Solution Manual For Thermal Radiation Heat Transfer, 6th Edition

solutions MANUAL FOR Thermal Radiation Heat Transfer by John R. Howell Robert Siegel M. Pinar Menguc solutionS MANUAL FOR Thermal Radiation Heat Transfer by John R. Howell Robert Siegel M. Pinar Menguc Boca Raton London New York CRC Press is an imprint of the Taylor & Francis Group, an informa business CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2010 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20150326 International Standard Book Number-13: 978-1-4665-9329-9 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com Solutions to Homework Problems Table of Contents Chapter 1 Introduction to Radiative Transfer…………..…………….…1.1 Chapter 2 Radiative Properties at Interfaces………….……….…….….2.1 Chapter 3 Radiative Properties of Opaque Materials……..…..……….3.1 Chapter 4 Configuration Factors for Diffuse Surfaces with Uniform Radiosity….…………….……..………………………4.1 Chapter 5 Radiation Exchange in Enclosures Composed of Black and/or Diffuse-Gray Surfaces………………….…………….5.1 Chapter 6 Exchange of Thermal Radiation among Nondiffuse Nongray Surfaces…..……………………………….…………6.1 Chapter 7 Radiation Combined with Conduction and Convection at Boundaries……..………………………..…………..……….7.1 Chapter 8 Inverse Problems in Radiative Heat Transfer………….….…8.1 Chapter 9 Properties of Absorbing and Emitting Media…………….….9.1 Chapter 10 Fundamental Radiative Transfer Relations…..…………….10.1 Chapter 11 Radiative Transfer in Plane Layers and Multidimensional Geometries………………………..………11.1 Chapter 12 Solution Methods for Radiative Transfer in Participating Media….………………………………..……. .12.1 Chapter 13 Conjugate Heat Transfer in Participating Media……….….13.1 Chapter 14 Electromagnetic Wave Theory………………….………..….14.1 Chapter 15 Absorption and Scattering by Particles and Agglomerates……………………………….…………….15.1 Chapter 16 Near-Field Thermal Radiation……………………..…..……16.1 Chapter 17 Radiative Effects in Translucent Solids, Windows, and Coatings………………..……….……………17.1 1.Introduction to Radiative Transfer SOLUTIONS- CHAPTER 1 1.1 What are the wave number range in vacuum and the frequency range for the visible spectrum (0.4 to 0.7 m)? What are the wave number and frequency values at the spectral boundaries between the near and the far infrared regions? SOLUTION: c0 = 2.99792458 x 108 m/s; 1 = 0.4 x 10-6 m, 2 = 0.7 x 10-6 m, 1 = 1 / 1 = 2.5 x 106 m-1, 2 = 1 / 2 = 1.428571 x 106 m-1, 1 = c0 / 1 = 2.99792458 x 108 / 0.4 x 10-6 = 7.494811 x 1014 s-1, 2 = c0 / 2 = 2.99792458 x 108 / 0.7 x 10-6 = 4.282749 x 1014 s-1, boundary = 25 x 10-6 m, boundary = 1 / boundary = 4 x 104 m-1, boundary = c0/boundary = 2.99792458 x 108 / 25 x 10-6=1.19917 x 1013 s-1 Answer: 2.5 x 106 to 1.4286 x 106 m-1; 4.2827 x 1014 to 7.4948 x 1014 s-1; 4 x 104 m-1; 1.1992 x 1013 s-1. 1.2 Radiant energy at a wavelength of 2.0 m is traveling through a vacuum. It then enters a medium with a refractive index of 1.24. (a) Find the following quantities for the radiation in the vacuum: speed, frequency, wave number. (b) Find the following quantities for the radiation in the medium: speed, frequency, wave number, and wavelength. SOLUTION: (a) c0 = 2.99792458 x 108 m/s; o = 2.0 x 10-6 m, n = 1.24, so speed in the vacuum = 2.99792458 x 108 m/s; 0 = c0 / 0 = 2.99792458 x 108 / 2.0 x 10-6 = 1.49896229 x 1014 s-1. 0 = 1 / 0 = 5 x 105 m-1; (b) Speed in the medium c m = c0 /n = 2.99792458 x 108 /1.24 = 2.417681113 x 108 m/s; m = 0 = 1.49896229 x 1014 s-1; m = cm / m = 2.417681113 x 108 / 1.49896229 x 1014 = 1.6129032 x 10-6 m; m = 1 / m = 6.20 x 105 m-1. Answer: (a) 2.9979 x108 m/s; 1.4990 x1014s-1; 5×105 m-1 (b) 2.4176 x 108 m/s; 1.4990 x 1014 s-1; 6.20 x 105 m-1; 1.6129 x 10-6 m 1.1 1.Introduction to Radiative Transfer 1.3 Radiation propagating within a medium is found to have a wavelength within the medium of 1.570 m and a speed of 2.500×108 m/s. (a) What is the refractive index of the medium? (b) What is the wavelength of this radiation if it propagates into a vacuum? SOLUTION: C0 = 2.99792458×108 m/s; m = 1.570×10-6 