Solution Manual for Radiative Heat Transfer – Michael Modest.pdf

Third Edition

RADIATIVE HEAT TRANSFER

Michael F. Modest University of California at Merced

Access full solution manual http://www.book4me.xyz/solution-manual-radiative-heat-transfer-modest/

SOLUTION MANUAL

Academic Press New York San Francisco London

PREFACE

This manual/web page contains the solutions to many (but not all) of the problems that are given at the end of each chapter, in particular for problems on topics that are commonly covered in a first (or, at least, second) graduate course on radiative heat transf er. Thus, solutions to problem s of Chapters 1 through 6, 9 through 11, 13, 14 and 18 are almost complete; for other chapters (7, 15, 16, 19) only around half of solutions are given, for problems on the more basic aspects covered in that chapter. Quite a few solutions, together with Fortran90 codes, are also given for Chapter 20, not so much intended as homework solutions, but perhaps useful for semester projects and /or to aid the researcher to write his/her first Monte Carlo code. At this date no problem solutions are provided for the remaining chapters. However, it is the intent to periodically update this 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 have prevented me from making an occasional error or overlooking an occasional typo. I would appreciate it very much 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 up solutions 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 a Bookmark column listing all chapters and problems. This enables the reader to immediately jump back and forth within the document to locally referenced items (chapters, problems, figures, equations). References to equations, figures, etc., in the book itself can, of course, not be reached in this way. Users of Adobe Acrobat can easily assemble individual problem set solutions from this manual. To make this also possible for instructors with access to only the shareware postscript /pdf viewer Ghostviewc , each individual problem is started at the top of a new page. State College, PA Michael F. Modest December 2002 Department of Mechanical Engineering The Pennsylvania State University University Park, PA 16802 e-mail: [email protected]

i

CONTENTS

Chapter 1

2

Problem 1.1 . . . . . . . . . . . . . . . . . . . . Problem 1.2 . . . . . . . . . . . . . . . . . . . . Problem 1.3 . . . . . . . . . . . . . . . . . . . . Problem 1.4 . . . . . . . . . . . . . . . . . . . . Problem 1.5 . . . . . . . . . . . . . . . . . . . . Problem 1.6 . . . . . . . . . . . . . . . . . . . . Problem 1.7 . . . . . . . . . . . . . . . . . . . . Problem 1.8 . . . . . . . . . . . . . . . . . . . . Problem 1.9 . . . . . . . . . . . .......... Problem 1.10 . . . . . . . . . . .......... Problem 1.11 . . . . . . . . . . .......... Problem 1.12 . . . . . . . . . . .......... Problem 1.13 . . . . . . . . . . .......... Problem 1.14 . . . . . . . . . . .......... Problem 1.15 . . . . . . . . . . ..........

.... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ..........

.... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ........... ........... ........... ........... ........... ........... ...........

2 3 4 5 6 7 8 9 10 11 12 13 15 16 17

Chapter 2 Problem 2.1 Problem 2.2 Problem 2.3 Problem 2.4 Problem 2.5 Problem 2.6

19 ........... ........... ........... ........... ........... ...........

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19 20 22 23 24 25

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.......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ..........

.......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ..........

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27 28 29 30 31 33 34 35 36 37 38 39 40 41 42 43 44 45 46

Chapter 3 Problem 3.1 . . . . . . . . . . . Problem 3.2 . . . . . . . . . . . Problem 3.3 . . . . . . . . . . . Problem 3.4 . . . . . . . . . . . Problem 3.5 . . . . . . . . . . . Problem 3.6 . . . . . . . . . . . Problem 3.7 . . . . . . . . . . . Problem 3.8 . . . . . . . . . . . Problem 3.9 . . . . . . . . . . . Problem 3.10 . . . . . . . . . . Problem 3.11 . . . . . . . . . . Problem 3.12 . . . . . . . . . . Problem 3.13 . . . . . . . . . . Problem 3.14 . . . . . . . . . . Problem 3.15 . . . . . . . . . . Problem 3.16 . . . . . . . . . . Problem 3.17 . . . . . . . . . . Problem 3.18 . . . . . . . . . . Problem 3.19 . . . . . . . . . .

27

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CONTENTS

Problem 3.20 Problem 3.21 Problem 3.22 Problem 3.23 Problem 3.24 Problem 3.25 Problem 3.26 Problem 3.27 Problem 3.28 Problem 3.29 Problem 3.30 Problem 3.31 Problem 3.32 Problem 3.33 Problem 3.34 Problem 3.35 Problem 3.36

.......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ..........

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.......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ..........

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47 48 49 50 52 53 54 55 56 57 58 59 60 61 62 63 65

Problem 4.1 . . . . . . . . . . . Problem 4.2 . . . . . . . . . . . Problem 4.3 . . . . . . . . . . . Problem 4.4 . . . . . . . . . . . Problem 4.5 . . . . . . . . . . . Problem 4.6 . . . . . . . . . . . Problem 4.7 . . . . . . . . . . . Problem 4.8 . . . . . . . . . . . Problem 4.9 . . . . . . . . . . . Problem 4.10 . . . . . . . . . . Problem 4.11 . . . . . . . . . . Problem 4.12 . . . . . . . . . . Problem 4.13 . . . . . . . . . . Problem 4.14 . . . . . . . . . . Problem 4.15 . . . . . . . . . . Problem 4.16 . . . . . . . . . . Problem 4.17 . . . . . . . . . . Problem 4.18 . . . . . . . . . . Problem 4.19 . . . . . . . . . . Problem 4.20 . . . . . . . . . . Problem 4.21 . . . . . . . . . . Problem 4.22 . . . . . . . . . . Problem 4.23 . . . . . . . . . . Problem 4.24 . . . . . . . . . . Problem 4.25 . . . . . . . . . . Problem 4.26 . . . . . . . . . .

.......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ..........

.......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ..........

.......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ..........

........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ...........

66 68 69 70 72 73 74 75 76 77 78 79 80 81 82 83 84 85 87 88 89 90 92 93 94 96

Chapter 4

66

Chapter 5 Problem 5.1 Problem 5.2 Problem 5.3 Problem 5.4 Problem 5.5 Problem 5.6 Problem 5.7 Problem 5.8

100 ............ ............ ............ ............ ............ ............ ............ ............

............... ............... ............... ............... ............... ............... ............... ...............

............... ............... ............... ............... ............... ............... ............... ...............

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100 101 103 105 106 107 108 109

iii

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CONTENTS

Problem 5.9 . . . . . . . . . . . . ............... Problem 5.10 . . . . . . . . . . . . . . .............. Problem 5.11 . . . . . . . . . . . . . . .............. Problem 5.12 . . . . . . . . . . . . . . .............. Problem 5.13 . . . . . . . . . . . . . . .............. Problem 5.14 . . . . . . . . . . . . . . .............. Problem 5.15 . . . . . . . . . . . . . . .............. Problem 5.16 . . . . . . . . . . . . . . .............. Problem 5.17 . . . . . . . . . . . . . . .............. Problem 5.18 . . . . . . . . . . . . . . .............. Problem 5.19 . . . . . . . . . . . . . . .............. Problem 5.20 . . . . . . . . . . . . . . .............. Problem 5.21 . . . . . . . . . . . . . . .............. Problem 5.22 . . . . . . . . . . . . . . .............. Problem 5.23 . . . . . . . . . . . . . . .............. Problem 5.24 . . . . . . . . . . . . . . .............. Problem 5.25 . . . . . . . . . . . . . . .............. Problem 5.26 . . . . . . . . . . . . . . .............. Problem 5.27 . . . . . . . . . . . . . . .............. Problem 5.28 . . . . . . . . . . . . . . .............. Problem 5.29 . . . . . . . . . . . . . . .............. Problem 5.30 . . . . . . . . . . . . . . .............. Problem 5.31 . . . . . . . . . . . . . . .............. Problem 5.32 . . . . . . . . . . . . . . .............. Problem 5.33 . . . . . . . . . . . . . . .............. Problem 5.34 . . . . . . . . . . . . . . .............. Problem 5.35 . . . . . . . . . . . . . . .............. Problem 5.36 . . . . . . . . . . . . . . .............. Problem 5.37 . . . . . . . . . . . . . . .............. Problem 5.38 . . . . . . . . . . . . . . .............. Problem 5.39 . . . . . . . . . . . . . . .............. Problem 5.40 . . . . . . . . . . . . . . ..............

