside the window Bsout = 280 [Wm−2 ]. 3.7 Use the data of Example 3.1 and plot the upward atmospheric flux as a function of z both inside and outside of the window. For the flux outside the window also plot the first 100 [m] of height and interpret the results.

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3.8 Take the model of Example 3.1 and plot as a function of kw and koutw the upward atmospheric flux separately inside the window and outside the window. Take values starting from those in Example 3.1 and go somewhat higher. Add the surface contribution (3.19) inside the window and comment. 3.9 Convert the volume fraction of water vapour in Table 3.2 into a mass fraction of 3 × 103 [ppm] (parts per million). Estimate the height of the water column when, hypothetically, all water vapour rains out and find it to be 3 [cm]. Take the average annual rainfall over earth as about 100 [cm]. What is your conclusion on the turnover rate of water in the atmosphere and on the sensitivity of models to a correct description of the water cycle? 3.10 Estimate τ , of Eq. (3.30) starting with calculating cs for a water column of h = 120 [m]. 3.11 Take the radiative forcing in 1750 as I = 0 and in 2005 as I = 1.6 [W m−2 ]. (a) Derive a smooth exponential curve for I(t) between 1750 and 2005. Assume this to have been the ‘real forcing’. (b) The forcing can be viewed as a big number of small sudden forcings d(I(t)). Write down an integral for T s (t = 2005) using Eq. (3.32). (c) Calculate the result, for example by using a simple code. (d) Assume the forcing does not increase after 2005 and plot the temperature between 2005 and 2050. (e) Repeat the calculation for the year 1850 and find the temperature increase to be 0.87 [◦ C]. (f) Check whether more recent IPCC reports give later data for the forcing or the heat capacity of the oceans and if so, recalculate the temperature increase. Hint: Rewrite the forcing as i(t) and the temperature increase as T(t). 3.12 The temperature difference between the Gulf Stream and the returning colder water is about 12 [◦ C]. Calculate the transported heat energy which is transported, using the data in the text. Assume for the Kuroshio a similar result and compare with Example 3.2. 3.13 Derive the hydrostatic equation (3.4) from Eqs. (3.35), (3.38) and (3.45) assuming that all vertical motion vanishes. 3.14 Find on internet a mapping of the isobars in your neighbourhood. Make a rough estimate of |∇ p|. Also calculate your value of the Coriolis parameter f . Find the geostrophic velocity and compare with the value given on your weather map. 3.15 For sea water one may take pH = 8. Use the equilibrium constants K 1 = [H3 O+ ][HCO3 − ]/[H2 CO3 ] = 4.3 × 10−7 and K 2 = [H3 O+ ][CO3 − − ]/[HCO3 − ] = 4.8 × 10−11 to calculate the fractions in sea water of [CO3 2 − ], [HCO3 − ] and [H2 CO3 ]. Students proficient with computers may plot the three fractions as a function of pH. 3.16 Scenario A1B forecasts a thermal expansion of sea water in the year 2100 of 0.13 [m] for a temperature rise of 1.8 [◦ C]. (Actually this is one of the low GCM forecasts). (a) Calculate the height of the oceans that is supposedly expanding, assuming that the height h expands and the rest does not. Use the cubic expansion coefficient for water α = 21 × 10–5 [K]. (b) How does your answer compare with Exercise 3.10. Explain. 3.17 The West Antarctic ice sheets (together some 0.83 million [km2 ] and only 6% of the total Antarctic ice surface) might lose their grip on the rock by warming and collapse into the sea. Estimate the resulting sea level rise. 3.18 Show (from tables) that air of 21 [◦ C] can contain about 6% more water vapour than air of 20 [◦ C].

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References [1] Solomon, S., Qin, D., Manning, et al. (eds) (2007) for the Intergovernmental Panel of Climate Change, The Physical Science Basis, IPCC: Cambridge University Press, Cambridge, UK. The IPCC papers contain a host of articles, with many authors. Readers should consult www.ipcc.ch to find the authors of the material we are referring to. [2] McIlveen, R. (2010) Fundamentals of Weather and Climate, 2nd edn, Oxford University Press. This is a very readable book with extensive meteorological discussions and mathematical derivations on a rather elementary level. [3] Hutter, K., Blatter, H. and Ohmura, A. (1990) Climate Change, Ice Sheet Dynamics and Sea Level Variations, Dept of Geography, ETH-Zurich. [4] Houghton, J.T., Jenkins, G.J. and Ephraims, J.J. (eds) (1990) Climate Change, The IPCC Scientific Assessment, Cambridge University Press, Cambridge, UK. [5] Campbell, I.M. (1977) Energy and the Atmosphere, A Physical-Chemical Approach, John Wiley, London. This is an interesting and easy to read introduction. [6] van Dorland, R. (1999) Radiation and Climate, From Transfer Modelling to Global Temperature Response, Thesis, Utrecht University, Utrecht, Netherlands. [7] http://savi.weber.edu/hi_plot/. The spectra to be plotted are based on the so-called Hitran data base, which regularly is updated. [8] Schlesinger, M. (1988) Quantitative analysis of feedbacks in climate model situations of CO2 induced warming, in: Physically-Based Modelling and Simulation of Climate and Climate Change, Part 2 (ed. M.E. Schlesinger), Kluwer, Dordrecht. [9] Sch¨onwiese, C.-D. and Diekmann, B. (1987) Der Treibhauseffekt, Deutsche Verlag Anstalt, Stuttgart. The warming effect as given in Table 3.2 originates from K.Y. Kondratiev and N.I. Moskalenko (1984). [10] Hartmann, D.L. (1994) Global Physical Climatology, Academic Press, San Diego. [11] Dijkstra, H.A. (1997) Oceaancirculatie en klimaatverandering. Nederlands Tijdschrift voor Natuurkunde 63, 79. [12] Houghton, J.T., Ding, Y., Griggs, D.J. et al. (eds) (2001) for the Intergovernmental Panel on Climate Change, Climate Change 2001, The Scientific Basis, Cambridge University Press, Cambridge, UK. [13] ESCOBA, A European Multidisciplinary Study of the Global Carbon cycle in Ocean, Atmosphere and Biosphere, European Commission 1999, EUR 16989 EN. [14] McGuffie, K. and Henderson-Sellers, A. (2005) A Climate Modelling Primer, 3rd edn, John Wiley, Chichester, UK. [15] Nakicenovic, N. and Swart, R. (eds) (2000) IPCC Special report on Emission Scenarios, Cambridge University Press, UK. Chapter 4.

4 Heat Engines It is characteristic for human beings to delegate hard physical labour. Before the industrial revolution animals and slaves were used to produce mechanical output in, for example, a treadmill. An estimate of the power that can be produced by a healthy human being can be made from an anecdote reported by the French physicist Coulomb [1]. He was told that French marines had reached the peak of Tenerife, with a height of 2923 metres, on foot in 7.25 hours. With an average body weight of 70 [kg] this corresponds to 77 [W] during their walk (Exercises 4.1 and 4.2). As a human being has to rest part of the day, the effective power averaged over 24 hours will be even lower. The industrial era in Europe started with the extensive use of water wheels and windmills to produce the mechanical power to drive the looms, soon followed by the use of steam engines for these and many other applications. In steam engines combustion of wood or coal produced heat, which then was converted into mechanical power. More generally, heat engines convert heat into power, irrespective of the origin of the heat. Physicists recognized that both heat and mechanical work are forms of energy. They developed the field of thermodynamics to describe the conversion process and found that thermodynamical variables are useful to understand and quantify many other processes, like chemical reactions, condensation and evaporation. Heat engines still form the heart of industrial society. This is illustrated in Figure 4.1. On the right in the figure one finds the end uses of power, from agriculture to transport. The power is provided from the top middle of the figure, which originates from the energy sources on the left. The sources are from stocks below the surface of the earth, which by mining or drilling get their concentrated forms, either as chemical energy (oil, gas, coal) or physical energy as in uranium. These may be used to produce intermediate heat, which a heat engine converts to power, for space heating in buildings or as process heat to speed up chemical reactions in industry. On the left of Figure 4.1 one also finds renewable energies (solar, biomass, wind and hydropower), some of which may be converted directly into mechanical or electrical power, and others which give intermediate heat and drive a heat engine. It is a challenge to scientists Environmental Physics: Sustainable Energy and Climate Change, Third Edition. Egbert Boeker and Rienk van Grondelle. © 2011 John Wiley & Sons, Ltd. Published 2011 by John Wiley & Sons, Ltd.

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Environmental Physics Flow energy = renewables Solar Wind Biomass Hydro

Concentrated energy Chemical Physical Uranium Oil 235U Gas Coal CxHy

Power Electric or mechanical

Heat engines

Application of power Agriculture Industry Services Domestic Transport

Intermediate heat

Mining drilling Energy stocks Oil fields U-ores Gas fields Coal beds

Heat end use Space heating Process heat

Figure 4.1 Energy supply in society. On the right one finds the end uses of power and the end uses of heat. On the left the energy sources, from stocks or from renewables, are displayed. They provide power either directly or by means of intermediate heat. In the latter case one needs heat engines to convert heat into power.

and engineers to apply their knowledge and design machines which loose as little energy as possible in all the steps of Figure 4.1. From Figure 4.1 it is obvious that heat engines play a key role in modern society. In this chapter we therefore start with heat itself. In Section 4.1 transfer and storage of heat is discussed, followed by the principles of thermodynamics in Section 4.2. In Section 4.3 examples of converting heat into motion are given, according to the literal meaning of the word thermodynamics: heat and motion. In power stations mechanical power from heat engines is converted into electrical power, which acts as an energy carrier to the final user. Therefore, in Section 4.4 electricity is discussed as an energy carrier, while Section 4.5 is devoted to the pollution aspects of heat engines. In Section 4.6 the subject matter is applied to the energy aspects of the private car. Finally, in a brief Section 4.7 the financial aspects of building and maintaining an expensive installation are described. In this chapter all properties summarized in Appendices B and C will be used.

