radiative heat transfer 3rd edition modest solutions manual · 2020. 11. 15. · graduate course on...

RADIATIVE HEAT TRANSFER

Third Edition

Michael F. ModestUniversity of California at Merced

SOLUTION MANUAL

Academic PressNew York San Francisco London

Radiative Heat Transfer 3rd Edition Modest Solutions ManualFull Download: http://alibabadownload.com/product/radiative-heat-transfer-3rd-edition-modest-solutions-manual/

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PREFACE

This manual/web page contains the solutions to many (but not all) of the problems that are given at the endof each chapter, in particular for problems on topics that are commonly covered in a first (or, at least, second)graduate course on radiative heat transfer. Thus, solutions to problems of Chapters 1 through 6, 9 through11, 13, 14 and 18 are almost complete; for other chapters (7, 15, 16, 19) only around half of solutions aregiven, for problems on the more basic aspects covered in that chapter. Quite a few solutions, together withFortran90 codes, are also given for Chapter 20, not so much intended as homework solutions, but perhapsuseful for semester projects and/or to aid the researcher to write his/her first Monte Carlo code. At this dateno problem solutions are provided for the remaining chapters. However, it is the intent to periodically updatethis web-based solution manual, at which time additional solutions may be posted.

Most solutions have been checked, and some have been rechecked. Still, only a major miracle could haveprevented me from making an occasional error or overlooking an occasional typo. I would appreciate it verymuch if any such errors in the manual (or, for that matter, in the text) would be reported to me. In addition,for the benefit of other instructors who will receive future editions of this manual, any reasonably written upsolutions to presently unsolved problems would also be appreciated.

This manual is posted as a pdf-file with hyperreferences enabled, including a Table of Contents and aBookmark column listing all chapters and problems. This enables the reader to immediately jump back andforth within the document to locally referenced items (chapters, problems, figures, equations). References toequations, figures, etc., in the book itself can, of course, not be reached in this way. Users of Adobe Acrobatr caneasily assemble individual problem set solutions from this manual. To make this also possible for instructorswith access to only the shareware postscript/pdf viewer Ghostview c©, each individual problem is started at thetop of a new page.State College, PADecember 2002

Michael F. ModestDepartment of Mechanical Engineering

The Pennsylvania State UniversityUniversity Park, PA 16802

e-mail: [email protected]

i

CONTENTS

Chapter 1 2Problem 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2Problem 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Problem 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Problem 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Problem 1.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6Problem 1.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Problem 1.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Problem 1.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Problem 1.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Problem 1.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11Problem 1.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Problem 1.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Problem 1.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15Problem 1.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Problem 1.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Chapter 2 19Problem 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19Problem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Problem 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Problem 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Problem 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24Problem 2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Chapter 3 27Problem 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Problem 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Problem 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Problem 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Problem 3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31Problem 3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33Problem 3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34Problem 3.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Problem 3.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36Problem 3.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Problem 3.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Problem 3.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39Problem 3.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40Problem 3.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41Problem 3.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42Problem 3.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43Problem 3.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44Problem 3.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45Problem 3.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

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Problem 3.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47Problem 3.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48Problem 3.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49Problem 3.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50Problem 3.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52Problem 3.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53Problem 3.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54Problem 3.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55Problem 3.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56Problem 3.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57Problem 3.30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58Problem 3.31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59Problem 3.32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60Problem 3.33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61Problem 3.34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62Problem 3.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63Problem 3.36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Chapter 4 66Problem 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66Problem 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68Problem 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69Problem 4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70Problem 4.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72Problem 4.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73Problem 4.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74Problem 4.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75Problem 4.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76Problem 4.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77Problem 4.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78Problem 4.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79Problem 4.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80Problem 4.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81Problem 4.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82Problem 4.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83Problem 4.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84Problem 4.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85Problem 4.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87Problem 4.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88Problem 4.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89Problem 4.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90Problem 4.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92Problem 4.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93Problem 4.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94Problem 4.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

Chapter 5 100Problem 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100Problem 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101Problem 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103Problem 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105Problem 5.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106Problem 5.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107Problem 5.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108Problem 5.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

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Problem 5.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110Problem 5.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112Problem 5.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113Problem 5.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114Problem 5.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115Problem 5.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118Problem 5.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120Problem 5.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122Problem 5.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124Problem 5.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125Problem 5.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126Problem 5.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127Problem 5.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129Problem 5.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130Problem 5.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131Problem 5.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132Problem 5.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134Problem 5.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135Problem 5.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136Problem 5.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137Problem 5.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139Problem 5.30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141Problem 5.31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143Problem 5.32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146Problem 5.33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154Problem 5.34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155Problem 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159Problem 5.36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160Problem 5.37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161Problem 5.38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164Problem 5.39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165Problem 5.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

Chapter 6 168Problem 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168Problem 6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169Problem 6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170Problem 6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172Problem 6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174Problem 6.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176Problem 6.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178Problem 6.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179Problem 6.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180Problem 6.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182Problem 6.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183Problem 6.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184Problem 6.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185Problem 6.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187Problem 6.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188Problem 6.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189Problem 6.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191Problem 6.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192Problem 6.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193Problem 6.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195Problem 6.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

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Problem 6.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200Problem 6.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202Problem 6.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203Problem 6.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205Problem 6.30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208Problem 6.31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210Problem 6.34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212Problem 6.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

Chapter 7 214Problem 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214Problem 7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216Problem 7.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218Problem 7.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219Problem 7.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221Problem 7.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224Problem 7.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225Problem 7.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227Problem 7.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232Problem 7.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234Problem 7.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

Chapter 8 241Problem 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241Problem 8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242Problem 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243Problem 8.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245Problem 8.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246Problem 8.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247Problem 8.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248Problem 8.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249Problem 8.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251Problem 8.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254Problem 8.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255Problem 8.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260

Chapter 9 265Problem 9.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265

Chapter 10 267Problem 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267Problem 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269Problem 10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270Problem 10.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272Problem 10.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273Problem 10.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275Problem 10.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276Problem 10.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278Problem 10.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279Problem 10.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280Problem 10.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281Problem 10.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283Problem 10.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285Problem 10.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287Problem 10.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288