............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ...............

.......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........

110 112 113 114 115 118 120 122 124 125 126 127 129 130 131 132 134 135 136 137 139 141 143 146 154 155 159 160 161 164 165 166

............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ...............

.......... .......... .......... .......... .......... .......... .......... .......... .......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........

168 169 170 172 174 176 178 179 180 182 183 184 185 187 188 189 191 192 193 195 198

Chapter 6 Problem 6.1 . . . . . . . . . . . . Problem 6.2 . . . . . . . . . . . . Problem 6.3 . . . . . . . . . . . . Problem 6.4 . . . . . . . . . . . . Problem 6.5 . . . . . . . . . . . . Problem 6.6 . . . . . . . . . . . . Problem 6.7 . . . . . . . . . . . . Problem 6.8 . . . . . . . . . . . . Problem 6.9 . . . . . . . . . . . . Problem 6.10 . . . . . . . . . . . . . . Problem 6.11 . . . . . . . . . . . . . . Problem 6.12 . . . . . . . . . . . . . . Problem 6.13 . . . . . . . . . . . . . . Problem 6.14 . . . . . . . . . . . . . . Problem 6.15 . . . . . . . . . . . . . . Problem 6.16 . . . . . . . . . . . . . . Problem 6.17 . . . . . . . . . . . . . . Problem 6.18 . . . . . . . . . . . . . . Problem 6.19 . . . . . . . . . . . . . . Problem 6.21 . . . . . . . . . . . . . . Problem 6.22 . . . . . . . . . . . . . .

168 ............... ............... ............... ............... ............... ............... ............... ............... ............... .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. ..............

CONTENTS

Problem 6.24 Problem 6.26 Problem 6.27 Problem 6.29 Problem 6.30 Problem 6.31 Problem 6.34 Problem 6.35

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200 202 203 205 208 210 212 213

Problem 7.1 . . . . . . . . . . . . Problem 7.2 . . . . . . . . . . . . Problem 7.5 . . . . . . . . . . . . Problem 7.6 . . . . . . . . . . . . Problem 7.8 . . . . . . . . . . . . Problem 7.10 . . . . . . . . . . . . . . Problem 7.11 . . . . . . . . . . . . . . Problem 7.13 . . . . . . . . . . . . . . Problem 7.14 . . . . . . . . . . . . . . Problem 7.15 . . . . . . . . . . . . . . Problem 7.16 . . . . . . . . . . . . . .

............... ............... ............... ............... ............... .............. .............. .............. .............. .............. ..............

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214 216 218 219 221 224 225 227 232 234 237

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241 242 243 245 246 247 248 249 251 254 255 260

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265

Chapter 7

214

Chapter 8 Problem 8.1 . . . . . . . . . . . . Problem 8.2 . . . . . . . . . . . . Problem 8.3 . . . . . . . . . . . . Problem 8.4 . . . . . . . . . . . . Problem 8.5 . . . . . . . . . . . . Problem 8.6 . . . . . . . . . . . . Problem 8.7 . . . . . . . . . . . . Problem 8.8 . . . . . . . . . . . . Problem 8.9 . . . . . . . . . . . . Problem 8.10 . . . . . . . . . . . . . . Problem 8.14 . . . . . . . . . . . . . . Problem 8.15 . . . . . . . . . . . . . .

241

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

265

Chapter 10

267

Problem 10.1 . . . . . . . . . . . . . . Problem 10.2 . . . . . . . . . . . . . . Problem 10.3 . . . . . . . . . . . . . . Problem 10.4 . . . . . . . . . . . . . . Problem 10.5 . . . . . . . . . . . . . . Problem 10.6 . . . . . . . . . . . . . . Problem 10.7 . . . . . . . . . . . . . . Problem 10.8 . . . . . . . . . . . . . . Problem 10.9 . . . . . . . . . . . . . . Problem 10.10 . . . . . . . . . . . . . . Problem 10.11 . . . . . . . . . . . . . .

.............. .............. .............. .............. .............. .............. .............. .............. .............. .............. ..............

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........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........

267 269 270 272 273 275 276 278 279 280 281

Problem 10.12 . . . . . . . . . . . . . . Problem 10.13 . . . . . . . . . . . . . . Problem 10.14 . . . . . . . . . . . . . . Problem 10.15 . . . . . . . . . . . . . .

.............. .............. .............. ..............

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283 285 287 288

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CONTENTS

Chapter 11 Problem 11.1 . . . . . . . . . . . . . . Problem 11.2 . . . . . . . . . . . . . . Problem 11.3 . . . . . . . . . . . . . . Problem 11.4 . . . . . . . . . . . . . . Problem 11.5 . . . . . . . . . . . . . . Problem 11.6 . . . . . . . . . . . . . . Problem 11.7 . . . . . . . . . . . . . . Problem 11.8 . . . . . . . . . . . . . . Problem 11.9 . . . . . . . . . . . . . . Problem 11.10 . . . . . . . . . . . . . . Problem 11.11 . . . . . . . . . . . . . . Problem 11.12 . . . . . . . . . . . . . . Problem 11.13 . . . . . . . . . . . . . . Problem 11.14 . . . . . . . . . . . . . . Problem 11.15 . . . . . . . . . . . . . . Problem 11.16 . . . . . . . . . . . . . . Problem 11.17 . . . . . . . . . . . . . . Problem 11.18 . . . . . . . . . . . . . . Problem 11.19 . . . . . . . . . . . . . . Problem 11.20 . . . . . . . . . . . . . . Problem 11.21 . . . . . . . . . . . . . . Problem 11.22 . . . . . . . . . . . . . . Problem 11.23 . . . . . . . . . . . . . . Problem 11.24 . . . . . . . . . . . . . . Problem 11.25 . . . . . . . . . . . . . . Problem 11.26 . . . . . . . . . . . . . . Problem 11.27 . . . . . . . . . . . . . . Problem 11.28 . . . . . . . . . . . . . . Problem 11.29 . . . . . . . . . . . . . . Problem 11.30 . . . . . . . . . . . . . . Problem 11.31 . . . . . . . . . . . . . . Problem 11.32 . . . . . . . . . . . . . . Problem 11.33 . . . . . . . . . . . . . . Problem 11.34 . . . . . . . . . . . . . . Problem 11.35 . . . . . . . . . . . . . . Problem 11.37 . . . . . . . . . . . . . . Problem 11.38 . . . . . . . . . . . . . . Problem 11.39 . . . . . . . . . . . . . . Problem 11.40 . . . . . . . . . . . . . . Problem 11.41 . . . . . . . . . . . . . . Problem 11.42 . . . . . . . . . . . . . .