m; cm = 2.500×108 m/s. (a) nm = c0/cm = 1.199 (b) 0= nmm = 1.882×10-6 m = 1.882 m. Answer: (a) 1.199; (b) 1.882 m. 1.4 What range of radiation wavelengths are present within a glass sheet that has a wavelength-independent refractive index of 1.33 when the sheet is exposed in vacuum to incident radiation in the visible range 0 =0.4 to 0.7 m? SOLUTION: nm = 1.33; m1 = 0.4/nm = 0.301 m; m2 = 0.7/nm = 0.526 m. Answer: 0.301 to 0.526 m. 1.5 A material has an index of refraction n(x) that varies with position x within its thickness. Obtain an expression in terms of co and n(x) for the transit time for radiation to pass through a thickness L. If n(x) = ni(1 + kx), where ni and k are constants, what is the relation for transit time? How does wave number (relative to that in a vacuum) vary with position within the medium? SOLUTION: n(x) = co / c(x), so c(x) = co / n(x) = dx/d. Then t ni L ni  1 L kL2  n ( x ) dx  ( 1  kx ) dx  L    c0 x 0 c0 x 0 c0  2  c = m ;  = c / m = co / o co / c = n = o / m; m / o = 1/n so m / o = n = ni(1 + kx). Answer : 2 ni kL L+ co 2 ; ni(1 + kx). 1.2 1.Introduction to Radiative Transfer Derive Equation 1.26 by analytically finding the maximum of the Eb/T5 vs. T 1.6 relation ( Equation 1.20). SOLUTION: Take the derivative of Eb/T5 with respect to T. To simplify notation, let   T .  d ( E b / T 5 ) 2 C1 d  d 5     2 C1    / exp(C2 /  )  1 5 d( ) d (  )     exp(C2 /  )  1  d()   2  exp(C2 /  )  6 5 C2 / (  )  2 C1 [ 5    ] / exp(C2 /  )  1      2 exp(C2 /  )  1         2 C1 C2 /   [ 5    exp(C2 /  )  1  1  exp(C2 /  )  6 Setting the result = 0 to find the maximum, C2 /  T max C2 / max  5 1  exp(C2 / max ) 1  exp  C2 /  T max    Clearly, the product (T)max at the maximum of the Planck curve is equal to a constant, which must be found by iteration. Using iteration or a root-finding program gives  T max  C3  2897.8m  K . 1.7 A blackbody is at a temperature of 1250 K (1250 K), and is in air. (a) What is the spectral intensity emitted in a direction normal to the black surface at = 3.75 m? (b) What is the spectral intensity emitted at = 45° with respect to the normal of the black surface at = 3.75 m? (c) What is the directional spectral emissive power from the black surface at  = 45° and = 3.75 m? (d) At what is the maximum spectral intensity emitted from this blackbody, and what is the value of this intensity? (e) What is the hemispherical total emissive power of the blackbody? SOLUTION: (a) At T = 4687.5 m K, I b,n  2C1  e 5 C2 / T  1 = 7823 W/m2•m•sr. (b) Because the intensity from a blackbody is independent of angle of emission, the result is the same as part (a). (c) Eb(3.75 m, 45°) = Ib cos 45° = 7823 x 0.7071 = 5532 W/m2•m•sr (d) From Equation 1.26, maxT = C3, and from Table A.4, C3 = 2897.8 m K. Thus, max = 2318.2 / 1250 = 2.3182 m. At max, 1.3 1.Introduction to Radiative Transfer I b,max  2C 1 C2 / maxT 5 max e  1 = 12499.08 W/m2•m•sr (e) From Equation 1.32 and Table A-4, Eb = T4 = 5.67040×10-8 W/m2•K4 x (1250)4 K4 = 138,437.5 W/m2 Answers: (a) 7,823W/(m2·m·sr); (b) 7,823W/(m2·m·sr); (c) 5532 W/(m2·m·sr); (d) 2.3182 m, 12499 W/(m2·m·sr); (e) 138,437 W/m2 . 1.8 Plot the hemispherical spectral emissive power Eb for a blackbody in air [W/(m2·m)] as a function of wavelength (m) for surface temperatures of 2000 and 6250K. SOLUTION: Eb = 2C1/ { 5[e(C2/T) – 1]} For a wavelength range of 0.01 to 5 m, following figures are obtained for different surface temperatures . 