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Chapter 11 289Problem 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289Problem 11.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290Problem 11.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291Problem 11.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292Problem 11.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293Problem 11.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294Problem 11.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295Problem 11.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296Problem 11.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297Problem 11.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298Problem 11.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299Problem 11.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300Problem 11.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301Problem 11.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302Problem 11.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303Problem 11.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304Problem 11.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305Problem 11.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306Problem 11.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307Problem 11.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309Problem 11.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310Problem 11.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311Problem 11.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313Problem 11.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314Problem 11.25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316Problem 11.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317Problem 11.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318Problem 11.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319Problem 11.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320Problem 11.30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321Problem 11.31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323Problem 11.32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324Problem 11.33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325Problem 11.34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327Problem 11.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328Problem 11.37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329Problem 11.38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330Problem 11.39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331Problem 11.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332Problem 11.41 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334Problem 11.42 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

Chapter 12 339Problem 12.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339Problem 12.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340Problem 12.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341Problem 12.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342Problem 12.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343Problem 12.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344Problem 12.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345Problem 12.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346Problem 12.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347Problem 12.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348Problem 12.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349

CONTENTS vii

Problem 12.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350Problem 12.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351Problem 12.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352Problem 12.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353Problem 12.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354Problem 12.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355Problem 12.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356Problem 12.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358Problem 12.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359

Chapter 14 360Problem 14.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360Problem 14.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361Problem 14.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362Problem 14.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364Problem 14.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366Problem 14.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367Problem 14.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368Problem 14.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369Problem 14.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370Problem 14.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371Problem 14.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372Problem 14.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373Problem 14.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374Problem 14.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376

Chapter 15 377Problem 15.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377Problem 15.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378Problem 15.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379Problem 15.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380Problem 15.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382Problem 15.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383Problem 15.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384Problem 15.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386Problem 15.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387Problem 15.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388Problem 15.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390Problem 15.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392Problem 15.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394Problem 15.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396Problem 15.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397Problem 15.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399Problem 15.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401Problem 15.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403

Chapter 16 405Problem 16.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405Problem 16.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406Problem 16.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407Problem 16.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408Problem 16.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410Problem 16.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412Problem 16.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413Problem 16.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

CHAPTER 0 1

Problem 16.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417Problem 16.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419Problem 16.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420Problem 16.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421Problem 16.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423Problem 16.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424Problem 16.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426Problem 16.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428

Chapter 17 429Problem 17.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429Problem 17.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430Problem 17.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431Problem 17.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432Problem 17.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434Problem 17.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436Problem 17.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438Problem 17.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439Problem 17.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441Problem 17.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443

Chapter 19 445Problem 19.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445Problem 19.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447Problem 19.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448Problem 19.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449Problem 19.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450Problem 19.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451Problem 19.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452

Chapter 20 454Problem 20.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454Problem 20.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455Problem 20.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456Problem 20.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457Problem 20.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459Problem 20.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464Problem 20.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466Problem 20.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468Problem 20.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469Problem 20.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470Problem 20.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472Problem 20.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474

Chapter 21 477Problem 21.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477Problem 21.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479

CHAPTER 11.1 Solar energy impinging on the outer layer of earth’s atmosphere (usually called “solar constant”) has been

measured as 1367 W/m2. What is the solar constant on Mars? (Distance earth to sun = 1.496 × 1011 m, Marsto sun = 2.28 × 1011 m).

SolutionThe total energy emitted from the sun Q = 4πR2

S σT4sun goes equally into all directions, so that

qSC(R) = Q/4πR2 = σT4sun

(RS

R

)2

qSC mars

qSC earth

=( RS

RM

)2/ (RS

RE

)2

=( RE

RM

)2

=

(1.8×1011

2.28×1011

)2

= 0.4305

qSC mars = 0.4305 × 1367 = 588 W/m2

2

CHAPTER 1 3

1.2 Assuming earth to be a blackbody, what would be its average temperature if there was no internal heatingfrom the core of earth?

SolutionWithout internal heating an energy balance for earth gives

Qabsorbed = Qemitted

Qabs = qsol × Aproj = qsolπR2E

Qem = σT4E A = 4πR2

E σT4E

Thus

TE =(qsol

4σ

)1/4=

(1367 W/m2

4×5.670×10−8 W/m2K4

)1/4

TE = 279 K = 6C

4 RADIATIVE HEAT TRANSFER

1.3 Assuming earth to be a black sphere with a surface temperature of 300 K, what must earth’s internal heatgeneration be in order to maintain that temperature (neglect radiation from the stars, but not the sun) (radiusof the earth RE = 6.37 × 106 m).

SolutionPerforming an energy balance on earth:

Q = Qemitted −Qabsorbed = σT4E A − qsol Aproj

= 4πR2E σT4

E − qsolπR2E = πR2

E

(4σT4

E − qsol

)= π ×

(6.37×106m

)2[4×5.670×10−8 W

m2K4 × 3004 K4− 1367

Wm2

]Q = 6.00 × 1016 W

CHAPTER 1 5

1.4 To estimate the diameter of the sun, one may use solar radiation data. The solar energy impinging onto theearth’s atmosphere (called the “solar constant”) has been measured as 1367 W/m2. Assuming that the sunmay be approximated to have a black surface with an effective temperature of 5777 K, estimate the diameterof the sun (distance sun to earth SES ' 1.496 × 1011 m).

SolutionThe sun’s energy travels equally into all directions. Thus, on a spherical “shell” around the sun (such thatearth’s orbit lies on that shell): total energy leaving sun

Qsun = qsol × 4πS2ES = σT4

sun 4πR2S.

Thus

RS = SES

(qsol

σT4sun

)1/2

= 1.496 × 1011 m(

1367 W/m2

5.670×10−8×57774 W/m2

)1/2

RS = 6.96 × 108 m


1.5 Solar energy impinging on the outer layer of earth’s atmosphere (usually called “solar constant”) has beenmeasured as 1367 W/m2. Assuming the sun may be approximated as having a surface that behaves like ablackbody, estimate its effective surface temperature. (Distance sun to earth SES ' 1.496 × 1011 m, radius ofsun RS ' 6.96 × 108 m).

SolutionThe sun’s energy travels equally into all directions. Thus, on a spherical “shell” around the sun (such that theearth’s orbit lies on that shell): total energy leaving sun

Qsun = qsol × 4πS2ES = σT2

sun 4πR2S.