289 .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. .............. ..............

............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ...............

........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........

289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 309 310 311 313 314 316 317 318 319 320 321 323 324 325 327 328 329 330 331 332 334 337

.............. .............. .............. .............. .............. .............. .............. .............. .............. .............. ..............

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339 340 341 342 343 344 345 346 347 348 349

Chapter 12 Problem 12.1 . . . . . . . . . . . . . . Problem 12.2 . . . . . . . . . . . . . . Problem 12.3 . . . . . . . . . . . . . . Problem 12.4 . . . . . . . . . . . . . . Problem 12.5 . . . . . . . . . . . . . . Problem 12.6 . . . . . . . . . . . . . . Problem 12.7 . . . . . . . . . . . . . . Problem 12.8 . . . . . . . . . . . . . . Problem 12.9 . . . . . . . . . . . . . . Problem 12.10 . . . . . . . . . . . . . . Problem 12.11 . . . . . . . . . . . . . .

339

CONTENTS

Problem 12.13 . . . . . . . . . . . . . . Problem 12.14 . . . . . . . . . . . . . . Problem 12.15 . . . . . . . . . . . . . . Problem 12.16 . . . . . . . . . . . . . . Problem 12.17 . . . . . . . . . . . . . . Problem 12.18 . . . . . . . . . . . . . . Problem 12.19 . . . . . . . . . . . . . . Problem 12.20 . . . . . . . . . . . . . . Problem 12.21 . . . . . . . . . . . . . .

.............. .............. .............. .............. .............. .............. .............. .............. ..............

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350 351 352 353 354 355 356 358 359

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360 361 362 364 366 367 368 369 370 371 372 373 374 376

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377 378 379 380 382 383 384 386 387 388 390 392 394 396 397 399 401 403

Chapter 14

360

Problem 14.1 . . . . . . . . . . . . . . Problem 14.2 . . . . . . . . . . . . . . Problem 14.3 . . . . . . . . . . . . . . Problem 14.4 . . . . . . . . . . . . . . Problem 14.5 . . . . . . . . . . . . . . Problem 14.6 . . . . . . . . . . . . . . Problem 14.7 . . . . . . . . . . . . . . Problem 14.8 . . . . . . . . . . . . . . Problem 14.9 . . . . . . . . . . . . . . Problem 14.10 . . . . . . . . . . . . . . Problem 14.11 . . . . . . . . . . . . . . Problem 14.12 . . . . . . . . . . . . . . Problem 14.13 . . . . . . . . . . . . . . Problem 14.14 . . . . . . . . . . . . . .

Chapter 15

377

Problem 15.1 . . . . . . . . . . . . . . Problem 15.2 . . . . . . . . . . . . . . Problem 15.3 . . . . . . . . . . . . . . Problem 15.4 . . . . . . . . . . . . . . Problem 15.5 . . . . . . . . . . . . . . Problem 15.6 . . . . . . . . . . . . . . Problem 15.7 . . . . . . . . . . . . . . Problem 15.8 . . . . . . . . . . . . . . Problem 15.9 . . . . . . . . . . . . . . Problem 15.10 . . . . . . . . . . . . . . Problem 15.11 . . . . . . . . . . . . . . Problem 15.12 . . . . . . . . . . . . . . Problem 15.13 . . . . . . . . . . . . . . Problem 15.14 . . . . . . . . . . . . . . Problem 15.15 . . . . . . . . . . . . . . Problem 15.16 . . . . . . . . . . . . . . Problem 15.18 . . . . . . . . . . . . . . Problem 15.19 . . . . . . . . . . . . . .

Chapter 16

405

Problem 16.1 Problem 16.2 Problem 16.3 Problem 16.4

.............. .............. .............. ..............

.............. .............. .............. ..............

............... ............... ............... ...............

........ ........ ........ ........

405 406 407 408

Problem 16.5 Problem 16.6 Problem 16.7 Problem 16.8

.............. .............. .............. ..............

.............. .............. .............. ..............

............... ............... ............... ...............

........ ........ ........ ........

410 412 413 415

vii

CHAPTER 0

Problem 16.9 . . . . . . . . . . . . . . Problem 16.10 . . . . . . . . . . . . . . Problem 16.11 . . . . . . . . . . . . . . Problem 16.12 . . . . . . . . . . . . . . Problem 16.13 . . . . . . . . . . . . . . Problem 16.14 . . . . . . . . . . . . . . Problem 16.15 . . . . . . . . . . . . . . Problem 16.18 . . . . . . . . . . . . . .

.............. .............. .............. .............. .............. .............. .............. ..............

............... ............... ............... ............... ............... ............... ............... ...............

........ ........ ........ ........ ........ ........ ........ ........

417 419 420 421 423 424 426 428

.............. .............. .............. .............. .............. .............. .............. .............. .............. ..............

............... ............... ............... ............... ............... ............... ............... ............... ............... ...............

........ ........ ........ ........ ........ ........ ........ ........ ........ ........

429 430 431 432 434 436 438 439 441 443

.............. .............. .............. .............. .............. .............. ..............

.............. .............. .............. .............. .............. .............. ..............

............... ............... ............... ............... ............... ............... ...............

........ ........ ........ ........ ........ ........ ........

445 447 448 449 450 451 452

Problem 20.1 . . . . . . . . . . . . . .

..............

...............

........

454

Problem 20.2 . . . . . . . . . . . . . . Problem 20.3 . . . . . . . . . . . . . . Problem 20.4 . . . . . . . . . . . . . . Problem 20.5 . . . . . . . . . . . . . . Problem 20.7 . . . . . . . . . . . . . . Problem 20.8 . . . . . . . . . . . . . . Problem 20.9 . . . . . . . . . . . . . . Problem 20.10 . . . . . . . . . . . . . . Problem 20.11 . . . . . . . . . . . . . . Problem 20.15 . . . . . . . . . . . . . . Problem 20.20 . . . . . . . . . . . . . .

.............. .............. .............. .............. .............. .............. .............. .............. .............. .............. ..............

............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ...............

........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........

455 456 457 459 464 466 468 469 470 472 474

.............. ..............

............... ...............

........ ........

477 479

Chapter 17

429

Problem 17.1 . . . . . . . . . . . . . . Problem 17.2 . . . . . . . . . . . . . . Problem 17.3 . . . . . . . . . . . . . . Problem 17.4 . . . . . . . . . . . . . . Problem 17.5 . . . . . . . . . . . . . . Problem 17.6 . . . . . . . . . . . . . . Problem 17.7 . . . . . . . . . . . . . . Problem 17.8 . . . . . . . . . . . . . . Problem 17.9 . . . . . . . . . . . . . . Problem 17.10 . . . . . . . . . . . . . .

Chapter 19 Problem 19.1 Problem 19.2 Problem 19.3 Problem 19.4 Problem 19.5 Problem 19.6 Problem 19.7

445

Chapter 20

454

Chapter 21 Problem 21.1 . . . . . . . . . . . . . . Problem 21.8 . . . . . . . . . . . . . .

477

1

CHAPTER 1 1.1 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, Mars to sun = 2.28 1011 m).