4.1

Heat Transfer and Storage

For some applications heat needs to be stored for a period of time. In these cases the loss of heat should be as small as possible. For other applications one needs heat at a different place to where it is produced and one has to know the ways in which heat energy may be transported. The hot exhaust pipe of a motor cycle illustrates the mechanisms of heat transport. When one touches the exhaust pipe, one will burn one’s fingers as heat is transported from the inside to the outside of the pipe by conduction. When one puts the

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T1 T2 x1

x2

x

q´´ d Figure 4.2 Heat flowing in a homogeneous sheet with surface area A over a distance d from a high temperature T1 to a lower temperature T2 , resulting in a heat current density q from left to right.

hand underneath the pipe one feels its radiation and when one puts the hand above the pipe one will feel both the rising hot air (convection) and radiation. Finally, heat is transported from the engine to the outside air with the flow of exhaust gases, mainly as sensible heat, partly as latent heat of water vapour. These mechanisms will be discussed below. 4.1.1

Conduction

Consider for simplicity a sheet of homogeneous material with two plane-parallel surfaces at a distance d, as sketched in Figure 4.2. It is assumed that the surface area A is big enough to ignore the effects of the edges. The temperature is indicated by T along the vertical axis and the position perpendicular to the planes by x. At position x1 the sheet has a temperature T 1 and at position x2 it has a temperature T 2 . The heat current density q is defined as the amount of heat in joules [J] which flows per second [s] through 1 [m2 ] perpendicular to its direction. Its dimension therefore is written as q [J s−1 m−2 = Wm−2 ]. The total heat current q may be written as q = Aq [W]

(4.1)

For a homogeneous material, as in Figure 4.2, Fourier’s law gives a simple relation between the heat current density q and the gradient of the temperature. In a one-dimensional situation, as shown in Figure 4.2, this may be written as dT (4.2) q = −k dx For a general three-dimensional case one uses the gradient vector (see Appendix B) and the expression becomes q = −k∇T (4.3) where k [Wm−1 K−1 ], the thermal conductivity is dependent on the kind of material, its temperature and density. Table 4.1 gives values of k for some common materials. For insulators one should use materials with small k and for highly conductive systems materials with a large k – especially metals.

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Table 4.1 Heat transport properties of common materials at T = 300 [K]. Thermal conductivity k/[Wm−1 K−1 ]

Density ρ/[kg m−3 ]

Insulators Air Glass fibre (loose fill) Urethane foam Cork Mineral wool granules Paper Glass Gypsum plaster

0.026 0.043 0.026 0.039 0.046 0.180 1.4 0.22

1.161 16 70 120 190 930 2500 1680

225 32 3.6 1.8

24 44 92

1.4 7.5 1.21

470 1620 635

Building materials Cement mortar Soft wood Hard wood Oak wood Brick Concrete

0.72 0.12 0.16 0.19 0.72 1.4

1860 510 720 545 1920 2300

4.96 1.71 1.77 1.46 4.49 6.92

1020 290 380 499 1075 1680

Metals Iron Aluminium Steel (C, Si) Copper

80.2 237 52 401

7870 2700 7800 8933

228 972 149 1166

17000 24000 13500 37000

0.27 0.52 0.13 0.06

1515 2050 1100 80

2.23 1.38 0.59 5.77

572 1400 540 80 1610 1120 525 105

Material

Miscellaneous Sand Soil Soft rubber Cotton Porcelain Human skin Linoleum Carpet wool Water

Fourier Contact coefficient Coefficient b/ a/[10−7 m2 s−1 ] [J m−2 K−1 s−1/2 ]

0.37

0.61

996

Data from [2]; the lowest four b-values from unpublished lecture notes Delft Technological University.

Figure 4.2 describes a stationary situation, therefore the heat flow q will be independent of x as for any plane the heat flow ‘in’ should equal the heat flow ‘out’. One finds with Eqs. (4.1) and (4.2):

q = −k A

dT = constant dx

(4.4)

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Therefore the derivative dT/dx is constant and T(x) is a straight line, as already shown in the figure. It follows that dT/dx = (T 2 − T 1 )/d and d = qR (4.5) kA Eq. (4.5) looks analogous to Ohm’s equation for an electric current I between potentials V 1 and V 2 , which reads V 1 − V 2 = iR, where the electric resistance is R. The heat resistance R [K W−1 ] follows from Eq. (4.5) and may be written as T1 − T2 = q

d (4.6) kA which is analogous to the expression for the electric resistance, in which case k would be the electric conductivity. This explains the name thermal conductivity for k in Eq. (4.2). In fact, Eq. (4.3) with T replaced by V and q by j is another expression of Ohm’s law. The analogy between conduction of heat and electricity reaches even further. This will become clear from looking at two sheets, with different thicknesses d1 and d2 and different thermal conductivities k1 and k2 as indicated in Figure 4.3. Again a stationary situation is considered, therefore no heat accumulates and one has the same current density q everywhere. Eq. (4.5) may be applied to each sheet and the two resulting equations may be added. This gives R=

d1 = qR1 k1 A d2 T2 − T3 = q = qR2 k2 A T1 − T3 = q(R1 + R2 ) = qR

T1 − T2 = q

(4.7)

where R = R1 + R2 may be interpreted as the substitution resistance of two resistances in series.

T1

T2 x1

T3 x3

x2 q´´ d1 k1

x

q´´ d2 k2

Figure 4.3 Two parallel homogeneous sheets in a stationary situation. Thickness d and thermal conductivity k may be different for the two sheets. The heat current density is the same everywhere; the thermal resistances may be added.

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Environmental Physics Table 4.2 Typical values for convection heat transfer coefficients h/[Wm−2 K−1 ]. Phase

Free convection

Forced convection

Gases Fluids

2–25 50–1000

25–250 50–20 000

Reproduced with permission of John Wiley, copyright 1990, from [2], p. 9. Phase change = evaporation or condensation: 2500–100 000.

4.1.2

Convection

When a hot body like a teapot of temperature T s is in contact with air with temperature T ∞ , the air particles will be heated, become less dense, move up and take part of the heat from the teapot with them. Newton’s law of cooling describes the heat current density of the resulting free convection as q  = h(Ts − T∞ )

(4.8)

where h [Wm−2 K−1 ] is the convection heat transfer coefficient. This process of heat transfer becomes more effective when air is blown around the pot with a fan, as more particles of air are passing the pot. This is called forced convection. Here, again Eq. (4.8) describes the process, but the coefficient h will be much larger. Table 4.2 gives some typical values of the convection coefficients. The range of values of h is big; also cooling with a fluid like water appears to be much more effective than with air, which agrees with experience. The bottom row of Table 4.2 will be discussed later. For a surface with area A it is possible to define a convection heat resistance analogous to Eq. (4.5). From (4.8) it follows that Ts − T∞ =

q  q = = qRconv h hA

(4.9)

with Rconv = 1/(hA). For a wall consisting of two parallel sheets with two outer surfaces, shown in Figure 4.3, the convection at the two outer surfaces adds to heat resistance. With convection coefficients h1 and h2 , the total heat resistance would become R=

1 d1 d2 1 + + + h 1 A k1 A k2 A h 2 A

(4.10)

This again reads as four resistances in series. Note that we came across convection in Figure 3.5 as one of the mechanisms by which the earth’s surface exchanges energy with the atmosphere. In the same figure we also met another mechanism of energy transfer, radiation. 4.1.3

Radiation

Any body in a radiation field may be characterized by three parameters, each depending on wavelength λ: the fraction α λ of incoming radiation to be absorbed, the fraction ρ λ to be reflected and the fraction tλ to be transmitted. As there are no other mechanisms than

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83

absorption, reflection and transmission it follows that α λ + ρ λ + tλ = 1. A black body by definition has no transmission or reflection at any wavelength, so ρ λ = tλ = 0 and α λ = 1. The emission from a black body depends on wavelength λ and temperature T but not on direction. At any (λ, T) no surface can emit more than a black body, of which the spectrum was already given in Eqs. (2.4) or (2.2), and shown in Figure 2.1. Integrated over all wavelengths one obtains formula (2.6), also used in Chapter 3, viz. q  = σ T 4

(4.11)

For a body which is not completely black one could write the spectrum as ελ times the black-body value (2.4) at wavelength λ. Here, ελ is called the emissivity of the surface with ε λ ≤ 1. Kirchhoff’s law states that, at any temperature T and for any surface, emissivity equals absorptivity ελ (T ) = αλ (T )

(4.12)

For a black body ελ (T) = α λ (T) = 1 and for a hypothetical grey body both are independent of λ and T, so ε = α. A grey body with surface temperature T s in an environment that is completely surrounding the body will emit q = εσ T s 4 . The surroundings may be assumed to behave like a black body as no radiation is reflected back. This environment, however, produces its own radiation, which may be described by an effective radiation temperature T surr . The absorption of this radiation by the grey body will be q = εσ T surr 4 . As ε = α it follows that the net heat exchange with the surroundings can be written as   4 4 q  = εσ Ts4 − ασ Tsurr (4.13) = εσ Ts4 − Tsurr In principle, one could define a heat resistance here as before, and add it to the resistance (4.10) (Exercise 4.3) 4.1.4

Phase Change

The bottom row of Table 4.2 refers to phase change, in particular the combination of convection with evaporation or condensation. An example is the way in which power stations dispose of their surplus heat. In so-called wet towers a spray of hot water from the condenser is cooled by rising air (forced convection). Furthermore, part of the water evaporates, extracting its evaporation heat from the spray as well, which together with the forced convection leads to effective cooling. Another example of phase change is found in the refrigerator, where a working fluid is circulated in a closed circuit. Inside the refrigerator the fluid is forced to evaporate, thereby cooling its surroundings. At the outside the vapour passes a condenser where the resulting condensation heat is exchanged with the surroundings. The resulting convection can be experienced by putting your hand above the grating at the back of the fridge. 4.1.4.1