1.4 1.Introduction to Radiative Transfer 1.9 For a blackbody at 2250 K that is in air, find : (a) the maximum emitted spectral intensity (kW /m2•m•sr ). (b) the hemispherical total emissive power (kW /m2 ). (c) the emissive power in the spectral range between o = 2 and 8m. (d) the ratio of spectral intensity at o = 2 m to that at o = 8 m. SOLUTION: (a) Ib,max = C4 T5 = 4.09570×10-12 x 22505 = 236.17 kW / m•m2•sr (b) Eb = T4 = 5.67040×10-8(W/m2•K4) x (2250)4 (K4) = 1453.26 kW/m2 (c) Using Equation 1.37, for 2T = 18,000 m•K, F0 2T = 0.97766 and for 1T = 4500 m•K, F0 1T = 0.56429 F2T  1T = 0.41336 Eb(12) = 0.3482 T4 = 600.730 kW/m2 2C (d) Using I b,n  5 C / 1T ,   e 2  1 [ Ib(=2) ] / [Ib(=8) ] = 1.5861×105/ 2.9695×103 = 53.4139 Answers: (a) 236.17 kW/m2•m•sr; (b) 1453.26 kW/m2; (c) 600.730 kW/m2; (d) 53.4139 1.5 1.Introduction to Radiative Transfer 1.10 Determine the fractions of blackbody energy that lie below and above the peak of the blackbody curve. SOLUTION: At the peak, maxT = 2893 mK. Using Equation 1.37 with C 14388(  m  K )  2   4.965 gives F02897.8  0.25005 . Therefore about 25 percent of T 2897.8(  m  K ) the blackbody energy is at wavelengths below the peak, and 75 percent is at longer wavelengths. 1.11 A blackbody at 1250K is radiating in the vacuum of outer space. (a) What is the ratio of the spectral intensity of the blackbody at = 2.0m to the spectral intensity at = 5m? (b) What fraction of the blackbody emissive power lies between the wavelengths of = 2.0m and = 5m? (c) At what wavelength does the peak energy in the radiated spectrum occur for this blackbody? (d) How much energy is emitted by the blackbody in the range 2.0   5 m? SOLUTION: (a) 1T = 2.0 x 1250 = 2500 m K; 2T = 5 x 1250 = 6250 m K (Note that these values bracket the peak in the blackbody curve.) 2C Using Ib  5 C / 1T ,   e 2  1 Ib /Ib = 11823.5 / 4237.4 = 2.7903 (b) From Equation 1.37, F01T = F02500 = 0.16136, F02T = F07250 = 0.75792. The fraction between 1T and 2T is then= ((0.75792- 0.16136) = 0.59656. (c) From Table A.4, C3 = 2897.77 m K, so max = 2897.77 / 1250 = 2.31824m. (d) From Table 1.2, the band emission is ( F02T – F01T ) T4 = 0.59656 x 5.67040×10-8(W/m2•K4) x (1250)4 (K4) = 82587.2 W/m2 Answer: (a) 2.7903; (b) 0.59656.; (c) 2.31824m; (d) 82587.2 W/m2. 1.12 Solar radiation is emitted by a fairly thin layer of hot plasma near the sun’s surface. This layer is cool compared with the interior of the sun, where nuclear reactions are occurring. Various methods can be used to estimate the resulting effective radiating temperature of the sun, such as determining the best fit of a blackbody spectrum to the observed solar spectrum. Use two other methods (below), and compare the results to the oft-quoted value of Tsolar = 5780 K. a) Using Wien’s Law and taking the peak of the solar spectrum as 0.50 m, estimate the solar radiating temperature. 1.6 1.Introduction to Radiative Transfer b) Given the measured solar constant in Earth orbit of 1368 W/m2, and using the “inverse square law” for the drop in heat flux with distance, estimate the solar temperature. The mean radius of the Earth’s orbit around the sun as 149×106 km and the diameter of the sun as 1.392×106 km. SOLUTION: a)  T max  C3  2898m  K; Tsolar  2898m  K  5796K 0.50m 1/ 4     2 2 RSun RSun qsolar 2 4   qsolar  1353(W / m )  qSun 2  TSun 2 : TSun  2   REarth orbit REarth orbit R   2 Sun   REarth orbit  b)     1368(W / m2 )  2 1.392  105 / 2 km2  8 2 4  5.6704  10 (W / m K ) 2  149  106 km2        1/ 4           5766K Answers: a) 5796K; b) 5766 K. 1.13 K. The surface of the sun has an effective blackbody radiating temperature of 5780 (a) What percentage of the solar radiant emission lies in the visible range  = 0.4 to 0.7 m? (b) What percentage is in the ultraviolet? (c) At what wavelength and frequency is the maximum energy per unit wavelength emitted? (d) What is the maximum value of the solar hemispherical spectral emissive power? SOLUTION: (a) 1T = 5780 K x 0.4m = 2312 m K ; 2T = 5780 K x 0.7m = 4046 m K F = F02T – F01T = 0.48916 – 0.12240 = 0.36676, or 36.7 % (b) From Fig. 1.2, the UV range is taken as 0.01 to 0.4 m. 1T = 5780 K x 0.01m = 57.8 m K; 2T = 5780 K x 0.4m = 2312 m K. F = F02T – F01T = 0.12240 – 0 = 0.12240, or 12.2 % 1.7 1.Introduction to Radiative Transfer (c) The maximum energy is at max, where, from Table A.4, C3 = 2897.8 m K, so max = 2897.8 / 5780 = 0.50134 m. The corresponding frequency is max = co / = 2.9979×108 (m/s) / 0.50134 x 10-6 (m) = 5.9798×1014 Hz (d) At max T = C3 = 2897.8 m K, using E max b  2C1 5 max e C2 / max T  1 gives Emax,b) =8.301×107 W/(m2·m). Answer: (a) 36.7%; (b) 12.2%; (c) 0.5013 m, 5.98×1014 Hz; (d) 8.301×107 W/(m2·m). 