×

×

Solution 4 The total energy emitted from the sun Q = 4 πR2S σTsun goes equally into all directions, so that 4 qSC (R) = Q/4πR2 = σTsun

qSCmars = qSCearth

RS

2

        RS RM

2

RS RE

2

=

qSCmars = 0.4305

R RE RM

2

=

1.8 1011 2.28 1011

× ×



2

= 0 .4305

× 1367 = 588W /m2

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2

CHAPTER 1

3

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

Without internal heating an energy balance for earth gives Qabsorbed = Qemitted Qabs = qsol

× Aproj = qsolπR2E

Qem = σTE4 A = 4 πR2E σTE4

Thus TE =

qsol 4σ

1367W /m2

1/4

= 4 5.670 10−8 W/m2 K4

  ×

×

TE = 27 9 K = 6 ◦ C

1/4



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4

RADIATIVE HEAT TRANSFER

1.3 Assuming earth to be a black sphere with a surface temp erature of 300 K, what must earth’s inter nal heat generation be in order to maintain that temperature (neglect radiation from the stars, but not the sun) (radius of the earth RE = 6 .37 106 m).

×

Solution

Performing an energy balance on earth: Q˙ = Qemitted

− Qabsorbed = σTE4 A − qsol Aproj = σTE4 − qsol πR2E = πR2E 4σTE4 − qsol 2 W W = π × 6.37 × 106 m 4 × 5.670 × 10−8 2 4 × 3004 K4 − 1367 2 m mK 4πR2E







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 the earth’s atmosphere (called the “solar constant”) has been measured as 1367 W/m2 . Assuming that the sun may be approximated to have a black surface with an e ffective temperature of 5777 K, estimate the diameter of the sun (distance sun to earth S ES 1.496 1011 m).



×

Solution

The sun’s energy travels equally into all directions. Thus, on a spherical “shell” around the sun (such that earth’s orbit lies on that shell): total energy leaving sun Qsun = q sol

4 × 4π S2ES = σTsun 4πR2S .

Thus 1/2

RS = SES

qsol 4 σTsun

  ×

= 1.496

1011 m

136 7 W /m2 5.670 10−8 57774 W/m2

× × RS = 6 .96 × 108 m



1/2

6

RADIATIVE HEAT TRANSFER

1.5 Solar energy impinging on the outer layer of earth’s atmosphere (usually called “solar constant”) has been measured as 1367 W/m2 . Assuming the sun may be approxim ated as having a surface that beha ves like a blackbody, estimate its effective surface temperature. (Distance sun to earth S ES 1 .496 1011 m, radius of sun RS 6.96 108 m).





×

×

Solution

The sun’s energy travels equally into all directions. Thus, on a spherical “shell” around the sun (such that the earth’s orbit lies on that shell): total energy leaving sun Qsun = q sol

2 × 4π S2ES = σTsun 4πR2S .

Thus 1/2

1/4

Tsun = =

SRES S

sol qσ

      136 7 W /m2 5.670 10−8 W/m2 K4

1/4

×

Tsun = 5777 K

1.496 1011 6.96 108

× ×



1/2

CHAPTER 1

7

1.6 A rocket in space may be approximated as a black cylinder of length L = 20 m and diameter D = 2m. It flies past the sun at a distance of 140 million km such that the cylinder axis is perpendicular to the sun’s rays. Assuming that (i) the sun is a blackbo dy 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; neglect radiation from earth and the stars). Solution 4 The total energy emitted from the sun Q = 4 πR2S σTsun goes equally into all directions, so that the flux arriving at the rocket a distance S RS away from the sun is 4 qR = Q/4πS2RS = σTsun

=

Energy emitted

TR4 =

4

2

=

  RS SRS

2

= flux/unit projected area of rocket

+

4

=

4

+

R AR σTR 2πR 2πRL σTR 2πR(R L) σT RS 2 4 = Energy absorbed = A p qR = 2 RL σTsun SRS

   

2RL RS T4 2πR(R + L) sun SRS

2

=

   × 

1 20 6.96 105 (5777K )4 π 1 + 20 140 106

TR = 30 2 K.

×

2

8

RADIATIVE HEAT TRANSFER

1.7 A black sphere of very high conductivity (i.e., isothermal) is orbiting earth. What is its temperature? (Consider the sun but neglect radiation from the earth and the stars). What would be the temperature of the sphere if it were coated with a material that behaves like a black body for wavelengths between 0 .4 µm and 3 µm, but does not absorb and emit at other wavelengths? Solution

Making an energy balance on the sphere (a) Qabsorbed = Qemitted Qabsorbed = qsolAproj = q sol πR2 Qemitted = σT4 A = σT4 4πR2

Thus T=

   qsol 4σ

1/4

=

1367W /m2 4 5.670 10−8 W/m2 K4

×

×



1/4

T = 279K = 6 ◦ C (b) If absorption and emission takes place only over the wavelength interval 0 .4 µm < λ < 3 µm, both Q abs and Qem will be reduced. Since the sphere is spectrally black, we have Qem = 4 πR2

Solar emission is black, i.e., q sol



3 µm





Ebλ dλ = 4 πR2 σT4 f (3 µm T) 0.4 µm

∝ Eb (Tsun), or

× − f (0.4 µm ×T)

λ

qsol,λ Ebλ (Tsun) = , 4 qsol σTsun

and Qabs = πR2



3 µm

qsol,λ dλ = πR2 0.4 µm



= πR2 qsol f (3 µm Tsun )

Thus,



4σT4 f (3 µm T)



×

4 σTsun

3 µm

Ebλ (Tsun) dλ 0.4 µm



− f (0.4 µm ×Tsun)

    × ×

− f (0.4 µm T)

T4 f (3 µmT)



qsol

2

= 1367 W /m

− f (0.4 µmT)

1367 (0.9788 0.1222) 4 5.670 10−8 = 5.1630 109 K4 =

−

×

 600 K (< 601 K) × [0.03934 − 0] = 5.0985 × 109

This nonlinear relation must be solved by iteration, leading to T 4

600

T = 60 0 K



− f (0.4 µm Tsun) f (17, 331) − f (2311)

= qsol f (3 µm Tsun )

CHAPTER 1

9

1.8 A 100 Watt light bulb may be considered to be an isothermal black sphere at a certain temperature. If the light flux (i.e., visible light, 0 .4 µm < λ < 0.7 µm) impinging on the floor directly (2.5 m) below the bulb is 42.6 mW/m2 , what is the light bulb’s e ffective temperature? What is its e fficiency? Solution

The total heat rate (100 W) leaving the light bulb will go at equal per-unit-area amounts through any (hypothetical) sphere around the bulb. Take a sphere whic h has the spot on the floor on its surface (radi us h = 2 .5 m)



qfloor,visible = qfv = f (λ2 T) f (λ2 T)



− f (λ1 T)

Q bulb 4πh2

− f (λ1 T) = 4πh2qfv/Qbulb = 4π 2.52m2 42.6 × 10−3 mW2 = 0.03346



By trial and error: T = 2500 K

f (λ2 T) T = 2500K;

→ λ1 T

100W

= 0.4 µm λ2 T = 0.7 µm

× 2500K = 1000 µm K × 2500K = 1750 µm K

− f (λ1 T) = 0.03369 − 0.00032 = 0.03337 η = 3 .34%

is converted into visible light

The answers for f (λ1 T ) and f (λ2 T) are the same whether Appendix C or program planck.exe are employed, since no interpolation is necessary.