Thermosyphon and Heat Pipe

Two other examples of phase change are shown in Figure 4.4. On the left a thermosyphon is shown, which consists of a closed pipe with liquid at the bottom. There heat is supplied from the outside, causing evaporation of the liquid. At the top of the syphon a cooling

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(a)

Heat in

Figure 4.4

Condensation

Heat out (b)

Vapour

Heat in

liquid

Cooling

Wick

Liquid

Heat out

Transport of heat by phase change in (a) a thermosyphon and (b) a heat pipe.

fluid is passing; consequently the vapour is cooled and it condensates, liberating the heat again. The heat is taken away by the cooling fluid. By gravity the condensed liquid again descends. The heat pipe, on the right in Figure 4.4 does not need gravitation to return the fluid to its original position. In this case there is a wick or gauze at the inner side of the pipe. The condensation takes place at the right of the figure and evaporation on the left. This causes a concentration gradient in the wick, resulting in a force to the left, to even out the concentration differences. 4.1.5

The Solar Collector

The solar collector, sketched in Figure 4.5 produces hot water from incoming solar radiation. The solar radiation is absorbed by an absorber, which heats water tubes that transport hot water to a container on the right. If necessary, some additional gas heating can be provided before the water is used for the tap or for home heating. Below, the mechanisms discussed above are used to answer some simple questions. For the sake of the argument it is assumed that under optimal conditions the water is heated to 80 [◦ C]. Reflection Insolation

Glazing Absorber Water tubes

Insulator

Hot water Home

Figure 4.5 A simple flat plate solar collector. Solar radiation is absorbed; it heats water tubes, which transport hot water to a container where, if necessary, additional heating is provided.

Heat Engines

Example 4.1

85

Maximum Incoming Energy Density

Give an estimate of the incoming energy density S [Wm−2 ] for which the collector should be designed. Answer The absolute maximum will be the solar constant S = 1366 [Wm−2 ], the energy density entering the upper atmosphere. From Figure 3.5 we conclude that on average only 47% reaches the surface, which is 650 [Wm−2 ]. At the poles the irradiation will be lower and at the equator higher. Therefore it seems reasonable to take a maximum energy density of S = 1000 [Wm−2 ] as a requirement for the collector design.

Example 4.2

Temperature Differences within the Collector

The absorber in Figure 4.5 is constructed from 10 [cm] thick high-quality steel. What will be the maximum temperature difference between the top of the absorber and the water tubes. Answer The thermal conductivity of steel is given in Table 4.1 as k = 52 [Wm−1 K−1 ]. From Eq. (4.5) one has T1 − T2 = q  ×

d 1000 [Wm−2 ] × 0.1 [m] = = 1.9 [K] k 52 [Wm−1 K−1 ]

This is small, but not negligible, so a thinner absorber would help. The construction should be insulated with, for example, glass fibre to reduce conduction losses to the surroundings of the device.

Example 4.3

The Glazing of the Collector

What is the function of the glazing on top of the collector? Answer Without glazing the absorber would be in direct connection with the atmosphere at, say T ∞ = 20 [◦ C]. The loss of heat by convection is found from Eq. (4.8) with h = 10 [Wm−2 K−1 ] as an average value from Table 4.2. This gives for the convection loss q  = h(Ts − T∞ ) = 10 [Wm−2 K−1 ] × 60 [K] = 600 [Wm−2 ] which is close to the maximal incoming power density. So, without glazing the temperature of 80 [◦ C] can never be reached. With the glazing there still will be convection

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underneath the glass. Assume that the resulting glass temperature is 30 [◦ C]. The heat loss from absorber to glass then would be q  = h(Ts − T∞ ) = 10 [Wm−2 K−1 ] × 10 [K] = 100 [Wm−2 ] which is still considerable. That is the reason why modern collectors work with a vacuum between the absorber and the ambient atmosphere.

Example 4.4 Radiation Losses from the Collector Argue why one would need coatings on the absorber with α λ = ελ ≈ 0.94 in the visible region and α λ = ελ ≈ 0.07 (or if possible less) in the IR region. What is the use of the glass cover in this respect? Answer Assume that the absorber emits the maximum possible, which is that of a black body q = σ T 4 . For T = 273 + 80 [K] the radiation loss becomes q = 881 [Wm−2 ], more than the incident radiation. This again would mean that the temperature of 80 [◦ C] cannot be reached. The clue is to use the fact that emissivity ελ and absorbtivity α λ are functions of λ and moreover that the solar spectrum to be absorbed and the infrared spectrum to be emitted (for T = 80 + 273 [K]) peak at very different wavelengths, see Figure 2.1. Therefore the trick is to use a coating on the absorber with a high value α λ ≈ 0.94 in the visible region around λ ≈ 0.5 (see Eq. (2.5)). In that region almost all of the incoming radiation will be absorbed. Of course ε λ ≈ 0.94 also in that region (Eq. (4.12)). However, blackbody emission for T = 80 + 273 [K] peaks at λ ≈ 8.2 [μm] (Eq. (2.5)) and is almost negligible in the region λ ≈ 0.5 [μm]; therefore one will not lose anything in the visible wavelength spectral regions. The coating must be such that in the infrared region around λ ≈ 8.2 [μm] the emission will be very small. In practice, values α λ = ελ ≈ 0.07 have been achieved, still implying a loss of 0.07 × 881 [Wm−2 ] = 62 [Wm−2 ]. The infrared radiation from the coating will be absorbed for a large part by the glass cover, which is transparent for visible light, but opaque for IR light. Even in the absence of convection inside the cover, the absorbed IR radiation will increase the cover’s temperature, resulting in some radiation loss to the outside.

Example 4.5 A Realistic Solar Collector A realistic solar collector has a glass cover with a high transmission tc = 0.97, a temperature T c = 30 [◦ C] and an emissivity in the IR region ε = 0.94. The ambient air has a temperature T air = 20 [◦ C] and the surroundings have a radiation temperature T surr = –10 [◦ C]. The convection coefficient may be taken as h = 10 [Wm−2 K−1 ]. (a) Calculate the heat flux into the collector. (b) A mass m [kg s−1 ] of water is passing the tubes per second and is heated from T in to T out . Calculate temperature increase

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of the water T out − T in for m = 0.01 [kg s−1 ] and a surface area of the collector A = 3 [m2 ]. Answer (a) As the glass cover does not transmit the IR from the absorber, the only interaction of the collector with the surroundings is the glass cover itself. One has the following relation for the net input q [Wm−2 ]:   4 q  = tc S  − εσ Tc4 − Tsurr − h(Tc − Tair ) = 336 [Wm−2 ] (4.14) (b) The specific heat of water is given in Appendix A as cp [J kg−1 K−1 ]. One has q = Aq  = mc p (Tout − Tin ) where the dimensions check, for [W] = [J s−1 ] = [kg s−1 ] [J kg−1 K−1 ] [K]. The answer becomes T out − T in = 24 [◦ C]. Note: On the internet one will find many modern types of solar collectors. It is interesting is that some models are using heat pipes to transport heat from the absorbers to a central tube, where water will take the heat to a central container.

4.1.6

The Heat Diffusion Equation

In general, the temperature T in a medium will depend on position r and time t so T = T(r, t). The heat diffusion equation describes the change of temperature that follows from conservation of heat, for the case where the heat transport in the medium is by conduction only. Conservation of heat means that for any time interval t to t + dt and for any volume element dV the following expression holds: increase of heat content [J] = net inflow by conduction [J] + internal heat production [J] (4.15) At the very right of Eq. (4.15) the heat production (which is negative in the case of a heat sink and positive for exothermal chemical reactions or radioactive decay) is determined by ˙ dt[J] to (4.15). The first ˙ t) [J s−1 m−3 ], which contributes qdV the heat production rate q(r, term on the right-hand side of Eq. (4.15), the net inflow, can be found by using Appendix B. There it is shown that the outflow of a field q [Wm−2 ] is described by div q [Wm−3 ]. The first term on the right of (4.15) is the inflow between t and t + dt and becomes − div q dVdt [J]. On the left-hand side of Eq. (4.15) the heat content of the volume element may be written as ρcp TdV [J], where ρ [kg m−3 ] is the density and cp [J kg−1 K−1 ] is the specific heat. Eq. (4.15) therefore gives ∂ ˙ dt (ρc p TdV)dt = −div q dV dt + qdV ∂t

(4.16)

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Environmental Physics

Using Fourier’s equation (4.3) leads to ∂T ρc p = div(k gradT ) + q˙ (4.17) ∂t This equation is known as the heat diffusion equation or in short the heat equation. It is, however, advisable not to use the short hand, as Eq. (4.17) does not describe convection or radiation of heat. If the thermal conductivity k is independent of position, Eq. (4.17) may be simplified to ∂T = k div(gradT ) + q˙ (4.18) ρc p ∂t or ∂T 1 q˙ (4.19) = aT + ∂t ρc p where  is the Laplace operator (Appendix B) and the material property a = k/(ρcp ) is called the Fourier coefficient and is, for a number of materials, listed in Table 4.1. When there are no heat sources or sinks present q˙ = 0 and Eq. (4.19) reduces to ∂T = aT (4.20) ∂t and in one dimension ∂T ∂2T =a 2 ∂t ∂x In the following we will give some examples of Eq. (4.21) taken from daily life. 4.1.6.1

(4.21)

Periodic Heat Wave Hitting a Semi-Infinite Medium

A periodic heat wave with frequency ω and amplitude T 0 coming from the left hits a very thick wall perpendicular to the wave. The coordinate perpendicular to the wall is called x; because of the thickness of the wall it is assumed to extend from x = 0 to x = ∞. The boundary conditions of this problem may be written as x = 0 : T = Tav + T0 cos ωt (4.22) x = ∞ : T = Tav where T av is the temperature at great depth in the wall, where the temperature variation never will be experienced. We guess that the solution within the medium will be a periodic wave travelling with velocity v and with an amplitude which will be damped as a function of x. This gives  x . (4.23) T = Tav + T0 e−Ax cos ω t − v This function indeed satisfies the boundary conditions (4.22). By substitution it is shown that (4.23) is a solution of Eq. (4.21) provided that A = v/(2a) and also A = ω/v, which gives √ v = √2aω (4.24) A = ω/2a The damping depth xe = 1/A is the depth at which the amplitude of the wave is reduced by a factor 1/e and the delay time τ = 1/v is the time required for the wave to travel 1 [m] (Exercise 4.8).