1.14 A blackbody radiates such that the wavelength at its maximum emissive power is 2.00 m. What fraction of the total emissive power from this blackbody is in the range = 0.7 to = 5 m? SOLUTION: Wien’s law gives max T = C3 = 2897.8 m K, so T = C3/max= 2897.8 / 2= 1448.9 K. Thus, 1T = 1448.9 K x 0.7m = 1014.2 m K; 2T = 1448.9 K x 5m = 7244.5 m K. Using Equation 1.37, F = F02T – F01T = 0.82137 – 0.00038= 0.82099, or 82.09 % Answer: 0.821. 1.15 A blackbody has a hemispherical spectral emissive power of 0.03500W/(m2·m) [0.0400W/(m2·m)] at a wavelength of 90 m. What is the wavelength for the maximum emissive power of this blackbody? SOLUTION: With Eb and  given, solve for T from Planck’s equation, Eb = 2C1/ { 5[e(C2/T) – 1]} which becomes T = C2/{ln [1+(2C1/5 Eb)]} Substituting numerical values gives T = 154.7 K, and Wien’s displacement law, maxT = C3, is used to find max = 18.73 m. Answer: 18.73 m. 1.16 A radiometer is sensitive to radiation only in the interval 3.6   8.5m. The radiometer is used to calibrate a blackbody source at 1000K. The radiometer records that the emitted energy is 4600 W/m2. What percentage of the blackbody radiated energy in the prescribed wavelength range is the source actually emitting? 1.8 1.Introduction to Radiative Transfer SOLUTION:  m 3.6 8.5 T, m K 3600 8500 F 0-T 0.4036 0.87417 F1T2T = 0.87417 – 0.4036 = 0.47058 Eb, = F1T2T T4 = 0.47058 x 5.67040×10-8 x 10004 = 26,684 W/m2 Percentage sensed = (4600/26684) x 100 = 17.239% Answer: 17.24 %. 1.17 What temperature must a blackbody have for 25 percent of its emitted energy to be in the visible wavelength region? SOLUTION: There are two solutions to this problem- one for the temperature at which the bulk of the energy is in the IR, and one for a higher temperature where the major portion is in the UV. We can assume a temperature and, by trial and error or by using a root solver, find the fraction in the visible (0.4   0.7), and continue until the fraction is 0.25. For example, for the lower temperature, try T = 4600 K. Then 1T = 4600 K x 0.4m = 1840 m K; 2T = 4600 K x 0.7m = 3220 m K. Using Equation 1.37, F = F02T – F01T = 0.32253 – 0.04422 = 0.27831, or 27.8 %. Try a lower temperature and proceed to convergence. Using a root solver in a computational software package, find T such that 0.7  E (T )d   0.25  0 0 b T 4 1.9 1.Introduction to Radiative Transfer This gives the two results T = 4,343 and 12,460 K. Answer: 4,343 K, 12,460 K. (Note, two solutions are possible!) 1.18 Show that the blackbody spectral intensity Ib increases with T at any fixed value of  SOLUTION: From Equation 1.15, Ib = 2C1/{5 [ e C2/T – 1 ] }. The change in Ib with T for a fixed  is: C 2  C  C2 2C1e T   22  T  I 2 C C e  b   T   1 2    2 2 6 C2 C2   T    5  T 2  T  e  1 T e  1     The values of C1, C2, and  are all positive, so that (Ib / T)is always positive. Hence, Ib must increase with T for every  There are other ways to carry out this proof, such as plotting Ib or showing that Ib itself increases as T increases by noting that the denominator of Planck’s spectral distribution of intensity decreases with increasing T at any fixed wavelength. 1.10 1.Introduction to Radiative Transfer 1.19 Blackbody radiation is leaving a small hole in a furnace at 1200 K (see the figure.) What fraction of the radiation is intercepted by the annular disk? What fraction passes through the hole in the disk? 1.5 cm 1 cm 3 cm Hole in furnace wall SOLUTION: From Table 1.2, the fraction of radiation to an annular disk is: 1.5 2 sin2  2  sin2 1  sin2  2  sin2 1  2 1 .  52 1 3 where sin 1 = 1/(32 + 12)1/2 ; sin 2 = 1.5/(32 + 1.52)1/2. Then Fannular disk = 1.52/(32 + 1.52) – 1/(32 + 12) = 0.2000 – 0.1000 = 0.1000. F hole = sin21 = 1/(32 + 12) = 0.1000 Answer: 0.1000; 0.1000. 1.20 A sheet of silica glass transmits 85% of the radiation that is incident in the wavelength range between 0.38 and 2.7m, and is essentially opaque to radiation having longer and shorter wavelengths. Estimate the percent of solar radiation that the glass will transmit. (Consider the sun as a blackbody at 5780 K.) If the garden in a greenhouse radiates as a black surface and is at 40 °C, what percent of this radiation will be transmitted through the glass? 