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10

RADIATIVE HEAT TRANSFER

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

≈

Radiation emitted from the plasma ( black disk) hits the spot below it generating an incoming radiative heat flux and a radiative pressure. From equation (1.36), for a black source 1cm

qin =

plasma

Ib cos θ dΩ θ

1cm

2π



= Ib

cos θ dΩ

θ max

Ω plasma

= 2π Ib



θmax

cos θ sin θ dθ 0

= Eb sin2 θmax = E b

2 −8 4 × 15 = 5.670 × 10 ×520, 000 W/m

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 be several MW/cm2 . The radiation pressure is found similarly, from equation (1.42) as p = =

2πIb c



θmax

cos2 θ sin θ dθ =

0

2 5.670 10−8 20, 0004 W/m2 [1 3 3 108 m/s

×

×

−

2Eb 1 3c

  −  

cos2 θmax =

2Eb 1 3c

4 5

N − 0.7155] = 5.735 Ws = 5 .735 2 m3 m

p = 5.735N /m2

3/2

CHAPTER 1

11

1.10 Solar energy incident on the surface of the earth may be broken into two parts: A direct component (traveling unimpeded through the atmosphere) and a sky component (reaching the surface after being scattered by the atmosphere). On a clear day the direct solar heat flux has been determined as qsun = 100 0 W/m2 (per unit area normal to the rays), while the intensity of the sky component has been found to be di ffuse (i.e., the intensity of the sky radiation hitting the surface is the same for all directions) and I sky = 70 W/m2 sr. Determine the total solar irradiation onto earth’s surface if the sun is located 60◦ above the horizon (i.e., 30◦ from the normal). Solution

Since earth’s surface is tilted away from the sun, less energy per unit surface area hits earth than is carried by the sunshine (per unit area normal to the rays), as seen from Fig. 1-8 (by a factor of cos θsun = cos30 ◦ ). Thus from equation (1.36)

qin = qsun cos θsun +

 

= qsun cos θsun +

Isky

Isky cos θ dΩ

2π 2π 0

π/2

cos θ sin θ dθ dψ 0

θ sun

= qsun cos θsun + πIsky =

qsun

1000 × cos30 ◦ + π × 70 = 1086 W/m2

60

°

qin = 108 6 W/m2

earth

12

RADIATIVE HEAT TRANSFER

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 horizo n. Estimate the amount of solar energy that (i) penetrates into the building, (ii) is absorbed by the window, and (iii) is reflected by the window. The window is made of (a) plain glass, (b) tinted glass, whose radiative properties may be approximated by

ρλ =

0.08 0.90 0 0.90 τλ = 0 τλ =

 

for all wavelengths (both glasses) , for 0 .35 µm < λ < 2 .7 µm (plain glass) for all other wavelengths for 0 .5 µm < λ < 1.4 µm (tinted glass) . for all other wavelengths

(c) By what fraction is the amount of visible light (0 .4 µm < λ < 0 .7 µm) reduced, if tinted rather that plain

glass is used? How would you modify this statement in the light of Fig. 1-11? Solution

If we want to determine how much sunshine penetrates through the window (for the entire window area, or per unit window area), we need to determine the angle between a normal to the window ( nˆ ) and a vector pointingtothesun( sˆ ),orcos β = nˆ sˆ . From geometric relations it follows that cos β = cos30 ◦ cos45 ◦ = 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 =



∞

τλ qin,λ dλ = 0.90

0





2.7 µm

qin,λ dλ 0.35 µm



− f (0.35 µm Tsun) = 0.90 × 612.4 W/m f (15, 598 µm K) − f (2022 µm K) = 0.90 × 612.4 × (0.9720 − 0.0702) = 497.0 W/m2 = 0.90 qin f (2.7 µm Tsun ) 2





It follows that qabs = q in

− qref − qtrans = 612.4 − 49.0 − 497.4 = 66.0 W/m2

(b) Tinted glass: Similarly,

 

× qin f (1.4 µm × 5777 K) − f (0.5 µm × 5777K) = 0.90 × 612.4 × f (8088) − f (2889) W /m2 = 0.90 × 612.4 × (0.8597 − 0.2481) = 337.1 W/m2 ,

qtrans = 0.90





qtrans = 337.1 W/m2 2

i.e., heat gain is reduced by 160 W/m (or 32%). (c) The tinted glass loses some visible light, since it is not transmissive for 0 .4 µm < λ < 0.5 µm. The fractional reduction is



− −



0.90 qin f (0.5 µm Tsun) f (0.4 µm Tsun) 0.90 qin f (0.7 µm Tsun) f (0.4 µm Tsun) f (2889) f (2311) 0.2481 0.1222 = = f (4044) f (2311) 0.4888 0.1222

fractional reduction =

− −

− −

fractional reduction = 34.3% Therefore, the tinted glass appears to cause an equivalent loss of visible light. However, Fig. 1-11 shows that the human eye is not very responsive to wavelengths below 0 .5 µm, so that the actual reduction of visible light 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 the earth after being scattered by the atmosphere may be approximated as I sky (θ) = Isky (θ = 0) cos θ, where θ is measured from the surface norm al. For a day with I sky (0) = 100 W/m2 sr determine the solar irradiation hitting a solar collector, if the collector is (a) horizontal, (b) tilted from the horizontal by 30 ◦ . Neglect radiation from the earth’s surface hitting the collector (by emission or reflection). s

Solution

n

s

(a) For a horizontal collector the solar irradiation is readily determined from equation (1.36) as

k =30

2π

ο

θ

π/2

j

ψ i

Earth s surface

qin =

  0

= 2π I0

(I0 cos θ) cos θ sin θ dθ dψ π/2 2π 2π I0 = cos2 θ sin θ dθ = 3 3 0 0

× 100 qin = 209.4 W/m2

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



I0 (kˆ sˆ )(nˆ sˆ ) dΩ,

·

·

n ˆ sˆ>0

·

where kˆ sˆ = cos θ gives the variation of sky intensity with sky polar angle θ, while nˆ sˆ = cos θ is the cosine of the angle between the surface normal and a unit vector pointing into the sky, ˆs. This unit vector may be expressed in terms of ˆı, ˆ, kˆ (unit vectors for an earth-bound coordinate system), with

·

·

ıˆ = cos αˆı

− sin αkˆ , ˆ = ˆ, nˆ = sin α ˆı + cos α kˆ ,

as ˆ sˆ = sin θ cos ψ ˆı + sin θ sin ψ ˆ + cos θ k = sin θ cos ψ ˆı + sin θ sin ψ ˆ + cos θ n ˆ.

Substituting for ˆı , ˆ and nˆ , ˆs(θ , ψ ) becomes sˆ = (sin θ cos ψ cos α + cos θ sin α) ˆı + sin θ sin ψ ˆ + (cos θ cos α

− sin θ cos ψ sin α) kˆ .