Heat Engines

4.1.6.2

89

A Sudden Change in Temperature

A similar half-infinite medium as above now experiences a sudden temperature change from T 0 to T 1 . Examples are the cooling of a hot chimney by a shower or the heating of a stone floor by a fire. The boundary conditions are T (x = 0, t ≥ 0) = T1 T (x > 0, t = 0) = T0

(4.25)

The equations imply that the outside temperature T 1 remains constant in the course of time, while inside the material a heat flow will start. The solution of the heat diffusion equation (4.21) with these conditions leads to the error function defined in Eq. (C7) of Appendix C:   x t ≥0 (4.26) T = A + Bx + Cerf √ 2 at This may be verified by substitution into (4.21). The error function should be differen√ tiated first to its argument x/(2 at) and then to x or t. The next step is to use Eqs. (4.25) and one will find   x t ≥0 (4.27) T = T1 + (T0 − T1 ) erf √ 2 at The heat current density q follows in one dimension from Eq. (4.2). This gives k(T1 − T0 ) −x 2 /(4at) e √ πat

(4.28)

k(T1 − T0 ) b T1 − T0 =√ √ √ π aπ t t

(4.29)

q  = At the boundary x = 0 one finds q  =

 where b = kρc p is called the contact coefficient and displayed in the last column of Table 4.1. Its name is understood from the following example. 4.1.6.3

Contact Temperature

One brings two half-infinite materials with temperatures T 1 and T2 (T 1 > T 2 ) into ideal thermal contact at time t = 0. The contact plane will acquire a contact temperature T c somewhere in between. This contact temperature is found by the requirement that the flow out of the hotter material (with contact coefficient b1 ) equals the flow into the colder (with contact coefficient b2 ). From Eq. (4.29) one finds b1 b2 √ (T1 − Tc ) = √ (Tc − T2 ) πt πt

(4.30)

or Tc =

b1 T1 + b2 T2 b1 + b2

(4.31)

For finite materials in contact this relation will be a good approximation for a short period of time when the heat flows have not changed the temperature of the objects involved, for

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example for a human foot walking on a colder floor. Its skin temperature will be about 30 [◦ C]. From experience it is known that on an oak wooden floor with a temperature of 15 [◦ C] a naked foot still feels comfortable. Apparently the heat flow out of the foot is small.

Example 4.6 A Pleasant Floor Temperature From the data given above deduce the temperatures where the naked foot still feels comfortable for a concrete floor and for a cork floor. Discuss the physics behind the difference. Answer The contact temperature of skin and wood follows from Eq. (4.31) and Table 4.1 as 25.4 [◦ C]. Eq. (4.31) may be rewritten as T 2 = ((b1 + b2 )Tc − b1 T 1 )/b2 which gives for concrete T 2 = 22 [◦ C] and for cork T 2 = −31 [◦ C]. The physical origin of the difference is that the low k value of cork results in a small heat current into the cork. The heat flow out of the foot then will be small, allowing a large temperature difference (Exercise 4.9).

4.1.7

Heat Storage

The solar collector of Figure 4.5 is coupled to a reservoir of hot water of about 150 [L], which at a temperature of 80 [◦ C] would be enough for a few hot showers at night when the sun no longer heats the reservoir. With good insulation the hot water may last till the next morning (Exercise 4.10). It is more complicated to collect hot water during the summer and use it during the winter. One obviously needs a much larger volume of water if one stores the water as sensible heat. An average household in mid-latitudes would need to store more than 100 [m3 ] water of 90 [◦ C] with very good insulation (Exercise 4.11). A better proposition could be to use the latent heat from a phase change at a temperature not far from room temperature. Water with phase change temperatures of 0 [◦ C] or 100 [◦ C] is not convenient. There are many suitable materials, each with their own strengths and weaknesses; much research is going on to match materials with a particular application [4]. We mention two examples of phase change materials. The first is octadecane, a paraffin with a melting point at 28 [◦ C] and heat of fusion of 189 [MJ/m3 ]. At temperatures below 28 [◦ C] it is a solid; for temperatures above, it becomes a liquid, extracting heat from the surroundings. When the heat is needed the temperature is lowered again. Difficulties are a low conductivity in the solid phase (k = 0.15 [Wm−1 K−1 ]) and a density change from solid to liquid. A second example is found in the class of hydrates such as Glauber salt Na2 SO4 .10H2 O. Below 32.4 [◦ C] the salt is solid. At that temperature it loses its water and becomes anhydrous sodium sulfate and a saturated solution of Na2 SO4 in water. The heat of fusion is 377 [MJ/m3 ]. Again there is a large density difference between the two phases. Moreover, part of the anhydrous sulfate settles down as bottom sediment; it is therefore difficult to get a reversible process.

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91

It appears (Exercise 4.11) that for the examples mentioned the enthalpy of fusion is too low for seasonal storage. Research therefore is looking for materials with fusion heats in the order of [GJ/m3 ]. It should be added that some phase change materials could be encapsulated inside building materials. During the day heat is absorbed by phase change keeping the building cool and during the cooler night the heat is given back to the building. In this way daily temperature changes are damped. The last example of heat storage we mention is to transfer the heat surplus of the summer into the ground water. There it is kept in aquifers, sand mixed with water, until the heat is needed in winter. There are two ways to transfer the heat. The simplest way is to have two boreholes, one to pump ground water up; the second to pump it down again after heat has been extracted. A better way is to have a closed system with one or several boreholes. In each hole a heat pump, to be discussed later in Eq. (4.64), takes care of the heat transfer. In this case there are two heat exchangers, one down in the ground and the second in a building. The heat pump will work in both directions: heating the building in winter and cooling it in summer. In using groundwater as storage medium, one will lose some heat to the surroundings and one has to be careful not to pollute the groundwater.

4.2

Principles of Thermodynamics

The two laws of thermodynamics provide a macroscopic understanding of energy conversion. For their application thermodynamic variables are introduced of which entropy S is the most important. 4.2.1

First and Second Laws

In the simplest case, a homogeneous quantity of matter with mass m in a single phase, the macroscopic state of the system is defined by two independent variables, usually the pressure p and either the volume V or the temperature T. For a mixture of chemical substances in ϕ perhaps several phases (solid, liquid, gas) the number of moles n i of chemical i in phase ϕ must also be specified ([5], [6]). The First Law of Thermodynamics expresses conservation of energy. This law has already been used for the specific case of 1 [kg] of air in Eq. (3.9). In general, the First Law may be written for an infinitesimal (= very small) change as δ Q = dU + δW

(4.32)

That is, the amount δQ of heat added to the system is used to increase its internal energy by dU and to perform work δW. In many handbooks ([5], [6]) δW is defined as work done on the system, in which case it will have a minus sign in Eq. (4.32). In practice, one will consider changes of a system from state 1 to state 2. In that case the work W 1→2 done by the system will depend on the path. Also the heat added Q1→2 will depend on the path. The fact that the amount of heat is dependent on the path is indicated by the symbol δ in Eq. (4.32). On the other hand, the change in internal energy only depends on the internal energy of the initial and final states U 1→2 = U 2 − U 1 . Such a function is called a state

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pA

dz

dV A V Figure 4.6

The work done by the gas equals pAdz = pdV.

function and its change is indicated by the symbol d as in dU (Eq. (4.32)). A finite change of a state function may be indicated by the symbol , for example U = U 2 − U 1 . For a quasistatic, reversible process where expansion is the only kind of work, the work done by the system may be written as δW = pdV. This is illustrated in Figure 4.6 for the simple case of a gas. The gas expands over a distance dz against the air pressure p. Its counterforce equals pA where A is the surface area of the piston. The work done (force × path) indeed equals pAdz = pdV. For the case of Figure 4.6 Eq. (4.32) becomes δ Q = dU + pdV

(4.33)

or in integral form 2 Q = U2 − U1 +

pdV

(4.34)

1

where the subscript 1 → 2 is omitted in the added heat Q. If the system also performs some other work, for example electromagnetic, that work is expressed by adding δW e to Eq. (4.33) and its integral to Eq. (4.34), giving δ Q = dU + pdV + δWe 2 2 Q = U2 − U1 + pdV + δWe 1

(4.35) (4.36)

1

The Second Law of Thermodynamics may be formulated by introducing the entropy function dS =

δQ T

(4.37)

where the reversibly added heat δQ is divided by the absolute temperature T. The integral of this function is independent of the path ([5], 179/180); S is a state function indicated by the symbol d in dS. More generally, one may include irreversible processes as well, which leads to the Clausius inequality dS ≥

δQ T

(4.38)

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p1,V1

93

p2,V2

Initial state

Final state

Throttle Figure 4.7 A certain mass expands through the throttle via narrow holes without heat exchange with the surroundings. The enthalpy is the same on the left and the right.