Thus

Tsun =(qsol

σ

)1/4 (SES

RS

)1/2

=

(1367 W/m2

5.670×10−8 W/m2K4

)1/4 (1.496× 1011

6.96×108

)1/2

Tsun = 5777 K

CHAPTER 1 7

1.6 A rocket in space may be approximated as a black cylinder of length L = 20 m and diameter D = 2 m. Itflies past the sun at a distance of 140 million km such that the cylinder axis is perpendicular to the sun’srays. Assuming that (i) the sun is a blackbody at 5777 K and (ii) the cylinder has a high conductivity (i.e.,is essentially isothermal), what is the temperature of the rocket? (Radius of sun RS = 696,000 km; neglectradiation from earth and the stars).

SolutionThe total energy emitted from the sun Q = 4πR2

S σT4sun goes equally into all directions, so that the flux arriving

at the rocket a distance SRS away from the sun is

qR = Q/4πS2RS = σT4

sun

( RS

SRS

)2

= flux/unit projected area of rocket

Energy emitted = ARσT4R =

(2πR2 + 2πRL

)σT4

R = 2πR(R + L) σT4R

= Energy absorbed = ApqR = 2RL σT4sun

( RS

SRS

)2

T4R =

2RL2πR(R + L)

T4sun

( RS

SRS

)2

=1π

201 + 20

(5777 K)4(

6.96×105

140×106

)2

TR = 302 K.


1.7 A black sphere of very high conductivity (i.e., isothermal) is orbiting earth. What is its temperature? (Considerthe sun but neglect radiation from the earth and the stars). What would be the temperature of the sphere ifit were coated with a material that behaves like a black body for wavelengths between 0.4µm and 3µm, butdoes not absorb and emit at other wavelengths?

SolutionMaking an energy balance on the sphere

(a)

Qabsorbed = Qemitted

Qabsorbed = qsolAproj = qsolπR2

Qemitted = σT4A = σT4 4πR2

Thus

T =(qsol

4σ

)1/4=

(1367 W/m2

4×5.670×10−8 W/m2K4

)1/4

T = 279 K = 6C

(b) If absorption and emission takes place only over the wavelength interval 0.4µm < λ < 3µm, both Qabsand Qem will be reduced. Since the sphere is spectrally black, we have

Qem = 4πR2∫ 3µm

0.4µmEbλ dλ = 4πR2 σT4 [

f (3µm×T) − f (0.4µm×T)]

Solar emission is black, i.e., qsol ∝ Ebλ(Tsun), or

qsol,λ

qsol=

Ebλ(Tsun)σT4

sun,

and

Qabs = πR2∫ 3µm

0.4µmqsol,λ dλ = πR2 qsol

σT4sun

∫ 3µm

0.4µmEbλ(Tsun) dλ

= πR2qsol[

f (3µm×Tsun) − f (0.4µm×Tsun)]

Thus,

4σT4 [f (3µm T) − f (0.4µm T)

]= qsol

[f (3µm Tsun) − f (0.4µm Tsun)

]= 1367 W/m2 [

f (17, 331) − f (2311)]

T4 [f (3µmT) − f (0.4µmT)

]=

13674×5.670×10−8 (0.9788−0.1222)

= 5.1630 × 109 K4

This nonlinear relation must be solved by iteration, leading to T ' 600 K (< 601 K)

6004× [0.03934 − 0] = 5.0985 × 109

T = 600 K

CHAPTER 1 9

1.8 A 100 Watt light bulb may be considered to be an isothermal black sphere at a certain temperature. If thelight flux (i.e., visible light, 0.4µm < λ < 0.7µm) impinging on the floor directly (2.5 m) below the bulb is42.6 mW/m2, what is the light bulb’s effective temperature? What is its efficiency?

SolutionThe total heat rate (100 W) leaving the light bulb will go at equal per-unit-area amounts through any (hy-pothetical) sphere around the bulb. Take a sphere which has the spot on the floor on its surface (radiush = 2.5 m)

qfloor,visible = qfv =[

f (λ2T) − f (λ1T)] Qbulb

4πh2

f (λ2T) − f (λ1T) = 4πh2qfv/Qbulb = 4π 2.52m2 42.6 × 10−3 Wm2

/100 W

= 0.03346

By trial and error:

T = 2500 K→ λ1T = 0.4µm × 2500 K = 1000µm Kλ2T = 0.7µm × 2500 K = 1750µm K

f (λ2T) − f (λ1T) = 0.03369 − 0.00032 = 0.03337

T = 2500 K; η = 3.34% is converted into visible light

The answers for f (λ1T) and f (λ2T) are the same whether Appendix C or program planck.exe are employed,since no interpolation is necessary.


1.9 When a metallic surface is irradiated with a highly concentrated laser beam, a plume of plasma (i.e., a gasconsisting of ions and free electrons) is formed above the surface that absorbs the laser’s energy, often blockingit from reaching the surface. Assume that a plasma of 1 cm diameter is located 1 cm above the surface, andthat the plasma behaves like a blackbody at 20,000 K. Based on these assumptions calculate the radiative heatflux and the total radiation pressure on the metal directly under the center of the plasma.

SolutionRadiation emitted from the plasma (≈ black disk) hits the spot below it generating an incoming radiative heatflux and a radiative pressure. From equation (1.36), for a black source

qin =

∫2π

Ib cosθ dΩ

1cm

1cm

plasma

θ

θ max= Ib

∫Ω plasma

cosθ dΩ

= 2π Ib

∫ θmax

0cosθ sinθ dθ

= Eb sin2 θmax = Eb ×15

=5.670 × 10−8

× 20, 0004 W/m2

5

qin = 1.814 × 109 W/m2 = 181.4 kW/cm2

This should be compared to the maximum flux at the center of an unobstructed laser beam, which may beseveral MW/cm2.The radiation pressure is found similarly, from equation (1.42) as

p =2πIb

c

∫ θmax

0cos2 θ sinθ dθ =

2Eb

3c

(1 − cos2 θmax

)=

2Eb

3c

[1 −

(45

)3/2]=

23

5.670×10−8 20, 0004 W/m2

3 × 108 m/s[1 − 0.7155] = 5.735

Wsm3 = 5.735

Nm2

p = 5.735 N/m2

CHAPTER 1 11

1.10 Solar energy incident on the surface of the earth may be broken into two parts: A direct component (travelingunimpeded through the atmosphere) and a sky component (reaching the surface after being scattered by theatmosphere). On a clear day the direct solar heat flux has been determined as qsun = 1000 W/m2 (per unit areanormal to the rays), while the intensity of the sky component has been found to be diffuse (i.e., the intensity ofthe sky radiation hitting the surface is the same for all directions) and Isky = 70 W/m2sr. Determine the totalsolar irradiation onto earth’s surface if the sun is located 60 above the horizon (i.e., 30 from the normal).