Equation (1.36) may be evaluated by evaluating kˆ sˆ in collector coordinates (i.e., integrating over θ and ψ ), or by determining nˆ sˆ in global coordinates (integrating over θ and ψ). Choosing the former, one obtains

·

qin =

·



I0 (cos θ cos α

− sin θ cos ψ sin α) cos θ sin θ dθ dψ

kˆ sˆ>0 nˆ sˆ>0

··

The integration limits imply that only directions above the collector (0 < θ < π/ 2) and above the horizontal kˆ sˆ = cos θ = cos θ cos α

·

− sin θ cos ψ sin α > 0

are to be considered. Solving the last expression for cos ψ yields cos ψ < cot α cot θ ,

14

RADIATIVE HEAT TRANSFER

and using symmetry

  π/2

qin = 2I0

0

π

(cos θ cos α cos−1 (cos α cos θ )

− sin θ cos ψ sin α)

× cos θ sin θ dψ dθ.

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

  π/2

qin = 2Ib

0

=

π

(cos θ cos α 0

2π 2π 3 I0 cos α = 3

− sin θ cos ψ sin α) cos θ sin θ dψ dθ 2

× 100 × cos30 ◦ = 181.4 W/m . qin = 181.4 W/m2

This number is a little bit too low because the additionally considered radiation from the earth has a kˆ ˆs = cos θ < 0.

·

15

CHAPTER 1

1.13 A 100 W light bulb is rate d to have a total light outpu t of 1750 lm. Assuming the light bulb to cons ist of a small, black, radiating body (the light filament) enclosed in a glass envelope (with a transmittance τ  = 0 .9 throughout the visible wavelengths), estimate the filament’s temperature. If the filament has an emittance of  f = 0 .7 (constant for all wavelengths and directions), how does it a ffect its temperature? Solution

The total heat rate leaving the light bulb by radiation (going equally into all directions) is 4 Qbulb = τ Qfil = τ (Afil fil σTfil ).

Thus the radiative heat flux on a large, hypothetical sphere of radius R surrounding the bulb is Qbulb Afil = τfil πIb (Tfil ) = I (R) dΩfil Asphere Asphere 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. q(R) =

From equation (1.47) the luminous flux at R is qlum (R) = Ld Ωfil = 286

lm W



0.7 µm

Iλ (R) dλ dΩfil 0.4 µm

With gray (i.e., constant) values for fil and τ (at least over the visible wavelengths), this makes Iλ (R) and





−





where Qlum is the total luminous flux leaving the bulb. Thus f (0.7 Tfil )

λ

lm I (R) dΩfil f (0.7 µm Tfil ) f (0.4 µm Tfil ) . W = qlum(R)Asphere lm = 286 Afil τfil πIb (Tfil ) f (0.7 µm Tfil ) f (0.4 µm Tfil ) W lm = 286 πτ Qfil f (0.7 µm Tfil ) f (0.4 µm Tfil ) , W

qlum (R) = 286 Qlum

∝ Ib (Tfil ),



−

−



Qlum 1750lm = 286(lm/W) πτ Qfil 286(lm/W) π 0.9 = 0.02164.

− f (0.4 Tfil) =

× × 100W

 2320 K f (0.7 × 2320) − f (0.4 × 2320) = 0.02165 − 0.00013 = 0 .02152. Tfil  232 0 K

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

This expression does not include the filament emittance:  f = 0 .7 would leave T fil unchanged (however,A fil would have to increase by a factor of 1 /0.7 = 1 .43, in order to emit 100 W of power).

16

RADIATIVE HEAT TRANSFER

1.14 A pyrometer is a device with which the temperature of a surface may be determined remotely by measuring the radiative energy falling onto a detector. Consider a black detector of 1 mm 1 mm area, that is exposed to a 1cm 2 hole in a furnace located a dist ance of 1 m away. The inside of the furnace is at 1500K and the intensity escaping from the hole is essentially blackbody intensity at that temperature. (a) What is the radiative heat rate hitting the detector? (b) Assuming that the pyrometer has been calibrated for the situation in (a), what temperature would the pyrometer indicate if the nonabsorbing gas between furnace and detector were replaced by one with an (average) absorption coefficient of κ = 0 .1 m−1 ?

×

Solution

(a) A total radiative intensity of I b (Tfurnace) leaves the hole, equally into all directions. From the definition of intensity the total heat rate hitting the detector—assuming the detector to be directly opposite the hole—is Q = A d

I (T ) dΩ

hole b

Qd =

=A hole detector

−

I (T ) hole b

Adetector

f

S2hd

Ahole σT4f A detector π S2hd 2

=

f

1 cm

× 5.670 × 10−12 W/cm2 K4 × 15004 K4 × 1 mm2 π 1 m2

× 10−6 W = 9.137 µW (b) The signal would now be attenuated by a factor of exp( −κShd ), or Q∗d = Qd exp(−κShd ) = 9 .137 µW × exp(−0.1 m−1 × 1 m) Qd = 9 .137

= 8.2675 µW

Since the d calibration is Qd

∝ T 4f we have Q∗d Qd = ∗4 4 Tf Tf T∗f =

  8.2675 9.137

∗

→ T∗f4 = QQdd T4f 1/4

1500 = 146 3 K,

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

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CHAPTER 1

17

×

1.15 Consider a pyrometer, which also has a detector area of 1mm 1mm, which is black in the wavelength range 1.0 µm λ 1 .2 µm, and perfectly reflecting elsewhere. In front of the detector is a focussing lens ( f = 10 cm) of diameter D = 2 cm, and transmissivity of τ l = 0.9 (around 1 µm). In order to meas ure the temperature inside a furnace, the pyrometer is focussed onto a hot black surface inside the furnace, a distance of 1 m away from the lens.

≤ ≤

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

1 1 1 = + ; f u v

M=

h (detector size) v = , u H spot size

 

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 (c) time? It turns out the furnace wall is not really black, but has an emittance of  = 0 .7 (around 1µm). Assuming there is no radiation reflected from the furnace surface reaching the detector, what is the true surface temperature 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 front of the lens, what furnace temperature would provide the same pyrometer reading as case (b)? Solution

(a) From geometric optics

1 1 = v f

1 1 9 − 1u = 10cm − 100cm = ; v = 11.1cm 100cm

u 100 = 1mm = 9mm. v 11.1 Thus, the spot seen by the pyrometer is 9mm 9mm in size.

×

H=h

×

(b) Energy leaving spot, intercepted by detector (or, by conservation of energy, energy intercepted by detector, coming from spot) is Qd = I b12 (T )AH ΩHl τl = I d12 Ah Ωhl ,



λ2

where I b12 (T) = λ Ibλ (T)dλ is emitted intensity from the spot on the furnace wall, I d12 is intensity absorbed 1 by the detector, and Ω Hl and Ω hl 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 = 2 R, we simply evaluate the solid angles from the center of spot H and detector h : πR2 πR2 ΩHl ; Ωhl = 2 , u2 v and 2 H 2 h .9 2 = πR2 = A h Ωhl = π 1cm2 = 2 .545 10−4 cm2 AH Ωhl = πR2 u v 100



  

 

×

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

σT4 [ f (λ2 T) π

−8

4

− f (λ1 T)] = 5.67 × 10π × 1200

W [ f (1440) m2  

   −    0.00961

W

=

280 m2

f (1200)]   0.00213

5

W  Ib (1.1µm, 1200k)∆λ = (Eb /T5 )(T5/π)∆λ = 1.7228 × 10−12 × 1200 0.2 2 π m λ

= 273

λ

W m2

18

RADIATIVE HEAT TRANSFER

and Qd = 280

W 2.545 m2

× 10−8 m2 × 0.9 = 6.41µW

(c) The energy hitting detector remains the same and, therefore, so does the intensity emitted from the spot: Ib12 (Ta )(actual) = I b12 (Tp = 1200K)(perceived)

or, if we assume the blackbody intensity over the detector range can be approximated by the value at 1.1 µm,  eC2 /λTa

− 1  eC

1 2 /λTp

− 1,

leading to

{

C2 ln 1 + [eC2 /λTp 1] λ 14, 388µmK = ln 1 + 0.7[e14,388/1.1×1200 1.1µm

Ta =

or

{

− }

− 1]}

Ta = 1241K .