1.11 1.Introduction to Radiative Transfer Solar radiation Glass Garden at 40 C SOLUTION: Part 1; 1T = 0.38m x 5780 K = 2196.4 m K; 2T = 2.7m x 5780 K= 15606m K. Using Equation 1.37, F = F02T – F01T = 0.96951 – 0.10022 = 0.86929. The percent transmitted is then 0.86929 x 0.85 x 100 = 73.9 %. Part 2: 1T = 0.38m x (40 + 273) K = 118.94 m K; 2T = 2.7m x 313 K= 845.1m K. Using Equation 1.37, F = F02T – F01T = 0.000037 – 0.00000 = 0.000037. The percent transmitted is then 0.000037 x 0.88 x 100 = 0.003 %. Answer: 73.9%; 0.003% 1.21 Derive Wien’s displacement law in terms of wave number by differentiation of Planck’s spectral distribution in terms of wave number, and show that T/max = 5099.4 m K. SOLUTION: From Equation 1.17, Eb = 2C13/[eC2/T – 1 ]. To obtain the maximum, differentiate with respect to , C2  CT2  3 6 C1  e  1  2 C1 C2 / T  e T  E  b       2  CT2    T  e  1   2 C2max  C2max  C    2 max  1  e T Setting this equal to zero gives 3 e T T   for which the solution is (T/max ) = constant. Substitution of (T/max) = 5099.4 m K shows this solution to be valid. 1.12 1.Introduction to Radiative Transfer 1.22 A student notes that the peak emission of the sun according to Wien’s displacement law is at a wavelength of about max = C3/5780 K = 2897.8/ 5780 = 0.501 m. Using max = 1/0.501 m, the student solves again for the solar temperature using the result derived in Homework Problem 1.21. Does this computed temperature agree with the solar temperature? Why? (This is not trivial– put some thought into why.) SOLUTION: Solving for T using max = 1/0.501 m in (T / max) = 5099.4 m K gives T = 5099.4 x 0.501 = 2554.8 K, which is much too small for the solar radiating temperature. The reason is that Eis the energy per unit interval, while E is the energy per unit  interval. The peaks in the curves of E vs  and E vs  do not occur at corresponding values of  and  as related by 1/. (See Homework Problem 1.24 for the correct relation.) 1.23 Derive the relation between the wave number and the wavelength at the peak of the blackbody emission spectrum. (You may use the result of Homework Problem 1.22.) SOLUTION: At a given temperature, Wien’s law in terms of wavelength (in SI units) is T = 2897.8 / max and, from the result of Homework Problem 1.18, in terms of wave number, T = 5099.4 max. At a given T, these may be equated to give  max = 2897.8 / ( 5099.4 max ) = 0.56826 / max where max will be in m-1. For wave number in cm-1 and wavelength in m, the relation is max (cm-1) = 5682.6 / max (m) Answer: max (cm-1) = 5682.6 / max (m). 1.24 A solid copper sphere 3 cm in diameter has on it a thin black coating. Initially, the sphere is at 850 K, and it is then placed in a vacuum with very cold surroundings. How long will it take for the sphere to cool to 200K? Because of the high thermal conductivity of copper, it is assumed that the temperature within the sphere is uniform at any instant during the cooling process. (Properties of copper: density,  = 8950 kg/m3; specific heat, c = 383 J/kg•K.) dT  T 4 A s , or SOLUTION: An energy balance on the sphere gives  cV d cV dT d   . Integrating and using V=D3/6 and As = D2, 4 A s T   cD  1 1  3  3   12430 s  3.453 hr . 18  Tf Ti  Answer: 3.453 hr. 1.13 1.Introduction to Radiative Transfer 1.25 A black rectangular sheet of metal, 6 cm by 10 cm in size, is heated uniformly with 2350 Watts by passing an electric current through it. One face of the rectangle is well insulated. The other face is exposed to vacuum and very cold surroundings. At thermal equilibrium, what fraction of the emitted energy is in the wave number range from 0.33 to 2 m-1? SOLUTION: The energy emitted must be equal to the electrical energy supplied: Qelec = TA4A, or TA = (Qelec/A)1/4 = [2350/(6x10x10-4×5.6704×10-8)]1/4 = 1621.2 K. The wavelengths corresponding to the wave numbers are 1 = 1/2 = 0.5 m; 2 = 1/0.33 = 3.03 m. Then 1T = 0.5×1621.2 = 810.58 m•K and 2T = 2×1621.2 = 4912.6m•K. Using the series expression Equation 1.37 for F0T gives the fraction of emitted energy as 0.6225. Answer: 0.6225. 