SolutionSince earth’s surface is tilted away from the sun, less energy per unit surface area hits earth than is carried bythe sunshine (per unit area normal to the rays), as seen from Fig. 1-8 (by a factor of cosθsun = cos 30). Thusfrom equation (1.36)

qin = qsun cosθsun +

∫2π

Isky cosθ dΩ

θ

earth

sun

60°

Isky

qsun

= qsun cosθsun +

∫ 2π

0

∫ π/2

0cosθ sinθ dθ dψ

= qsun cosθsun + πIsky

= 1000 × cos 30 + π × 70 = 1086 W/m2

qin = 1086 W/m2


1.11 A window (consisting of a vertical sheet of glass) is exposed to direct sunshine at a strength of 1000 W/m2.The window is pointing due south, while the sun is in the southwest, 30 above the horizon. Estimate theamount of solar energy that (i) penetrates into the building, (ii) is absorbed by the window, and (iii) is reflectedby the window. The window is made of (a) plain glass, (b) tinted glass, whose radiative properties may beapproximated by

ρλ = 0.08 for all wavelengths (both glasses),

τλ = 0.90 for 0.35µm < λ < 2.7µm

0 for all other wavelengths (plain glass)

τλ = 0.90 for 0.5µm < λ < 1.4µm

0 for all other wavelengths (tinted glass).

(c) By what fraction is the amount of visible light (0.4µm < λ < 0.7µm) reduced, if tinted rather that plainglass is used? How would you modify this statement in the light of Fig. 1-11?

SolutionIf we want to determine how much sunshine penetrates through the window (for the entire window area, orper unit window area), we need to determine the angle between a normal to the window (n) and a vectorpointing to the sun (s), or cos β = n·s. From geometric relations it follows that cos β = cos 30×cos 45 = 0.6124.Thus sunshine on the window, per unit window area, is qin = 612.4 W/m2. Of that the fraction 0.08 is reflected,i.e.,

qref = 0.08 × 612.4 = 49.0 W/m2 of window area.

(a) Plain glass: The transmitted sunshine may be calculated as

qtrans =

∫∞

0τλ qin,λ dλ = 0.90

∫ 2.7µm

0.35µmqin,λ dλ

= 0.90 qin[

f (2.7µm Tsun) − f (0.35µm Tsun)]

= 0.90 × 612.4 W/m2 [f (15, 598µm K) − f (2022µm K)

]= 0.90 × 612.4 × (0.9720 − 0.0702) = 497.0 W/m2

It follows thatqabs = qin − qref − qtrans = 612.4 − 49.0 − 497.4 = 66.0 W/m2

(b) Tinted glass: Similarly,

qtrans = 0.90 × qin[

f (1.4µm × 5777 K) − f (0.5µm × 5777 K)]

= 0.90 × 612.4 ×[

f (8088) − f (2889)]

W/m2

= 0.90 × 612.4 × (0.8597 − 0.2481) = 337.1 W/m2,

qtrans = 337.1 W/m2

i.e., heat gain is reduced by 160 W/m2 (or 32%).

(c) The tinted glass loses some visible light, since it is not transmissive for 0.4µm < λ < 0.5µm. The fractionalreduction is

fractional reduction =0.90 qin

[f (0.5µm Tsun) − f (0.4µm Tsun)

]0.90 qin

[f (0.7µm Tsun) − f (0.4µm Tsun)

]=

f (2889) − f (2311)f (4044) − f (2311)

=0.2481 − 0.12220.4888 − 0.1222

fractional reduction = 34.3%

Therefore, the tinted glass appears to cause an equivalent loss of visible light. However, Fig. 1-11 shows thatthe human eye is not very responsive to wavelengths below ' 0.5µm, so that the actual reduction of visiblelight is considerably less.

CHAPTER 1 13

1.12 On an overcast day the directional behavior of the intensity of solar radiation reaching the surface of theearth after being scattered by the atmosphere may be approximated as Isky(θ) = Isky(θ = 0) cosθ, where θis measured from the surface normal. For a day with Isky(0) = 100 W/m2sr determine the solar irradiationhitting a solar collector, if the collector is (a) horizontal, (b) tilted from the horizontal by 30. Neglect radiationfrom the earth’s surface hitting the collector (by emission or reflection).

Solution(a) For a horizontal collector thesolar irradiation is readily deter-mined from equation (1.36) as

qin =

∫ 2π

0

∫ π/2

0(I0 cosθ) cosθ sinθ dθ dψ

= 2π I0

∫ π/2

0cos2 θ sinθ dθ =

2π3

I0 =2π3× 100

j

i

k

Earth s surface

ψ

s

s

θ

n

=30 ο

qin = 209.4 W/m2

(b) To evaluate qin for a tilted collector, equation (1.36) becomes

qin =

∫n·s>0

I0(k · s)(n · s) dΩ,

where k · s = cosθ gives the variation of sky intensity with sky polar angle θ, while n · s = cosθ′ is the cosineof the angle between the surface normal and a unit vector pointing into the sky, s. This unit vector may beexpressed in terms of ı, , k (unit vectors for an earth-bound coordinate system), with

ı′ = cosαı − sinαk, ′ = , n = sinα ı + cosα k,

as

s = sinθ cosψ ı + sinθ sinψ + cosθ k= sinθ′ cosψ′ ı′ + sinθ′ sinψ′ ′ + cosθ′ n.

Substituting for ı′, ′ and n, s(θ′, ψ′) becomes

s = (sinθ′ cosψ cosα + cosθ′ sinα) ı + sinθ′ sinψ′

+ (cosθ′ cosα − sinθ′ cosψ′ sinα) k.