Here the equality sign (=) would hold for reversible processes. The Second Law says that for a closed system (where the added heat δQ = 0) the entropy does not decrease dS ≥ 0

(4.39)

From the four state functions (p, V, U, S) others can be derived for specific applications: enthalpy H, free energy F and Gibbs free energy G all with the unit [J]. The enthalpy H is defined by H = U + pV

(4.40)

For an infinitesimal change of state it follows that dH = dU + pdV + V d p = δ Q + V d p

(4.41)

The last equality is restricted to processes where Eq. (4.33) holds. Many processes (phase change, chemical reactions) are taking place at constant pressure dp = 0. Then the added heat is equal to the increase in enthalpy of the system. Enthalpy changes are tabulated as H ([kJ kg−1 ] or [kJ mol−1 ]). The significance of the term pV in Eq. (4.40) can be understood by the example of the throttle shown in Figure 4.7. A certain amount of gas expands through little holes from the left to the right without exchange of heat with the surroundings Q = 0. The pressures p1 on the left and p2 on the right are kept constant (p1 > p2 ). Eq. (4.34) may be written as 0 0 = U2 − U1 +

V2 pd V +

V1

0 pdV = U2 − U1 + p1

dV + p2 V1

0

V2

= U2 − U1 − p1 V1 + p2 V2

dV 0

(4.42)

or H1 = U1 + p1 V1 = U2 + p2 V2 = H2

(4.43)

which means that the enthalpy of the initial state equals the enthalpy of the final state. In Eq. (4.42) p2 V 2 − p1 V 1 is the work done by the system, so pV may be interpreted as a work function. Another useful function is the free energy F of the system defined by F = U − TS

(4.44)

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Its relevance follows from the reversible infinitesimal change dF = dU − TdS − SdT = dU − δ Q − SdT = −δW − SdT

(4.45)

where Eqs. (4.37) and (4.32) were used. For a reversible isothermal change with dT = 0, it follows that –δW = dF or in words: the work done on the system is equal to its increase in free energy. Conversely, for an isothermal process the work δW that can be delivered equals the decrease in free energy, hence the name ‘free’. Note that this applies to the total work, expansion and nonexpansion work together. A similar meaning can be attached to the last state function to be considered, the Gibbs free energy (often shortened to free energy, so be careful) G = H − TS

(4.46)

For a reversible, infinitesimal process one finds, using Eqs. (4.40), (4.35) and (4.37) dG = dH − T dS − SdT = dU + pdV + Vd p − TdS − SdT = δ Q − δWe + Vd p − δ Q − SdT = Vd p − SdT − δWe

(4.47)

Consequently, for isobaric, isothermal processes (dp = dT = 0) one has dG = −δWe

(4.48)

This relation shows that nonexpansion work δW e reversibly done by the system will decrease its Gibbs free energy by the same amount (dp = dT = 0). The Gibbs free energy is often used to discuss a system consisting of different compounds, for example, ni moles of chemical i with i = 1,2,3. . . Then one could write G = G(p, T, ni ) and its differential as



 ∂G

∂G

∂G



dp + dT + dn i (4.49) dG = ∂p

∂T

∂n

T,n i

p,n i

i

i p,T,n j

where the partial derivatives on the right assume all variables except one to be constant. Eqs. (4.49) and (4.47) should hold for all values of dp, dT, dni , therefore their coefficients should be equal. Thus the first two terms in Eq. (4.49) are equal to the first two terms of Eq. (4.47), while the last term may be written as  ∂G



dn i = μi dn i (4.50) −∂ We =

∂n i p,T,n j i i which defines the chemical potentials μi = ∂G/∂ni [J mol−1 ]. The nonexpansion work δW e done by the system if dp = dT = 0 equals the right-hand side of Eq. (4.50) with an extra minus sign. In general one has  μi dn i (4.51) dG = Vd p − SdT + i

Similarly one will find (Exercise 4.12) dU = − pdV + TdS +

 i

μi dn i

(4.52)

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95

The chemical potential μi [J mol−1 ] gives the increase of internal energy of the system when one adds dni moles of substance i while keeping all other variables constant. In deriving Eqs. (4.45) and (4.48) the fact that for reversible processes the definition of entropy (4.37) gives δQ = TdS was used. For irreversible processes the Clausius inequality (4.38) gives TdS ≥ δQ. This implies (Exercise 4.13) that for isothermal processes (dT = 0) δW ≤ −dF

(4.53)

and for isothermal isobaric processes (dp = dT = 0) δWe ≤ −dG

(4.54)

Eq. (4.53) shows that for isothermal (dT = 0) irreversible processes (the inequality sign) the work is smaller than the decrease in free energy; similarly if dp = dT = 0 the nonexpansion work that can be done is smaller than the decrease in Gibbs free energy. It must be stressed that Eqs. (4.51) and (4.52) hold for all processes, reversible or irreversible. The difference in G– or U–values between two states can be found from tabulated values of the chemical potentials, as they are independent of the path. 4.2.2

Heat and Work; Carnot Efficiency

An ideal heat engine operates between two reservoirs of heat, one at high temperature T H and one at a low temperature T C . It is assumed that the reservoirs are so big that addition or extraction of heat does not change the temperatures. For each reservoir the finite change of entropy S may be found from Eq. (4.37) as Q δQ = (4.55) S = T T where Q is the total amount of heat added to the reservoir at constant temperature T. The Clausius inequality (4.38) becomes Q (4.56) T The two reservoirs and the engine form a closed system for which the Second Law becomes, from Eq. (4.39) S ≥

S ≥ 0

(4.57)

The ideal heat engine, sketched in Figure 4.8a, operates in a cycle where the initial and final states of the system are the same. In the cycle an amount of heat QH is taken from the high temperature reservoir, produces work W and the remaining heat QC is delivered to the cold reservoir. The net heat put into the engine equals QH − QC . For a completed cycle the change in internal energy of the system is zero. The First Law (4.32) then gives QH − QC = W. The entropy change S has a contribution (4.55) from the hot reservoir and one from the cold reservoir. The entropy change of the engine is zero as the initial and final states are the same. The hot reservoir has −QH as heat added and the cold reservoir +QC . This gives as total entropy change (4.57) −

QC QH QH − W QH + =− + ≥0 TH TC TH TC

(4.58)

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TH

TH QH

TH = Tamb QH

QH Win

Wout QC

QC

TH = Tamb

TC = T amb

(a) Heat engine

TH = Tamb QH Win

Wout QC

QC TC

(b) Heat pump

(c) ‘Cold’ heat engine

TC (d) Refrigerator

Figure 4.8 Ideal ways of converting heat into work and vice versa. The heat engines (a) and (c) deliver work, the heat pump (b) and refrigerator (d) use work to convert low temperature heat into high temperature heat. (Reproduced by permission from IEEE Publishing, copyright 1986, from [3], p. 76.)

or

  TC W ≤ QH 1 − TH

(4.59)

The best one can achieve is to perform all changes reversibly which gives the maximum work W max from the equality sign in Eq. (4.59)   TC (4.60) Wmax = Q H 1 − TH The thermal efficiency η of a heat engine is defined by η=

W work output = heat input QH

(4.61)

For the reversible heat engine, working between two reservoirs this gives the Carnot efficiency ηCarnot = 1 −

TC TH

(4.62)

The usual heat engine operates between the ambient temperature T C and a much higher temperature T H , as shown in Figure 4.8a. In principle, an engine could also operate using the ambient temperature as T H and a much lower temperature as T C , illustrated in Figure 4.8c. Heat engines could also use work as input in order to transfer heat from a lower to a higher temperature. The first example is the heat pump depicted in Figure 4.8b. It may be used to convert low temperature heat in, for example, groundwater to the higher temperatures needed in a living room. The coefficient of performance COP of installations is generally defined as COP =

useful output required input

(4.63)

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97

for a complete cycle. For the heat pump the useful output is the high temperature heat QH . The input is the work W = QH − QC , where again the first law (4.32) is applied. For a complete cycle the only entropy changes are taking place in the reservoirs. This gives from (4.57) QC QH TH QH − ≥0→ ≥ TH TC QC TC With this inequality the COP can be calculated as COP =

    useful output Q C −1 TC −1 QH QH −1 = 1− ≤ 1− = ηCarnot = = required input W QH − QC QH TH (4.64)

This COP is bigger than 1 and in practice may be as high as 5 or 6. The heat pump is sometimes called a thermodynamic lever. To come back to Section 4.1.7 on heat storage, it is clear that heat storage in summertime in the ground or in groundwater may be done by a heat pump. In summer the low temperature T C is in the ground and the higher temperature T H in a building; in winter it is the other way round. Figure 4.8d shows the second example where work is used to transport heat QC from T C to a higher temperature T H . The lower temperature is inside the refrigerator or freezer and the higher temperature is the ambient one, consequently the ‘useful output’ of Eq. (4.63) becomes QC and the required input remains W = QH − QC . The COP can be calculated as COP =

QC QC useful output = = = required input W QH − QC



QH −1 QC

−1

 ≤

−1

TH −1 TC

(4.65)

Some examples of refrigerators will follow in Section 4.3.5. 4.2.3

Efficiency of a ‘Real’ Heat Engine

The equality signs for efficiencies or COPs in the previous section apply to cases where all changes are reversible. In this section we consider a heat engine which is closer to reality ([7], 33–37), as the unavoidable irreversible processes are described explicitly. One distinguishes between external reservoirs with temperatures T H and T C and hypothetical internal reservoirs with temperatures θ H and θ C . These internal reservoirs remain the same during the process; they do not keep heat, just transmit it. In this more realistic situation heat QH is transported from the external to the internal reservoir by conduction (irreversibly). Next, that heat drives a Carnot engine between the two internal reservoirs; this process is assumed to be reversible as the irreversibility has already been taken into account. Between the internal reservoirs work W is produced; heat QC leaves the cooler internal reservoir and is transported irreversibly to the external cold reservoir. The equations representing this process are easy to write down. Conduction is proportional to the difference in temperature, see Eq. (4.5), which we rewrite as Q H = aH (TH − θH ) Q C = aC (θC − TC )