1.26 Spectral radiation at = 2.445 m and with intensity 7.35 kW/(m2•m•sr) enters a gas and travels through the gas along a path length of 21.5 cm. The gas is at uniform temperature 1000 K and has an absorption coefficient 2.445 m = 0.557 m-1. What is the intensity of the radiation at the end of the path? Neglect scattering, but include emission by the gas. SOLUTION: Taking into account both absorption and emission gives I (S )  I (0)e S  Ib 1  e  S and eS  e0.5570.215  0.8871   For Ib at 2.445 m and 1000 K, Equation 1.15 gives Ib=3.803 kW/(m2 m sr) Substituting into the original equation gives I  7.35  0.8871  3.803(1  0.8871)  6.950kW / (m2  m  sr ) . Answer: 6.950 kW/( m2•m•sr ). 1.27 Radiation from a blackbody source at 2000 K is passing through a layer of air at 12,000 K and 1 atm. Considering only the transmitted radiation (that is, not accounting for emission by the air), what path length is required to attenuate by 35% the energy at the wavelength corresponding to the maximum emission by the blackbody source? (At this , take = 1.200 x 10-1 cm-1 for air at 12,000 K and 1 atm. SOLUTION: From Wien’s displacement law, max  C3 2897m  K   1.4489m T 2000K I S   0.65  e S  e 0.12S I  0  Solving for S yields S = 3.589 cm. Answer: 3.59 cm. Then 1.14 1.Introduction to Radiative Transfer 1.28 A gas layer at constant pressure P has a linearly decreasing temperature across the layer, and a constant mass absorption coefficient m (no scattering). For radiation passing in a normal direction through the layer, what is the ratio I2/I1 as a function of T1, T2, and L? The temperature range T2 to T1 is low enough that emission from the gas can be neglected. The gas constant is R. SOLUTION: For a linear temperature distribution, T1 T T dT   1 2 dx , and for L I1 T(x) I2 x T2 L absorption, dI   ( x )I ( x )dx   m  ( x )I ( x )dx   m P I ( x )dx Integrating from x = 0 to RT ( x ) T2 dT  P L dx  mPL dI  m    I1 I ( x ) R 0 T ( x ) R T1  T2  T1 T ( x ) x = L,  I2  PL T   m ln 1  I  T   mPL I R T T T ln  2  or 2  e  1 2   2  Then, ln  2   I1  I1  R T1  T2   T1   m PL  T1   ln  I R T T T Answer: 2  e  1 2   2  I1 . 1.30 Hawking (1974) predicts that black holes should emit a perfect blackbody radiation spectrum in proportion to the surface gravity of the black hole. Further predictions are that the surface gravity of a black hole is inversely proportional to its mass. Thus, very small black holes should emit very large blackbody radiation. If a black hole with 1 Earth mass has a predicted blackbody temperature of 10-7 K, what fraction of an Earth mass black hole will emit at T = 2.7K (equal to the background temperature of space) and thus possibly be detectable? SOLUTION: The relation between black hole mass and its emitted blackbody temperature is, from the problem statement,  107  T1 m2  3.7  108 mEarth  , so m2  mEarth    T2 m1  2.7  (The Earth mass is about 5.97  1024 kg, so the smaller black hole will have an effective mass of 2.2  1017 kg.) 1.15 1.Introduction to Radiative Transfer 1.30 Observers at NASA noticed an unexpected small deceleration of the early deep space probes Pioneer 10 and 11 as they moved away from Earth. This was known as “the Pioneer Anomaly.” After many speculations and false starts, researchers determined that thermal radiation pressure exerted by radiation from thermoelectric generators heated by small nuclear power sources plus radiated waste heat from the electronics was the culprit. Assuming a projected radiating area of 0.2 m2 facing along the spacecraft trajectory, a spacecraft mass of 258 kg, and a mean radiating temperature on a the outward facing surface (effective emissivity = 0.3) of 375 K, what deceleration was exerted on the spacecraft? [Much more detailed analysis can be found in Turyshev et al. (2012). The detailed thermal model