Equation (1.36) may be evaluated by evaluating k · s in collector coordinates (i.e., integrating over θ′ and ψ′),or by determining n · s in global coordinates (integrating over θ and ψ). Choosing the former, one obtains

qin =

∫k·s>0n·s>0

I0(cosθ′ cosα − sinθ′ cosψ′ sinα) cosθ′ sinθ′ dθ′ dψ′

The integration limits imply that only directions above the collector (0 < θ < π/2) and above the horizontal

k · s = cosθ = cosθ′ cosα − sinθ′ cosψ′ sinα > 0

are to be considered. Solving the last expression for cosψ′ yields

cosψ′ < cotα cotθ′,


and using symmetry

qin = 2I0

∫ π/2

0

∫ π

cos−1(cosα cosθ′)(cosθ′ cosα − sinθ′ cosψ′ sinα)

× cosθ′ sinθ′ dψ′ dθ′.

Any radiation from earth’s surface hits the collector with a polar angle of θ′ > 60, making a negligiblecontribution to qin. Thus, as a first approximation, the lower limit for ψ′ may be replaced by 0, such that

qin = 2Ib

∫ π/2

0

∫ π

0(cosθ′ cosα − sinθ′ cosψ′ sinα) cosθ′ sinθ′ dψ′ dθ′

=2π3

I0 cosα =2π3× 100 × cos 30 = 181.4 W/m2.

qin = 181.4 W/m2

This number is a little bit too low because the additionally considered radiation from the earth has a k · s =cosθ < 0.

CHAPTER 1 15

1.13 A 100 W light bulb is rated to have a total light output of 1750 lm. Assuming the light bulb to consist of asmall, black, radiating body (the light filament) enclosed in a glass envelope (with a transmittance τ1 = 0.9throughout the visible wavelengths), estimate the filament’s temperature. If the filament has an emittance ofε f = 0.7 (constant for all wavelengths and directions), how does it affect its temperature?

SolutionThe total heat rate leaving the light bulb by radiation (going equally into all directions) is

Qbulb = τQfil = τ(Afil εfil σT4fil).

Thus the radiative heat flux on a large, hypothetical sphere of radius R surrounding the bulb is

q(R) =Qbulb

Asphere=

Afil

Asphereτεfil πIb(Tfil) = I(R) dΩfil

where I(R) indicates that the flux at R is coming from a very small solid angle normal to the surface and is,therefore, simply a normal intensity multiplied by the small solid angle.

From equation (1.47) the luminous flux at R is

qlum(R) = LdΩfil = 286lmW

∫ 0.7µm

0.4µmIλ(R) dλ dΩfil

With gray (i.e., constant) values for εfil and τ (at least over the visible wavelengths), this makes Iλ(R) ∝ Ibλ(Tfil),and

qlum(R) = 286lmW

I(R) dΩfil[

f (0.7µm Tfil) − f (0.4µm Tfil)].

Qlum = qlum(R)Asphere

= 286lmW

Afilτεfil πIb(Tfil)[

f (0.7µm Tfil) − f (0.4µm Tfil)]

= 286lmWπτQfil

[f (0.7µm Tfil) − f (0.4µm Tfil)

],

where Qlum is the total luminous flux leaving the bulb. Thus

f (0.7 Tfil) − f (0.4 Tfil) =Qlum

286(lm/W)πτQfil=

1750 lm286(lm/W)π × 0.9 × 100 W

= 0.02164.

By trial and error, one finds from Appendix C, for Tfil ' 2320 K

f (0.7 × 2320) − f (0.4 × 2320) = 0.02165 − 0.00013 = 0.02152.

Tfil ' 2320 K

This expression does not include the filament emittance: ε f = 0.7 would leave Tfil unchanged (however, Afilwould have to increase by a factor of 1/0.7 = 1.43, in order to emit 100 W of power).


1.14 A pyrometer is a device with which the temperature of a surface may be determined remotely by measuringthe radiative energy falling onto a detector. Consider a black detector of 1 mm × 1 mm area, that is exposedto a 1 cm2 hole in a furnace located a distance of 1 m away. The inside of the furnace is at 1500 K andthe intensity escaping from the hole is essentially blackbody intensity at that temperature. (a) What is theradiative heat rate hitting the detector? (b) Assuming that the pyrometer has been calibrated for the situationin (a), what temperature would the pyrometer indicate if the nonabsorbing gas between furnace and detectorwere replaced by one with an (average) absorption coefficient of κ = 0.1 m−1?

Solution(a) A total radiative intensity of Ib(Tfurnace) leaves the hole, equally into all directions. From the definition ofintensity the total heat rate hitting the detector—assuming the detector to be directly opposite the hole—is

Qd = AholeIb(T f ) dΩhole−detector = AholeIb(T f )Adetector

S2hd

Qd =Ahole σT4

f Adetector

πS2hd

=1 cm2

× 5.670 × 10−12 W/cm2 K4× 15004 K4

× 1 mm2

π 1 m2

Qd = 9.137 × 10−6 W = 9.137µW

(b) The signal would now be attenuated by a factor of exp(−κShd), or

Q∗d = Qd exp(−κShd) = 9.137µW × exp(−0.1 m−1× 1 m)

= 8.2675µW

Since the d calibration is Qd ∝ T4f we have

Qd

T4f

=Q∗dT∗4f→ T∗4f =

Q∗dQd

T4f

T∗f =(8.2675

9.137

)1/4

1500 = 1463 K,

i.e., the measurement would be wrong by 37 K.

CHAPTER 1 17

1.15 Consider a pyrometer, which also has a detector area of 1mm × 1mm, which is black in the wavelength range1.0µm ≤ λ ≤ 1.2µm, and perfectly reflecting elsewhere. In front of the detector is a focussing lens ( f = 10 cm)of diameter D = 2 cm, and transmissivity of τl = 0.9 (around 1µm). In order to measure the temperatureinside a furnace, the pyrometer is focussed onto a hot black surface inside the furnace, a distance of 1 m awayfrom the lens.

(a) How large a spot on the furnace wall does the detector see? (Remember that geometric optics dictates

1f

=1u

+1v

; M =h (detector size)

H(spot size

) =vu,

where u = 1m is the distance from lens to furnace wall, and v is the distance from lens to detector.)(b) If the temperature of the furnace wall is 1200 K, how much energy is absorbed by the detector per unit

time?(c) It turns out the furnace wall is not really black, but has an emittance of ε = 0.7 (around 1µm). Assuming

there is no radiation reflected from the furnace surface reaching the detector, what is the true surfacetemperature for the pyrometer reading of case (b)?