At these wavelengths Wien’s law holds almost exactly, i.e., we maydrop the two “1”s from theaboveequat ion, and Tp C2 C2 /λ = Ta ln(eC2 /λTp ) = C λT 2 λ + ln  1 + p ln 





λTp

C2

which again leads to 1241k. (d) Similar to (c) we get τ f I b12 (Ta ) = I b12 (Tp ) and, since τ f =  (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 2 2.1 Show that for an electromagnetic wave traveling through a dielectric ( m1 = n1 ), impinging on the interface with another, optically less dense dielectric ( n2 < n 1 ), light of any polarization is totally reflected for incidence angles larger than θ c = sin−1 (n2 /n1 ). Hint: Use equations (2.105) with k 2 = 0. Solution

Equations (2.105) become for k 2 = 0, η n sin θ = w  sin θ , 0 1 2 2 1 2 t2 wt2 w t = η 0 n2 ,

−

wt w t cos θ2 = 0 .

The last of these relations dictates that either wt = 0 or cos θ2 = 0 ( wt = 0 is not possible since—from the first relation—this would imply η0 n1 sin θ1 = 0 which is known not to be true).

 w t = 0 : Substituting this into the second relation leads to wt = η 0 n 2 , and the first leads to n1 sin θ1 = n 2 sin θ2 .

≤

Since n 1 > n 2 this is a legitimate solution only for sin θ1 ( n2 /n1 ), or θ 1 cos θ2 = 0 ( θ2 = π/ 2): Substituting the first relation into the second gives w t = η0



≤ θc = sin−1(n2 /n1).

n21 sin2 θ1

− n22 , i.e., a legitimate nonzero solution for n21 sin2 θ1 − n22 ≥ 0 or θ1 ≥ θ c . Inspection of therefle ction coefficients, equations (2.109), shows that

r =



2  in21 w t + n2 w i cos θ1

in21 w t

− n22 wi cos θ1 ,

r⊥ =

wi cos θ1 + iw t wi cos θ1

− iwt

   

Since, in both reflection coe fficients, there are no sign changes within the real and imaginary parts, it follows readily that ρ = r r∗ = ρ ⊥ = r⊥ r∗⊥ = 1 .

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19

20

RADIATIVE HEAT TRANSFER

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  sˆ iw nˆ ; this implies that E 0 is not a vector normal to sˆ . It is best to assume E 0 = E  eˆ  + E⊥ eˆ ⊥ + Es sˆ .

−

−

Solution

Inside the absorbing medium wt = w t iwt = w t sˆ t iwt nˆ , and the electric field vector does not lie in a plane normal to ˆs. Thus, we assume a general three-dimensional representation, or

−

−

E0 = E  eˆ  + E⊥ eˆ ⊥ + Es sˆ .

Following the development for nonabsorbing media, equations (2.77) through (2.88), then leads to νµH0 = w

× E0 = (w sˆ − iw nˆ ) × (E eˆ  + E⊥eˆ ⊥ + Essˆ ).

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 E s vanishes for incident and refle cted wave. Using the same vector relations as given for the nonabsorbing media interface, one obtains

−

νµH0 = w  (E eˆ ⊥

− E⊥eˆ  ) − iw (E eˆ ⊥ cos θ − E⊥ˆt + Eseˆ ⊥ sin θ).

For the interface condition (with νµ the same everywhere) νµH0

× nˆ = w (E ˆt + E⊥eˆ ⊥ cos θ) − iw (E cos θˆt + E⊥eˆ ⊥ + Es sin θˆt).

Thus, from equation (2.78) wi (Ei ˆt + Ei⊥ eˆ ⊥ cos θ1 ) wi (Er ˆt + Er⊥ eˆ ⊥ cos θ1 ) ˆ = w t (Et tˆ + Et⊥ eˆ ⊥ cos θ2 ) iw ˆ ⊥ + Ets sin θ2 ˆt) t ( Et cos θ2 t + Et⊥ e

−

−

or wi (Ei − Er )  wi (Ei⊥ Er⊥ )cos θ1

= (wt =

− iwt cos θ2 ) Et − iwt sin θ2 Ets  (w cos θ2 iw) Et⊥ t

t

−

−

Similarly, from equation (2.77), E0

(2.2-A) (2.2-B)

× nˆ = (E eˆ  + E⊥eˆ ⊥ + Essˆ ) × nˆ = −Eeˆ ⊥ cos θ + E⊥ˆt − Eseˆ ⊥ sin θ

and (Ei + Er ) cos θ1 = Et cos θ2 + Ets sin θ2 Ei⊥ + Er⊥ = Et⊥

(2.2-C) (2.2-D)

These four equations have 5 unknowns (Er , Ers , Et , Et⊥ , and E ts ), 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 sˆ

·

− iwt nˆ ) · (Et eˆ t + Et⊥eˆ ⊥ + Etssˆ ) − Ets cos θ2 ).

= wt Ets + iw t ( Et sin θ2



Eliminating Et⊥ from equations (2.2-B) and (2.2-D), with r⊥ = E r⊥ /Ei⊥ , gives w (1 − r⊥ ) cos θ1 = ( w cos θ2 − iw)(1 + r⊥ ), t

i

or



r⊥ =

which is identical to equation (2.109).

wi cos θ1

t



− (wt cos θ2 − iwt ) , − iwt )

wi cos θ1 + (wt cos θ2

(2.2-E)

CHAPTER 2

21

Now, eliminatingEts 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 cos θ1 (Ei + Er ) = wt Et .

(2.2-F)

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

iw t E t sin θ2 wt iw t cos θ2

−

−



(Ei + Er ) cos θ1 = E t cos θ2

2



− wiw−tiwsin cosθ2θ2 t

t



= E t

wt cos θ2 iw t . wt iw t cos θ2

−

−

Using this to eliminate E t from equation (2.2-F), with r = E r /Ei , gives



 − iw cos θ2 t − r) + iwt cos θ1(1 + r) = w t cos θ1(1 + r) wwt cos θ2 − iw t t wi (wt cos θ2 − iw t )(1 − r )    = wt (wt − iw t cos θ2 ) − iwt ( wt cos θ2 − iwt ) cos θ1 (1 + r ) wi (1

=





η 20 m22







cos θ1 (1 + r ),



r =

wi (wt cos θ2 wi (wt cos θ2





− iwt ) − η20 m22 cos θ1 , − iwt ) + η20 m22 cos θ1

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)

or

n21 (Ei eˆ i nˆ + Er eˆ r nˆ ) = m22 (Et eˆ t nˆ + Ets sˆ nˆ )

·

n21 (Ei

·

− Er ) sin θ1

·

= m22 (Et sin θ2

·

− Ets cos θ2), etc.