predicts a deceleration of 7 to 10 x10-10 m/s2.] SOLUTION: Taking a = F/m =(PA)/m, the radiation pressure due to emission from the surface is needed. The radiation pressure by emission is (Equation 1.65): P = 2εT4/3c=[3×0.3×5.67×10-8 x 3754(W/m2)/3×2.998×108 (m/s)] =1.12×10-6 Pa. Thus, the deceleration is a = [1.12×10-6 (Pa) x 0.2 (m2)/ 258 (kg)] = 8.68 x 10-10 m/s. 1.16 2. Radiative Properties at Interfaces SOLUTIONS-CHAPTER 2 2.1 A material has a hemispherical spectral emissivity that varies considerably with wavelength but is fairly independent of surface temperature (see, for example, the behavior of tungsten in Figures 3.31 and 3.32. Radiation from a gray source at Ti is incident on the surface uniformly from all directions. Show that the total absorptivity for the incident radiation is equal to the total emissivity of the material evaluated at the source temperatureTi . SOLUTION: For a gray source, the incident radiation is proportional to blackbody radiation at the source temperature; that is, CEb(Ti) d in the spectral range d. From Equation 2.31 the total hemispherical absorptivity is    (TA )dQ,id   0  0 dQ,id Substituting for the incident energy,     0   (TA )CEb  Ti  d   0 CEb  Ti  d   ε E  T  d From Table 2.2,  (T ) = ε (T ) = ε, so that    A  0 A  b Ti i 4   ε E  T  d    For properties independent of surface temperature, ε  0  b Ti 2.2 K. i 4 Using Figure 3.31, estimate the hemispherical total emissivity of tungsten at 2600  SOLUTION: Use a numerical or graphical integration to find ε   ε  E d  0 b T 4 A careful numerical integration with ε (>2.65 m ) = 0.1 gives ε = 0.284. Answer: 0.284. 2.3 Suppose that εis independent of (gray-body radiation). Show that F0T represents the fraction of the total radiant emission of the gray body in the range from 0 to T. 2.1 2. Radiative Properties at Interfaces SOLUTION: The emission in a wavelength interval from  = 0 to is   * 0 ε  Eb d  * and, for all wavelengths the emission is    0 ε  Eb d  . The fraction of energy emitted for the range 0  is, if ε is independent of , Fraction (0   )  ε    * 0 ε  Eb d  *   0 Eb d     E d  *  E d  * 0  b 0 b From Equation 1.33, this is F0T. 2.4 For a surface with hemispherical spectral emissivity ε  , does the maximum of the E distribution occur at the same  as the maximum of the Eb distribution at the same temperature? (Hint: examine the behavior of dE/d.) Plot the distributions of Eas a function of  for the data of Figure 2.9 at 600 K and for the property data at 700 K. At what is the maximum of E? How does this compare with the maximum of Eb? SOLUTION: E  ε  Eb   2 C1ε  d 5   dE d   5   exp C2 / T   1     2 C1ε t   d d d   exp C2 / T   1   C2 /  2  exp C2 /    5 6  5  =2 C1ε t   2    exp C /  T  1   exp C2 / T   1  2    =   C2 /   2 C1ε t  5    1  exp C2 / T    6 exp C2 / T   1  where ε t is the total hemispherical emittance. (C2 /  ) 5 1  exp(C2 / T ) (λT)max=2897.8 m.K. So for the same temperature, the maximum occurs at the same point as Eλb. Only the intensity of this power is reduced by the numerical value of emittance. For data in Figure 2.9, the following figure is obtained Setting the result = 0 to find the maximum; 2.2 2. Radiative Properties at Interfaces 2.3 2. Radiative Properties at Interfaces Eλ is maximum at 2897.8/600=4.83 m which overlaps with the plot. Eλ is maximum at 2897.8/700=4.14 m which overlaps with the plot. 2-5 Find the emissivity at 400 K and the solar absorptivity of the diffuse material with the measured spectral emissivity shown in the figure. SOLUTION: The emissivity using Equation 2.10 is   ε  T  E T  d  ε (T )   b 0  0.83  1.90  0 T 4 Eb T  d  T 4  0.50  2.80  1.93 Eb T  d  T 4  0.17     2.80 Eb T  d  T 4  0.83F01.90T  0.50F1.90T 2.80T  0.17F2.80T   0.83F01.90T  0.50  F02.80T  F01.90T   0.17 1  F02.80T  so only two blackbody fractions need be found. Using Equation 1.37 for the F values, ε (T  5780K )  0.83F01.905780  0.50  F02.805780  F01.905780   0.17 1  F0 2.805780   0.83  0.9316  0.50   0.9745  0.9316   0.17  (1 0.9745)  0.7990. Answer: ε (T=400 K) = 0.17; s(T=5780 K) = 0.7990 2.4 2. Radiative Properties at Interfaces 2.6 The surface temperature-independent hemispherical spectral absorptivity of a surface is measured when it is exposed to isotropic incident spectral intensity, and the results are approximated as shown below. What is the total hemispherical emissivity of this surface when it is at a temperature of 1000 K? 1.0 0.75  0.50 0.25 0 1.2 2.0 (2.8)   Eb  TA  d  SOLUTION: ε  =  from Table 2.2 gives ε  0  so that E T d   b  A  m 0 ε = 0.75 F01200 + 0.5 F 12002000 = 0.75 x 0.002134 + 0.5 x (0.06673 – 0.002134) = 0.03390. Answer: 0.0339. 