(d) To measure higher temperatures pyrometers are outfitted with filters. If a τ f = 0.7 filter is placed in frontof the lens, what furnace temperature would provide the same pyrometer reading as case (b)?

Solution(a) From geometric optics

1v

=1f−

1u

=1

10cm−

1100cm

=9

100cm; v = 11.1cm

H = huv

= 1mm ×10011.1

= 9mm.

Thus, the spot seen by the pyrometer is 9mm × 9mm in size.

(b) Energy leaving spot, intercepted by detector (or, by conservation of energy, energy intercepted by detector,coming from spot) is

Qd = Ib12(T)AHΩHlτl = Id12AhΩhl,

where Ib12(T) =∫ λ2

λ1Ibλ(T)dλ is emitted intensity from the spot on the furnace wall, Id12 is intensity absorbed

by the detector, and ΩHl and Ωhl are the solid angles with which the lens is seen from H and h, respectively.Since H and h are small compared to f and D = 2R, we simply evaluate the solid angles from the center ofspot H and detector h:

ΩHl 'πR2

u2 ; Ωhl =πR2

v2 ,

and

AHΩhl = πR2(H

u

)2

= πR2(

hv

)2

= AhΩhl = π1cm2(.9

100

)2

= 2.545 × 10−4cm2

Therefore, Ib12(T)τl = Id12, i.e., intensity hitting detector is the same as emitted from furnace, except for theattenuation through the lens: total energy is concentrated on a smaller spot by increasing the solid angle.Finally,

Ib12 =σT4

π[ f (λ2T) − f (λ1T)] =

5.67 × 10−8× 12004

πWm2 [ f (1440)︸︷︷︸

0.00961

− f (1200)︸︷︷︸0.00213

]

= 280Wm2

' Ibλ(1.1µm, 1200k)∆λ = (Ebλ/T5)(T5/π)∆λ = 1.7228 × 10−12×

12005

π0.2

Wm2

= 273Wm2


and

Qd = 280Wm2 2.545 × 10−8m2

× 0.9 = 6.41µW

(c) The energy hitting detector remains the same and, therefore, so does the intensity emitted from the spot:

εIb12(Ta)(actual) = Ib12(Tp = 1200K)(perceived)

or, if we assume the blackbody intensity over the detector range can be approximated by the value at 1.1µm,

ε

eC2/λTa − 1'

1eC2/λTp − 1

,

leading to

Ta =C2

λ

/ln1 + ε[eC2/λTp − 1]

=14, 388µmK

1.1µm

/ln1 + 0.7[e14,388/1.1×1200

− 1]

orTa = 1241K .

At these wavelengths Wien’s law holds almost exactly, i.e., we may drop the two “1”s from the above equation,and

Ta 'C2

λ

/ln(εeC2/λTp ) =

C2/λC2λTp

+ ln ε=

Tp

1 +λTp

C2ln ε

which again leads to 1241k.

(d) Similar to (c) we get τ f Ib12(Ta) = Ib12(Tp) and, since τ f = ε(c), the answer is the same, i.e,

Ta = 1241K .

Thus, a much stronger filter is needed to really extend the pyrometer’s range.

CHAPTER 22.1 Show that for an electromagnetic wave traveling through a dielectric (m1 = n1), impinging on the interface

with another, optically less dense dielectric (n2 < n1), light of any polarization is totally reflected for incidenceangles larger than θc = sin−1(n2/n1).Hint: Use equations (2.105) with k2 = 0.

SolutionEquations (2.105) become for k2 = 0,

η0 n1 sinθ1 = w′t sinθ2,

w′2t − w′′2t = η20 n2

2,

w′t w′′t cosθ2 = 0.

The last of these relations dictates that either w′′t = 0 or cosθ2 = 0 (w′t = 0 is not possible since—from the firstrelation—this would imply η0 n1 sinθ1 = 0 which is known not to be true).

w′′t = 0: Substituting this into the second relation leads to w′t = η0 n2, and the first leads to n1 sinθ1 = n2 sinθ2.

Since n1 > n2 this is a legitimate solution only for sinθ1 ≤ (n2/n1), or θ1 ≤ θc = sin−1(n2/n1).cosθ2 = 0 (θ2 = π/2): Substituting the first relation into the second gives

w′′t = η0

√n2

1 sin2 θ1 − n22,

i.e., a legitimate nonzero solution for n21 sin2 θ1−n2

2 ≥ 0 orθ1 ≥ θc. Inspection of the reflection coefficients,equations (2.109), shows that

r‖ =in2

1 w′′t + n22 w′i cosθ1

in21 w′′t − n2

2 w′i cosθ1, r⊥ =

w′i cosθ1 + iw′′tw′i cosθ1 − iw′′t

Since, in both reflection coefficients, there are no sign changes within the real and imaginary parts, itfollows readily that

ρ‖ = r‖r∗‖ = ρ⊥ = r⊥r∗⊥ = 1.

19


2.2 Derive equations (2.109) using the same approach as in the development of equations (2.89) through (2.92).Hint: Remember that within the absorbing medium, w = w′ − iw′′ = w′s − iw′′n; this implies that E0 is not avector normal to s. It is best to assume E0 = E‖e‖ + E⊥e⊥ + Ess.

SolutionInside the absorbing medium wt = w′t − iw′′t = w′t st − iw′′t n, and the electric field vector does not lie in a planenormal to s. Thus, we assume a general three-dimensional representation, or

E0 = E‖e‖ + E⊥e⊥ + Ess.

Following the development for nonabsorbing media, equations (2.77) through (2.88), then leads to

νµH0 = w × E0 = (w′s − iw′′n) × (E‖e‖ + E⊥e⊥ + Ess).

This formulation is valid for the transmitted wave, but also for the incident wave (w′′i = 0) and reflected wave(w′′r = 0, w′r = −w′i ). The contribution from Es vanishes for incident and reflected wave. Using the samevector relations as given for the nonabsorbing media interface, one obtains

νµH0 = w′(E‖e⊥ − E⊥e‖) − iw′′ (E‖e⊥ cosθ − E⊥ t + Ese⊥ sinθ).

For the interface condition (with νµ the same everywhere)

νµH0 × n = w′(E‖ t + E⊥e⊥ cosθ) − iw′′ (E‖ cosθt + E⊥e⊥ + Es sinθt).