(4.66)

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Inside the engine everything happens reversibly, so we use Eq. (4.59) with the equality sign and find η=

QH − QC QC θC W = =1− =1− QH QH QH θH

(4.67)

Eq. (4.67) gives the opportunity to express both QC /QH and θ C /θ H in (1 – η). With the two equations (4.66) there are four equations from which one may eliminate QC , θ C and θ H . One finds   1 TC Q H = a TH − 1−η aH aC (4.68) a= aH + aC The delivered work becomes

 W = ηQ H = ηa TH −

1 TC 1−η

 (4.69)

Note that for η = 0 or η = (1 − T C /T H ), which is the Carnot efficiency of the complete device, the work will vanish. For constant a, T H , T C the maximum work is found by taking ∂ W/∂η = 0. This gives a quadratic equation in η with two roots, one of which becomes

TC (4.70) η =1− TH The other root has η > 1, which is not physical. The maximum efficiency is found as in √ Eq. (4.70), which is smaller than the Carnot efficiency by a factor 1 + TC /TH > 1. 4.2.4

Second Law Efficiency

The thermal efficiency defined in Eq. (4.61) is connected with the conversion of heat into mechanical energy. More generally, efficiency for an energy transfer process could be defined by η=

energy output (heat or work) by a device energy input required

(4.71)

In this way, besides the thermal efficiency (4.61) the COP (4.63) is also included in the definition, and the efficiency (4.71) may be either smaller or bigger than 1. It must be stressed that a high efficiency applies to the performance of a device and does not imply an intelligent use of energy. Heating of a human being inside a house, for example, may be achieved in several ways: (a) By heating the house with electric heaters with an efficiency close to 1 (b) By heating the house with heat pumps, electrically driven with COP > 1 (c) By gas heaters in each room with η = 0.8, but only burning at full capacity when somebody uses the room (d) by heating the house with a gas driven central heating system (e) By wrapping the person concerned in an electric blanket.

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99

If the aim is to make a person comfortable with as little energy consumption as possible – and if option (e) is discarded, all aspects should be taken into account: the way in which the electricity or gas is produced, the losses during transport to the building and so on. In comparing options (c) and (d) it is clear that option (c) will be the most economic as it is geared to the task of keeping a person comfortable. The point of the foregoing is that in studying energy efficiencies one should look at the task to be performed and not solely at the device. To do so, the Second Law efficiency ε for a single output (either work or heat) is defined as follows ε=

useful output (heat or work) by a device/system maximum possible output by any device/system for the same energy input (4.72)

This Second Law efficiency is never bigger than 1 and focuses on the task (hence ‘useful output’). The maximum in the denominator refers to the maximum permitted by the laws of thermodynamics. A concept related to the Second Law efficiency is the available work B of a system or fuel. It is defined as B = the maximum work that can be provided by a system or a fuel as it proceeds by any path to a specified final state (4.73) in thermodynamic equilibrium with the atmosphere The atmosphere is taken as the obvious reservoir in applications, although other reservoirs may be introduced in the definition. In the process of providing work the system may perform some work against the atmosphere. That, however, cannot be recovered and therefore is not counted as available work. Note that definition (4.73) applies to work, which can be considered as the highest form of energy. Work can be converted into heat with 100% efficiency. That is not possible the other way round. By comparing Eqs. (4.72) and (4.73) it will be clear that the available work will be related to the denominator of Eq. (4.72). A simple example of available work is that of a mass m raised to a height h. The available work then equals mgh. A second example is that of a quantity of heat Q extracted from a hot reservoir with temperature T H . With the atmospheric temperature taken as T 0 the available work becomes   T0 (4.74) B = Q 1− TH Here the Carnot efficiency (4.62) was used, as the maximum work corresponds with reversible changes everywhere in the process. It is clear that high-temperature heat comprises a high value of available work. Example 4.7

Second Law Efficiency of a Gas Heater

A living room is heated by a gas heater, which delivers an amount Q of heat with temperature T H during a day. Compare the efficiency η of Eq. (4.71) with the second law efficiency ε of Eq. (4.72) using the concept of available work.

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Answer In order to deliver the required amount of heat Q, an amount of gas is needed with an enthalpy change on combustion |H |. The efficiency (4.71) becomes η=

Q |H |

(4.75)

Let the available work in the same amount of gas be B. This is used to drive a heat pump between the low temperature T 0 of the atmosphere outside and the required temperature in the living room T H . The maximum amount of heat that can be delivered to the living room by the heat pump will be B × COP. When one inserts Eq. (4.64) for the coefficient of performance COP the Second Law efficiency becomes   Q T0 Q Q = 1− (4.76) = ε= B × COP B/ (1 − T0 /TH ) B TH In practice the available work B in the gas and the enthalpy change |H | are close. If one ignores the difference the Second Law efficiency ε = (1 − T 0 /T H ) × η is much smaller than η. With for example T 0 = 265 [K] and T H = 296 [K] one finds ε = 0.10 × η. Conclusion: the Second Law efficiency of the gas heater is only 10% of the efficiency (4.75). For the heat pump Q = B × COP; its Second Law efficiency obviously is 100% as this device was used as a reference. 4.2.4.1

Available Work = Exergy

In order to express the available work B in terms of entropy changes, one considers a system characterized by its energy U, volume V and entropy S ([8], 35–38, [9]). The atmosphere has temperature T 0 and pressure p0 . The system may exchange heat and work with the atmosphere (reversibly!) and will reach equilibrium with the atmosphere at energy Uf , entropy Sf and volume Vf . The atmosphere is so big that it does not change. Define a function B, which is to become the available work by B = (U − U f ) + p0 (V − V f ) − T0 (S − S f )

(4.77)

In order to show that this B obeys definition (4.73) one considers a finite change of the system by U, V and S. There is a heat flow Q to the atmosphere; useful work W is done on other systems and nonuseful work W  = p0 V on the atmosphere. The ‘other systems’ are assumed to be passive receivers of work, possibly in an irreversible way; they do not otherwise interact with the system or the atmosphere. The First Law (4.32) may be written as − Q  = U + W + W 

(4.78)

The change in entropy (4.37) is calculated for the atmosphere, for which the temperature T 0 does not change (S)system+atmosphere − (S)system ≡ (S)s+a − (S) =

Q T0

(4.79)

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101

Below the system + atmosphere is denoted by the index s + a, but the changes for the system itself do not have an index attached. From Eqs. (4.78) and (4.79) one finds W = −Q  − U − W  = = −T0 (S)s+a + T0 S − U − p0 V = = −T0 (S)s+a − (U + p0 V − T0 S) = −T0 (S)s+a + B

(4.80)

In the last equality definition (4.77) was used while recognizing that U = Uf − U, V = Vf − V and S = Sf − S. From Eq. (4.80) it also follows that B = W + T0 (S)s+a

(4.81)

‘System + atmosphere’ do not exchange heat with the ‘other systems’, so from Eq. (4.39) one has (S)s+a ≥ 0. The last term on the right in Eq. (4.81) therefore is non-negative and the maximum of useful work W equals B, which therefore correctly is called ‘available work’, according to Eq. (4.73). The term T 0 (S)s+a in Eq. (4.81) may be called the lost work due to the entropy increase of system + atmosphere. Because Eq. (4.73) expresses the available work in state functions one may consider it as a state function itself, called the exergy of the system. It is now possible to rewrite the Second Law efficiency (4.72) if one produces work W. Then the numerator in (4.72) will be W and the denominator will be the available work (4.81). This gives ε=

W W + T0 (S)s+a

(4.82)

Definition (4.77) of the available work may be compared with the change in Gibbs function G for isobaric, isothermal processes (p = T = 0). In this case Eq. (4.47) for a finite change becomes G = U + pV − T S (4.83)    since pdV = p dV = pV and T dS = T dS = T S and dU = U . It follows that B and G are equivalent if the change of the system occurs at atmospheric conditions: p = p0 and T = T 0 . The concept of Second Law efficiency is useful as an intellectual tool to compare different applications of the same kind of fuels. It helps in deciding how to make the best use of the available work, like in Example 4.7. 

4.2.5



Loss of Exergy in Combustion

In this section we will show that the loss of exergy in combustion is of the order of 30% ([8], 42–46). Consider a fuel/air mixture with atmospheric pressure p0 , atmospheric temperature T 0 and volume V 1 . After adiabatic combustion (process α) these quantities become p0 , Tc and Vc , where Tc is the high flame temperature. The combustion products cool down (process β) with intermediate temperatures T. During cooling the products deliver work between temperatures T and T 0 , for example by driving a piston in an internal combustion engine. From temperature T to temperature T + dT (with dT < 0) isobaric expansion work is performed while sensible heat δQ = −ncp dT > 0 is extracted from the hot ‘reservoir’ of

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combustion products at temperature T; some average specific heat cp is taken for n moles of the mixture. The atmosphere is acting as a cold reservoir and we assume that the work δW is delivered with the Carnot efficiency (4.62). Then   T0 (4.84) δW = δ Q 1 − T The total work performed until the combustion products have cooled down is T0 δW =

W =



Tc

Tc

nc p

T0 1− T

 dT

T0

(4.85)

= nc p (Tc − T0 ) − nc p T0 ln(Tc /T0 ) The second term on the right originates from the fact that the ‘hot reservoir’ does not keep a constant temperature, but cools down. The first term on the right reflects the enthalpy change for the isobaric process α Eq. (4.41). After processes α and β are done, one ends up with the same p0 ,T 0 as in the beginning except that chemical energy contained in the Gibbs function was liberated during combustion. In many cases |G| ≈ |H| and also, as remarked before, |H| ≈ |B|. Therefore the term ncp (Tc − T 0 ) in (4.85) will be roughly equal to the exergy before combustion. Following this interpretation of Eq. (4.85) the term –ncp T 0 ln(Tc /T 0 ) then would be roughly the lost exergy or lost work. If this is correct, it should also follow from the term T 0 (S)s+a in (4.81). Here (S)s+a is the entropy increase of system + atmosphere. For process β the Carnot efficiency (4.84) was taken, so it is assumed to have been reversible and its contribution to (S)s+a will vanish. Process α(combustion) was adiabatic, so the atmosphere remains unchanged and the major entropy increase (S)s+a originates from the temperature rise of only the system with δ Q  = nc p dT dS = δ Q  /T = nc p dT/T

(4.86) (4.87)

at each temperature T. The lost work T 0 (S)s+a becomes Tc T0 S = T0

δ Q  = nc p T0 ln(Tc /T0 ) T

(4.88)

T0

which indeed is lost from the work in Eq. (4.85). The relative amount of lost exergy or lost work becomes |B| nc p T0 ln(Tc /T0 ) T0 ln(Tc /T0 ) ≈ = B nc p (Tc − T0 ) (Tc − T0 )

(4.89)

With T 0 = 300 [K] and Tc = 2240 [K] one finds a loss of 30%, due to the irreversibility of the combustion process. In reality, the Carnot efficiency of Eq. (4.84) is too optimistic because of irreversible processes; in practice, the loss will be higher.