22

RADIATIVE HEAT TRANSFER

2.3 Find the normal spectral reflectivity at the interface between two absorbing media. [Hint: Use an approach similar 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 colinear with the surface normal]. Solution

Equations (2.19) and (2.20) remain valid for incident, reflected and transmitted waves, with w = w  iw = (w iw ) n ˆ for all three cases. Fromequation(2.31) w w = ( w iw )2 nˆ nˆ = η 20 m2 it follows that w iw = η0 m. Thus

−

·

wi wr

− iwi = η0 m1 , − iwr = − η0 m1 = t η0 m 2 . wt − iw

−

·

−

(reflected wave is moving in a direction of

−

±

− nˆ ),

From equations (2.23) and (2.24), it follows that the electric and magnetic field vectors are normal to nˆ , i.e., tangential to the surface, say E0 = E 0 ˆt. Then, from equation (2.77) (Ei + Er ) ˆt or

× nˆ = Et ˆt × nˆ ,

Ei + Er = E t

From equation (2.25) νµH0 = w × E0 = ( w − iw ) E n ˆ × ˆt, and from equation (2.78) n1 (Ei



− Er ) = m 2 Et .

Substituting for Et and dividing by Ei , with r = E r /Ei : m1 (1

or

and



ρn =

m1 + m2

  

(m1 − m2 )(m1 − m2 )∗ ρn = rr∗ = (m1 + m2 )(m1 + m2 )∗





− r) = m2 (1 + r) m1 − m2 r = =

−

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

−

−

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

−



2

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CHAPTER 2

23

2.4 A circularly polarized wave in air is incident upon a smooth dielectric surface ( n = 1 .5) with a direction of 45◦ o ff normal. What are the normalized Stokes’ parameters before and after the reflection, and what are the degrees of polarization? Solution

From the definition of Stokes’ parameters the incident wave has degrees of polarization Qi Ui = = 0, Ii Ii

Vi = 1, Ii

±

the sign giving the handedness of the circular polarization. With E r = Ei r and E r⊥ = Ei⊥ r⊥ , from equations (2.50) through (2.53): Ir = E i E∗i r2 + Ei⊥ E∗i⊥ r2⊥ = E i E∗i (ρ + ρ⊥ ) = 1 (ρ + ρ⊥ ) Ii 2

Since E i E∗i = E i⊥ E∗i⊥ [from equation (2.51)] and ρ = r 2 . Similarly,

Qr = Ei E∗i r2 Ei⊥ E∗i⊥ r2⊥ = E i E∗i (ρ ρ⊥ ) Ur = Ei E∗i⊥ r r⊥ + Ei⊥ E∗i r⊥ r = U i r r⊥ = 0

−

Vr = i(Ei E∗i⊥

−

−

− Ei⊥E∗i ) r r⊥ = Vi r r⊥

ρ ρ⊥ Qr = , Ir ρ + ρ⊥

2r r⊥ Vi Vr = . Ir ρ + ρ⊥ Ii

From Snell’s law sin θ2 =

sin θ1 ; cos θ2 = n2

− 1

sin 2 θ1 = n22

−  1

0 .5 = 1.52

7 , 9

and from equations (2.89) and (2.90)

−

r =

n1 cos θ2 n2 cos θ1 = n1 cos θ2 + n2 cos θ1

r⊥ =

n1 cos θ1 n2 cos θ2 = n1 cos θ1 + n2 cos θ2

−

−

√7/9 − 1.5 √1/2 √7/9 + 1.5 √1/2 = −0.0920, ρ = 0.0085 √1/2 − 1.5 √7/9 √ √ = −0.3033, ρ⊥ = 0.0920 1/2 + 1.5 7/9

Qr Ur 0.0085 0.0920 = = 0.8315, =0 Ir 0.0085 + 0.0920 Ir Vr = Ir

−

0.0920 × 0.0085 = ±0.5556 ±20×.0085 + 0.0920

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

24

RADIATIVE HEAT TRANSFER

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 the y-z-plane? Solution

From equations (2.50) through (2.53) it follows that Qr /Ir = 1 , Ur = V r = 0 (i.e., linear polarization), if either Er or Er⊥ vanish. From Fig. 2-9 it follows that r⊥  0 and, therefore E r⊥  0 for all incidence directions, while r = 0 for θ = θ p (Brewster’s angle), or n2 = tan −1 1.5 = 56.31◦ . θp = tan −1 n1 The resulting wave is purely perpendicularpolarized, i.e., eˆ ⊥ must lie in the y z plane, or eˆ  must be in the x z plane. Therefore, the surface may be expressed in terms of its surface normal as ˆ cos θp = ( ˆı 1.5kˆ ) 3.25. nˆ = ˆı sin θp k

−

−

−

−

x

Ei

z

n

p p

Er

√

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CHAPTER 2

25

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

−

(a) Determine the material’s reflectivity in the vicinity of λ = 2 µm (for platinum at 2 µm mPt = 5 .29 6.71 i, for MgO m MgO = 1 .65 0.0001 i). (b) Estimate the thickness of MgO required to reduce the average reflectivity in the vicinity of 2 µm 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 relevant reflection coefficient. From equation (2.122) r12 =

−

 11 −+ nn22 = 11 −+ 11..65 = −0.2453 65

1 m2 1 + m2

since k 2 1, and r 12 = 0 .2453. r23 may also be calculated from equation (2.122) or, more conveniently, from equation (2.126):





r223 =



(1.65 5.29)2 + 6.712 = 0 .6253 or r 23 = 0 .7908. (1.65 + 5.29)2 + 6.712

−

Since the real part of r12 < 0 it follows that δ 12 = π , while tan δ23 =

2(1.65 6.71 5.29 10−4 ) = 0.3150. 1.652 + 10−8 (5.292 + 6.712 )

×

− −

×

−



 × × cos[ δ12 ± (δ23 − ζ12 )] = cos[ π ± (2.8364 − 10.3673)] = −0.3175. Also κ2 d = 4 π × 10−4 × 1 µm/2 µm = 2π × 10−4 and τ = e − d = 0.9994  1. Thus 0.24532 + 2 × 0.2453 × 0.7908 × (−0.3175) + 0.79082 R=

Since the (r23 ) > 0 (numerator) and (r23 ) < 0 (denominator) δ 23 lies in the second quadrant, π/2 < δ23 < π, or δ 23 = 2 .8364. Also ζ12 = 4 π 1.65 1 µm/2 µm = 10 .3673, and

κ2

1 + 2 0.2453 0.7908 ( 0.3175) + 0.24532 0.79082

×

×

(b) The cos in the numerator fluctuates between dropping the cos-term. Then Rav =

or

r212 + r223 τ2

1 + r212 r223 τ2

×− −1 < cos < +1.

R = 0.6149.

× The average value for R is obtained by

,

− r212 = 0.4 − 0.24532 = 0.5782, − r212 ) 0.79082 (1 − 0.24532 ) 1 1 − ln 0.5782 = 43.6 µm. d = − ln τ = − ln τ2 = κ2 2 κ2 4π × 10−4 µm−1

τ2 =

Rav

r223 (1

More accurate is the averaged expression, equation (2.129) Rav = ρ 12 +

ρ23 (1 ρ12 )2 τ2 1 ρ12 ρ23 τ2

−

−

or τ2 = =

Rav − ρ12 Rav − ρ12 − ρ12)ρ12 + (1 − ρ12 )2 = ρ23 1 − (2 − Rav)ρ12 0.4 − 0.24532 = 0 .6013 0.79082 [1 − 1.6 × 0.24532 ]



ρ23 (Rav

 



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