Thus, from equation (2.78)

w′i (Ei‖ t + Ei⊥e⊥ cosθ1) − w′i (Er‖ t + Er⊥e⊥ cosθ1)= w′t(Et‖ t + Et⊥e⊥ cosθ2) − iw′′t (Et‖ cosθ2 t + Et⊥e⊥ + Ets sinθ2 t)

or

w′i (Ei‖ − Er‖) = (w′t − iw′′t cosθ2) Et‖ − iw′′t sinθ2Ets (2.2-A)w′i (Ei⊥ − Er⊥) cosθ1 = (w′t cosθ2 − iw′′t ) Et⊥ (2.2-B)

Similarly, from equation (2.77),

E0 × n = (E‖e‖ + E⊥e⊥ + Ess) × n = −E‖e⊥ cosθ + E⊥ t − Ese⊥ sinθ

and

(Ei‖ + Er‖) cosθ1 = Et‖ cosθ2 + Ets sinθ2 (2.2-C)Ei⊥ + Er⊥ = Et⊥ (2.2-D)

These four equations have 5 unknowns (Er‖, Ers, Et‖, Et⊥, and Ets), and an additional condition is needed,e.g., equation (2.23) or equation (2.64). Choosing equation (2.23) we obtain, inside the absorbing medium,

w · E0 = 0 = (w′t s − iw′′t n) · (Et‖et‖ + Et⊥e⊥ + Etss)= w′tEts + iw′′t (Et‖ sinθ2 − Ets cosθ2). (2.2-E)

Eliminating Et⊥ from equations (2.2-B) and (2.2-D), with r⊥ = Er⊥/Ei⊥, gives

w′i (1 − r⊥) cosθ1 = (w′t cosθ2 − iw′′t )(1 + r⊥),

or

r⊥ =w′i cosθ1 − (w′t cosθ2 − iw′′t )

w′i cosθ1 + (w′t cosθ2 − iw′′t ),

which is identical to equation (2.109).

CHAPTER 2 21

Now, eliminating Ets from equations (2.2-A) and (2.2-C) [multiplying equation (2.2-C) by iw′′t and adding]:

w′i (Ei‖ − Er‖) + iw′′t cosθ1(Ei‖ + Er‖) = w′tEt‖. (2.2-F)

Eliminating Ets from equations (2.2-C) and (2.2-E) leads to

Ets =−iw′′t Et‖ sinθ2

w′t − iw′′t cosθ2

(Ei‖ + Er‖) cosθ1 = Et‖

cosθ2 −iw′′t sin2 θ2

w′t − iw′′t cosθ2

= Et‖w′t cosθ2 − iw′′tw′t − iw′′t cosθ2

.

Using this to eliminate Et‖ from equation (2.2-F), with r‖ = Er‖/Ei‖, gives

w′i (1 − r‖) + iw′′t cosθ1(1 + r‖) = w′t cosθ1(1 + r‖)w′t − iw′′t cosθ2

w′t cosθ2 − iw′′tw′i (w

′

t cosθ2 − iw′′t )(1 − r‖)=

[w′t(w

′

t − iw′′t cosθ2) − iw′′t (w′t cosθ2 − iw′′t )]

cosθ1(1 + r‖)= η2

0 m22 cosθ1(1 + r‖),

r‖ =w′i (w

′

t cosθ2 − iw′′t ) − η20 m2

2 cosθ1

w′i (w′

t cosθ2 − iw′′t ) + η20 m2

2 cosθ1,

which is the same as equation (2.109).

It is a simple matter to show that other conditions give the same result. For example, from equation (2.64)

n21(Ei‖ei‖ · n + Er‖er‖ · n) = m2

2(Et‖et‖ · n + Etss · n)or

n21(Ei‖ − Er‖) sinθ1 = m2

2(Et‖ sinθ2 − Ets cosθ2), etc.


2.3 Find the normal spectral reflectivity at the interface between two absorbing media. [Hint: Use an approachsimilar to the one that led to equations (2.89) and (2.90), keeping in mind that all wave vectors will be complex,but that the wave will be homogeneous in both media, i.e., all components of the wave vectors are colinearwith the surface normal].

SolutionEquations (2.19) and (2.20) remain valid for incident, reflected and transmitted waves, with w = w′ − iw′′ =(w′−iw′′) n for all three cases. From equation (2.31) w·w = (w′−iw′′)2n·n = η2

0m2 it follows that w′−iw′′ = ±η0 m.Thus

w′i − iw′′i = η0 m1,

w′r − iw′′r = −η0 m1 (reflected wave is moving in a direction of − n),w′t − iw′′t = η0 m2.

From equations (2.23) and (2.24), it follows that the electric and magnetic field vectors are normal to n, i.e.,tangential to the surface, say E0 = E0 t. Then, from equation (2.77)

(Ei + Er) t × n = Et t × n,or

Ei + Er = Et

From equation (2.25) νµH0 = w × E0 = (w′ − iw′′) E n × t, and from equation (2.78)

n1(Ei − Er) = m2 Et.

Substituting for Et and dividing by Ei, with r = Er/Ei:

m1(1 − r) = m2(1 + r)or

r =m1 −m2

m1 + m2

and

ρn = rr∗ =(m1 −m2)(m1 −m2)∗

(m1 + m2)(m1 + m2)∗=

∣∣∣∣∣ (n1 − n2) + i(k1 − k2)(n1 + n2) + i(k1 + k2)

∣∣∣∣∣2

ρn =(n1 − n2)2 + (k1 − k2)2

(n1 + n2)2 + (k1 + k2)2

CHAPTER 2 23

2.4 A circularly polarized wave in air is incident upon a smooth dielectric surface (n = 1.5) with a direction of45 off normal. What are the normalized Stokes’ parameters before and after the reflection, and what are thedegrees of polarization?

SolutionFrom the definition of Stokes’ parameters the incident wave has degrees of polarization

Qi

Ii=

Ui

Ii= 0,

Vi

Ii= ±1,

the sign giving the handedness of the circular polarization. With Er‖ = Ei‖r‖ and Er⊥ = Ei⊥r⊥, from equa-tions (2.50) through (2.53):

Ir = Ei‖E∗i‖r2‖

+ Ei⊥E∗i⊥r2⊥ = Ei‖E∗i‖(ρ‖ + ρ⊥) =

12

(ρ‖ + ρ⊥) Ii

Since Ei‖E∗i‖ = Ei⊥E∗i⊥ [from equation (2.51)] and ρ = r2.