Heat Engines

4.3

103

Idealized Cycles

In Figure 4.8 four idealized ways were presented of converting heat into work and vice versa. In practice any engine will work in a cycle, where the initial and final states of the system are the same. In this section, four, still idealized, cycles are described where heat is converted into work and one cycle for refrigeration, where work is used to extract heat from a cold reservoir. In describing the cycles extensive use will be made of the thermodynamic functions p, V, S, T, H since that makes it easy to see what variables are kept constant during parts of the cycle. In the first two cycles to be discussed, the Carnot cycle and the Stirling cycle, heat is reversibly transported from a high temperature reservoir to a low temperature one. This should result in the Carnot efficiency (4.62), which indeed is the case. It must be stressed that in a heat engine the source of heat does not matter. The heat may come from solar heating or from waste heat in a nuclear power station, or it may use the flame of the combustion of oil, gas, wood or coal. These combustion processes may be conducted with due precautions against pollution. Finally, the heat may be transported over some distance by means of heat pipes (Section 4.1.4). 4.3.1

Carnot Cycle

The cycle is sketched in Figure 4.9; because all changes occur reversibly, all parts of the cycle have definite thermodynamic variables and the cycle is indicated by drawn lines. For simplicity we take heats QH and QC as positive in the considerations below. For heat leaving the engine a minus sign is added, if required. In part 1 → 2 of the cycle a gas is compressed adiabatically (δQ = 0). That is immediately clear from the diagram on the right, as the entropy is conserved with the value S1 = S2 . In the next step 2 → 3 the gas expands isothermally and picks up an amount of heat QH = T H (S3 – S2 ). This relation follows from Eqs. (4.37) or (4.55) because the temperature T H remains constant. In step 3 → 4 the gas expands adiabatically. From 4 → 1 it is compressed at the constant temperature T C ; it picks up a negative amount of heat T C (S1 – S4 ). It is T p

QH 2

TH

3

2

3

TC

1

4

1

4

QC S

V

S2

S3

Figure 4.9 Idealized Carnot cycle with on the left its pV diagram and on the right its ST diagram. From the latter one observes that in parts 1 → 2 and 3 → 4 of the cycle the entropy is conserved, hence those parts are adiabatic; in part 2 → 3 heat QH is taken from the high-temperature reservoir while in part 4 → 1 heat QC is released.

104

Environmental Physics

more convenient to describe this as the release of QC = T C (S4 − S1 ) = T C (S3 − S2 ). The initial and final states are the same (U = 0) so the work done follows from the First Law (4.32) as W = Q H − Q C = (TH − TC )(S3 − S2 )

(4.90)

The thermal efficiency (4.61) is found as QH − QC TH − TC TC W = = =1− (4.91) η= QH QH TH TH which is precisely the Carnot efficiency (4.62) as expected. The work W can also be expressed by means of the pV diagram on the left of Figure 4.9. For 2 → 3, for example, the work done by the system would be 3 W2→3 =

pdV

(4.92)

2

and for the complete cycle would be 2

3 pdV +

W = 1

4 pdV +

2

1



pdV + 3

pd V = 4

pdV

(4.93)

cycle

If one looks at the figure it is clear that the contribution from the steps 2 → 3 and 3 → 4 is positive, and together represent the total area below steps 2 → 3→ 4. For the other two steps the integration goes to the left and together they represent the total area underneath steps 4 → 1 → 2, but with a negative sign. The four integrals together are equal to the area within the diagram, indicated by the right-hand side of Eq. (4.93). Carnot cycles are used for demonstration only; they are not used in practice. The reason is that for an ideal gas the isotherm obeys Eq. (3.3) pV = constant and the adiabat follows the Poisson equation (3.12) pV κ = constant. Because κ = 1.4 the curves in Figure 4.9 are close together, so the work done during a cycle is relatively small, and friction and other losses count heavily. 4.3.2

Stirling Engine

Much research and development has been done on the Stirling engine, which is named after a Scottish theologian and inventor in the 1830s. The idealized cycle is shown in Figure 4.10 with on the left the pV diagram and on the right the corresponding ST diagram. T p

QH

3

3

TH 2

4

TC

1

V

4

QR Q R’ 2

S2

1

S1

QC S3

S4

S

Figure 4.10 Idealized cycle for a Stirling engine. During two of the steps the volume is kept constant, during the other two the temperature is kept constant.

Heat Engines

105

In step 1 → 2 a gas (air, helium or hydrogen) is isothermally compressed, releasing heat QC . In step 2 → 3 the gas is pressurized without changing volume, the temperature rises with absorption of heat QR . In step 3 → 4 the gas expands with constant temperature absorbing heat QH . Finally, in step 4 → 1 the pressure falls with constant volume and heat QR is released. by looking at the pV diagram on the In practice QR ≈ QR , which may be understood  left of Figure 4.10. In both cases the heat equals cV dT , the temperature change and range is about the same, while the specific heat cV will not be very different at the two volumes 3 4 V 1 and V 2 concerned. If QR ≈ QR , then also 2 T dS ≈ 1 T dS. If the shapes of the two curves 2 → 3 and 1 → 4 in the TS diagram are similar, it follows that S3 − S2 ≈ S4 − S1 or S4 − S3 ≈ S1 − S2 . An essential part of the Stirling engine is played by the so-called regenerator. In this device 98% of the released heat QR is stored and fed back as QR . Therefore the thermal efficiency (4.61) is not found from η = W/(QH + QR ), but from η = W/QH . This gives η=

Q H + Q R − Q C − Q R QH − QC W = = = QH QH QH

(4.94)

QC TC (S1 − S2 ) TC = 1− =1− =1− QH TH (S4 − S3 ) TH

which indeed is the Carnot efficiency, as we are essentially working between two reservoirs. 4.3.3

Steam Engine

A mixture of water and water vapour is used to perform the cycle of a steam engine. In Figure 4.11 a simplified version of the so-called Rankine cycle is given. On the right one finds the TS diagram, where underneath the dashed line an equilibrium mixture exists of liquid with vapour. To the left of that line one has the liquid phase, to the right the vapour phase, while above the dashed line the phase difference disappears. The numbers 1 to 4 on the left correspond to 1 to 4 on the right. At point 1 of the cycle one has liquid water, which is compressed by a pump, doing work Wp resulting in a higher temperature of the water (point 2). Between points 2 and 3 (in the boiler on the left) heat QH is added isobarically at pressure p2 , first increasing the 3

Turbine

4

W

QH

QH Condensor

Boiler

T QC

Pump

2

1

3

Liquid

Vapour

2

Liquid+ vapour

1

4

QC

S Wp

Figure 4.11 Idealized Rankine cycle for a steam engine. The points indicated by numbers 1, 2, 3, 4 on the left correspond with points in the TS diagram on the right. The dashed line in that diagram represents the phase diagram underneath which a liquid–vapour equilibrium mixture exists.

106

Environmental Physics

temperature until the liquid reaches the dashed line. Then the temperature remains constant while all liquid evaporates; when that is finished the temperature rises again until point 3. After point 3 the vapour is allowed to expand adiabatically in a turbine (or an oldfashioned cylinder with piston) performing work W until point 4 is reached. There at low pressure and temperature all vapour condenses, releasing heat QC until point 1 is reached. The cycle may be repeated with the same water or with fresh water. The cycle is best described by using the enthalpy function (4.40). We first discuss the two steps 1 → 2 and 3 → 4, where work is involved, and then the two other steps where heat is exchanged. In step 1 → 2 one has δQ = 0, but the enthalpy of the system will increase because of the work Wp > 0 done on it by the pump, which increases its enthalpy: Wp = H 2 − H 1 . This relation holds because in the discussion following Figure 4.7 it was concluded that the term pV in the definition of enthalpy H = U + pV corresponds to work. The work by the system will be denoted by W 1→2 = −Wp . Finally one uses the fact that putting pressure on water indeed increases the pressure, but hardly changes the volume. This gives 2 − W1→2 = W p = H2 − H1 =

2 d( pV ) =

1

( pdV + V d p) 1

2 =

V d p = V1 ( p2 − p1 )

(4.95)

1

In step 3 → 4 the system performs positive work W 3→4 itself, losing enthalpy: W3→4 = H3 − H4

(4.96)

The net work performed by the system becomes W = W1→2 + W3→4 = H1 − H2 + H3 − H4

(4.97)

For the heat exchange we turn to the other two steps. Step 2 → 3 is performed at constant pressure p2 and step 4 → 1 at a constant, but lower pressure p1 . For these isobaric processes it follows from Eq. (4.41) that the added heat equals the enthalpy increase. So Q H = H3 − H2 > 0 Q C = H4 − H1 > 0

(4.98) (4.99)

According to the first law (4.32) W = QH − QC = H 3 − H 2 – H 4 + H 1 , which agrees with (4.97). The maximal thermal efficiency (4.61) of the process becomes η=

H3 − H2 − H4 + H1 H4 − H1 QC W = =1− =1− QH H3 − H2 H3 − H2 QH

(4.100)

The absorbed heat QH is made as high as possible by boiling under high pressure p2 and high temperatures. The heats QC and QH are difficult to measure directly, therefore one uses tabulated functions of enthalpies instead.

Heat Engines

4.3.4

107

Internal Combustion

In an internal combustion engine air is mixed with a vaporized fuel that is ignited by a spark (such as for the Otto cycle) or is ignited by the temperature rise resulting from compression (the diesel cycle). The engines are not heat engines in the sense discussed before: there are no external heat reservoirs, nor is there a single gas changing its thermodynamic state during the cycle. In fact, the chemical energy of the fuel/air mixture is converted into work, exhaust gases are expelled, fresh air supplied and fuel injected. In practice, however, the air/fuel ratio in mass is around 15 for an Otto cycle and around 20 for a diesel engine, which implies that air dominates in the ‘cycle’. Also the nitrogen in the air remains essentially unchanged, which makes it acceptable to approximate the cycle of an internal combustion engine in a pV or TS diagram. The lower temperature in the diagram may be taken as the ambient air temperature and the higher temperature as the temperature of the fuel/air/exhaust mixture after ignition. 4.3.4.1

Otto Cycle

In Figure 4.12 the pV and TS diagrams of an idealized Otto cycle are sketched, together with a picture of the moving piston. At point 1 the mixture of n1 moles of air with gasoline vapour enters the cylinder. In step 1 → 2 the mixture, approximated as an ideal gas, experiences adiabatic compression. The temperature rises, there is no heat exchanged with Fuel/air

2

Ignition 2 3

1 2 3

1 4 Exhaust

Flywheel transmission T

3

3

QH

p 2

4 1

V2

V1

T2 T1 V

2

4

QC

1

S

Figure 4.12 The idealized Otto cycle of an internal combustion engine. On top a cylinder with piston and flywheel and numbers 1, 2, 3, 4, which correspond to the pV and TS diagrams below.

108

Environmental Physics

the surroundings (δQ = 0) and the entropy remains constant. This is represented by a vertical line in the TS diagram. For adiabatic compression of an ideal gas the Poisson relation pV κ = constant

(4.101)

holds, with κ = cp /cV , the ratio of the specific heat with constant pressure cp and the specific heat with constant cV volume (see Eq. (3.12)). From the ideal gas law pV = n1 RT one may deduce that p = n1 RT/V. Substitution in the Poisson relation gives TV κ –1 = constant, and it follows that T1 V1κ−1 = T2 V2κ−1

(4.102)

In step 2 → 3 the pressure rises with constant volume; heat QH is absorbed, simulating the explosion in the gasoline cylinder. Assuming a constant cV we find Q H = cV (T3 − T2 )

(4.103)

In step 3 → 4 the n1 moles of air expand adiabatically up to the initial volume V 4 = V 1 . Similar to Eq. (4.102) we have T3 V3κ−1 = T4 V4κ−1 = T4 V1κ−1

(4.104)

In step 4 → 1 heat QC is released to a cold reservoir, simulating the exhaust part of the cycle, while the exhaust gases are replaced by fresh air and fuel. Similar to Eq. (4.103) Q C = cV (T4 − T1 )

(4.105)

The thermal efficiency (4.61) of the cycle becomes η=

QH − QC QC T4 − T1 W = =1− =1− QH QH QH T3 − T2

(4.106)

Define a compression ratio r = V 1 /V 2 ; then from (4.104) one will find T 3 = T 4 κ–1 ; also T 2 = T 1 κ–1 . Substitution in Eq. (4.106) results in η =1−

1 r κ−1

(4.107)

It follows that a bigger compression ratio r will lead to higher engine efficiency. An upper limit for r is given by the fact that too high a compression will pre-ignite the engine or cause pinging or knocking. This effect can be prevented by ‘doping’ the fuel. Before environmental regulations outlawed this, lead was used for this purpose. In practice, the upper limit of the compression ratio for internal combustion engines is around r = 13. 4.3.4.2

Diesel Engine

In a diesel engine pre-ignition is prevented by making step 2 → 3 in the pV-diagram horizontal, as shown in Figure 4.13. Step 1 → 2 again is adiabatic compression, but this time of the air only. Eq. (4.102) again holds true. At point 2 the compression is so high that when the fuel is added at this point the mixture ignites. This happens isobarically, so in step 2 → 3 the piston moves out and the heat absorbed equals Q H = c p (T3 − T2 )

(4.108)

Heat Engines

109

T 2

T3

3

p

T4 T2 2 T1 1

4 1

V2 V3

V1

3

QH

V

4

QC S

Figure 4.13 Idealized diesel cycle. One essential difference from the Otto cycle is that step 2 → 3 is isobaric; the other difference is that fuel is injected at point 2, the end of the compression, preventing pre-ignition.

At point 3 the fuel addition is cut off and the expansion stops. The volume V 3 defines a cut-off ratio rc f =

V3 V2

(4.109)

Step 3 → 4 again is adiabatic expansion, for which Eq. (4.104) holds and in step 4 → 1 heat Q C = cV (T4 − T1 )

(4.110)

is released to the outside atmosphere. The thermal efficiency (4.61) now becomes η=

κ W QH − QC QC 1 T4 − T1 1 1 rc f − 1 = =1− =1− =1− QH QH QH κ T3 − T2 κ r κ−1 rc f − 1

(4.111)

The proof of the last equality is left to the reader (Exercise 4.19). Apart from the factor containing the cut-off ratio rcf , this expression is very similar to the efficiency (4.107) of the Otto cycle. The compression ratio r of the diesel engine is between 18 and 25, which makes it possible to be more efficient than the Otto spark ignition engine. The drawback with the diesel engine lies in the last factor of Eq. (4.111); it has the cut-off ratio both in the numerator and the denominator. For rcf > 1 the ratio (rcκf − 1)/(rc f − 1) slowly increases with rcf (Exercise 4.20). A small value of rcf would increase the efficiency, but reduce the work per cycle, as the surface area inside the curves on the left of graph (4.13) would become smaller, requiring more cycles per second and possibly larger losses. In practice, therefore, both parameters r and rcf are used to optimize the performance of the engine. The traditional diesel fuel consists of carbon compounds somewhat heavier than gasoline, which is mostly C8 H18 . Diesel fuel emits much more particulate matter than gasolinepowered vehicles. These particles are small and easily inhaled. They consist of a solid carbon core on which many toxic compounds may adsorb ([21], 41). In a diesel engine a variety of fuels may be used besides diesel, such as methanol (CH3 OH) or ethanol (C2 H5 OH) ([4], 94; [3], 138–143). In all cases the engine designer has to be aware of the strict environmental regulations.

110

Environmental Physics

Condensor

High-pressure vapour

Expansion valve (throttle)

Evaporator

Low-pressure vapour

Vapour compression: 1. Compressor Absorption: 1. Absorb vapour in liquid while removing heat 2. Elevate pressure of liquid with pump 3. Release vapour by applying heat

Figure 4.14 Two refrigeration systems, vapour compression and absorption. In both cases the fluid is evaporated at low temperature and pressure, picking up heat from the refrigerator, then compressed in different ways. At high temperature and pressure the vapour condenses, delivering heat to the surroundings after which it will expand via a throttle. (Reproduced from Principles of Energy Conservation, A Culp, table 9.1, pg 482, 1991, with permission from McGraw-Hill)

4.3.5

Refrigeration

In refrigeration work is used to transport heat from a reservoir of low temperature to a reservoir of high temperature (see Figure 4.8d). In practice a working fluid will perform a cycle between two reservoirs. The low-temperature reservoir should be somewhat lower than the refrigerator to be cooled, otherwise no heat would flow, −10 [◦ C] say, and the high temperature reservoir should be somewhat higher in temperature than room temperature for the same reason, +35 [◦ C], say. In Figure 4.14 the cycle is sketched. It uses the thermodynamic properties of a phase change to absorb or eject a big amount of heat with a relatively small mass and volume of working fluid. At low temperature and pressure the fluid is evaporated, taking heat away from the refrigerator. Next, the fluid is compressed in one of two ways to be discussed below, then the fluid is condensed at high temperature and pressure, giving off heat to the surroundings; finally, by way of a throttle, the fluid expands to the initial pressure again. 4.3.5.1

The Vapour-Compression Cycle

In the vapour-compression cycle, applied in households and small- or medium- sized shops, the vapour is compressed by an ordinary pump. In Figure 4.15 the cycle is sketched in a pH diagram. Just as in a conventional pV diagram, the dashed line indicates where one has pure liquid to the left of the left branch, or pure vapour right of the right branch. On the branch itself the liquid (left) or vapour (right) is saturated. In between, going from left to right the temperature remains constant and the fraction of vapour increases [10]. We follow Figure 4.15, starting at point 1. (a) At point 1 the working fluid is just saturated vapour. In step 1 → 2 it is compressed adiabatically by a pump. As there is no exchange of heat with the surroundings, the work done by the pump equals the increase of enthalpy W = H 2 – H 1 . Note that here we used the connection between enthalpy and work, discussed below Eq. (4.43).

Heat Engines

111

Critical point Saturated vapour

p

Saturated liquid

3

2

Throttle 1

4

H3=H4

H1

H2

H

Figure 4.15 pH diagram of a commercial refrigerator