Similarly,

Qr = Ei‖E∗i‖r2‖− Ei⊥E∗i⊥r2

⊥ = Ei‖E∗i‖(ρ‖ − ρ⊥)Ur = Ei‖E∗i⊥r‖r⊥ + Ei⊥E∗i‖r⊥r‖ = Uir‖r⊥ = 0Vr = i(Ei‖E∗i⊥ − Ei⊥E∗i‖) r‖r⊥ = Vir‖r⊥Qr

Ir=ρ‖ − ρ⊥ρ‖ + ρ⊥

,Vr

Ir=

2r‖r⊥ρ‖ + ρ⊥

Vi

Ii.

From Snell’s law

sinθ2 =sinθ1

n2; cosθ2 =

√1 −

sin2 θ1

n22

=

√1 −

0.51.52 =

√79,

and from equations (2.89) and (2.90)

r‖ =n1 cosθ2 − n2 cosθ1

n1 cosθ2 + n2 cosθ1=

√7/9 − 1.5

√1/2

√7/9 + 1.5

√1/2

= −0.0920, ρ‖ = 0.0085

r⊥ =n1 cosθ1 − n2 cosθ2

n1 cosθ1 + n2 cosθ2=

√1/2 − 1.5

√7/9

√1/2 + 1.5

√7/9

= −0.3033, ρ⊥ = 0.0920

Qr

Ir=

0.0085 − 0.09200.0085 + 0.0920

= −0.8315,Ur

Ir= 0

Vr

Ir= ±

2×0.0920×0.00850.0085 + 0.0920

= ±0.5556

Since the perpendicular polarization is much more strongly reflected, the resulting wave is no longer circularlypolarized, but to a large degree linearly polarized (in the perpendicular direction).


2.5 A circularly polarized wave in air traveling along the z-axis is incident upon a dielectric surface (n = 1.5).How must the dielectric-air interface be oriented so that the reflected wave is a linearly polarized wave in they-z-plane?

SolutionFrom equations (2.50) through (2.53) it follows thatQr/Ir = 1, Ur = Vr = 0 (i.e., linear polarization), ifeither Er‖ or Er⊥ vanish. From Fig. 2-9 it follows thatr⊥ , 0 and, therefore Er⊥ , 0 for all incidence di-rections, while r‖ = 0 for θ = θp (Brewster’s angle),or

θp = tan−1 n2

n1= tan−1 1.5 = 56.31.

The resulting wave is purely perpendicular-polarized, i.e., e⊥ must lie in the y− z plane, ore‖ must be in the x−z plane. Therefore, the surfacemay be expressed in terms of its surface normal asn = ı sinθp − k cosθp = (ı − 1.5k)

/√3.25.

Ei

Er

p

p

n

xz

CHAPTER 2 25

2.6 A polished platinum surface is coated with a 1µm thick layer of MgO.

(a) Determine the material’s reflectivity in the vicinity of λ = 2µm (for platinum at 2µm mPt = 5.29 − 6.71 i,for MgO mMgO = 1.65 − 0.0001 i).

(b) Estimate the thickness of MgO required to reduce the average reflectivity in the vicinity of 2µm to 0.4.What happens to the interference effects for this case?

Solution(a) The desired overall reflectivity must be calculated from equation (2.124) after determining the relevantreflection coefficient. From equation (2.122)

r12 =1 −m2

1 + m2'

1 − n2

1 + n2=

1 − 1.651 + 1.65

= −0.2453

since k2 1, and r12 = 0.2453. r23 may also be calculated from equation (2.122) or, more conveniently, fromequation (2.126):

r223 =

(1.65 − 5.29)2 + 6.712

(1.65 + 5.29)2 + 6.712 = 0.6253 or r23 = 0.7908.

Since the real part of r12 < 0 it follows that δ12 = π, while

tan δ23 =2(1.65×6.71 − 5.29×10−4)

1.652 + 10−8 − (5.292+6.712)= −0.3150.

Since the =(r23) > 0 (numerator) and<(r23) < 0 (denominator) δ23 lies in the second quadrant, π/2 < δ23 < π,or δ23 = 2.8364. Also ζ12 = 4π × 1.65 × 1µm/2µm = 10.3673, and

cos [δ12 ± (δ23 − ζ12)] = cos [π ± (2.8364 − 10.3673)] = −0.3175.

Also κ2d = 4π × 10−4× 1µm/2µm = 2π × 10−4 and τ = e−κ2d = 0.9994 ' 1. Thus

R =0.24532 + 2×0.2453×0.7908×(−0.3175) + 0.79082

1 + 2×0.2453×0.7908×(−0.3175) + 0.24532×0.79082

R = 0.6149.

(b) The cos in the numerator fluctuates between −1 < cos < +1. The average value for R is obtained bydropping the cos-term. Then

Rav =r2

12 + r223τ

2

1 + r212r2

23τ2,

or

τ2 =Rav − r2

12

r223(1 − r2

12)=

0.4 − 0.24532

0.79082(1 − 0.24532)= 0.5782,

d = −1κ2

ln τ = −1

2κ2ln τ2 =

− ln 0.57824π × 10−4 µm−1 = 43.6µm.

More accurate is the averaged expression, equation (2.129)

Rav = ρ12 +ρ23(1 − ρ12)2τ2

1 − ρ12ρ23τ2

or

τ2 =Rav − ρ12

ρ23[(Rav − ρ12)ρ12 + (1 − ρ12)2] =

Rav − ρ12

ρ23[1 − (2 − Rav)ρ12

]=

0.4 − 0.24532

0.79082[1 − 1.6 × 0.24532]= 0.6013


and

d =− ln 0.6013

4π × 10−4 µm−1 = 40.47µm

For such a large d, it follows that ζ2 ' 40×10.3673 ' 450. A full interference period is traversed if ζ2 ' 450±π.Around λ = 2µm this implies a full period is traversed between 2µm ± 0.014µm. Such interference effectswill rarely be observed because (i) the detector will not respond to such small wavelength changes, and (ii)the slightest inaccuracies in layer thickness will eliminate the interference effects.

Note: since incoming radiation at λ0 = 2µm has a wavelength of λ = λ0/n1 = 2/1.65 = 1.2. µm, mPt shouldreally be evaluated at 1.21µm.

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