2.7 (a) Obtain the total absorptivity of a diffuse surface with properties given in the figure for incident radiation from a blackbody with a temperature of 6200 K. (b) What is the total emissivity of the diffuse surface with properties given in the figure if the surface temperature is 500 K? 1.0 0.95 =ε 0.15 0 1.5 m 2.5 2. Radiative Properties at Interfaces SOLUTION:  (a) Using      q d  with q,i = Eb(6200K) and Equation 1.37 for the blackbody  q d  0 ,i  0 ,i fractions gives =0.95 F01.5×6200+0.15 (1-F01.5×6200) = 0.95 x 0.8975+0.15(1-0. 0.8975)= 0.8680 (b) Similarly, for emission, ε=0.95 F01.5×500+0.15 (1-F01.5×500) = 0.95 x (0)+0.15[1-(0)] = 0.15. Answer: (a) 0.8680; (b) 0.15. 2.8 For the spectral properties given in the figure for a diffuse surface: (a) what is the solar absorptivity of the surface (assume the solar temperature is 5800 K)? (b) what is the total hemispherical emissivity of the surface if the surface temperature is 700 K?  1.0 0.9  = ε 0.1 0 1.2 3 Wavelength, m  SOLUTION: The definitions of ε and  in terms of ε  and  are used as in Problems 2.5 and 2.6. Using Equation1.37, (a) λ = 0.9F0-1.2×5800 + 0.1(F0-3×5800 – F0-1.2×5800) + 1(1 – F0-3×5800 ) = 0.72505 + 0.01704 + 0.02399 = 0.76608 For a diffuse surface, Table 2.2 gives ε  = Then (b) ε = 0.9F0-1.2×700 + 0.1(F0-3×700 – F0-1.2×700) + 1(1-F0-3×700) = 0 + 0.00830 + 0.91695 = 0.92528. Answer: (a) 0.766; (b) 0.925. 2.6 2. Radiative Properties at Interfaces 2.9 A white ceramic surface has a hemispherical spectral emissivity distribution at 1600 K as shown. What is the hemispherical total emissivity of the surface at this surface temperature? 1.0 0.8 0.6 ε  0.4 0.2 0 0 2 4 6 8 10 12 , m SOLUTION: Numerical or graphical integration is necessary, as no analytical integration appears possible even for this simple variation in spectral emissivity. From Equation 2,10 with Eλb=Iλb   ε  1600K  E 1600K  d  ε 1600K    0 b  1600  4 Now, ε 1600K       1   1 1 1 2  ε E d  ε ε 1  ε 2  ε 1    Eb d  4  0 1  b 4   TA TA  2  1    1 3 ε E d TA4  2 2 b where ε 1 =0.2, ε 2 =0.8, λ1=3 m, λ2=7 m, λ3= 10 m Numerical Romberg integration of the ε (1600 K) equation gives 0.28128. Answer: 0.281 2.7 2. Radiative Properties at Interfaces 2.10 A surface has the following values of hemispherical spectral emissivity at a temperature of 800K. ε (800 K)  m 8 0 0 0.2 0.4 0.6 0.8 0.8 0.8 0.7 0.6 0.4 0.2 0 0 (a) What is the hemispherical total emissivity of the surface at 800 K? (b) What is the hemispherical total absorptivity of the surface at 800 K if the incident radiation is from a gray source at 1800 K that has an emissivity of 0.815? The incident radiation is uniform over all incident angles. SOLUTION: (a) 0.8 0.6 ε (750K) 0.4 0.2 0 0 1 2 3 4 m From Equation 2.10, 2.8 5 6 7 8 2. Radiative Properties at Interfaces   ε  800K  E 800K  d  ε (800K )  b 0 T 4 A    1  1 2  ε  ε 2  ε 1  E  800K  d  4    1 1 2  1  b  TA     2  1 3  ε  ε 3  ε 2  E  800K  d  4    2 2 3  2  b  TA     3  1 4  ε  ε 4  ε 3  E  800K  d  4    3 3 4  3  b  TA  where TA = K, ε 1=0, ε 2=0.8, ε 3=0.8, ε 4=0, 1=1 m, 2 = 3 m, 3 = 4 m, 4 = 8 m. Accurate numerical integration of the ε ( K) equation gives 0.43826. (b) From Equation 2.25     800K  0.815E 1800K  d    ε  800K  E 1800K  d   (800K )    1800  0.815E 1800K  d  b 0 0  b 4 b 0  2    1  1 ε  ε 2  ε 1  E 1800K  d  4    1 1 2  1  b  1800   3    2  1 ε  ε 3  ε 2  E 1800K  d  4    2 2 3  2  b  1800   4    3  1 ε  ε 4  ε 3  E 1800K  d  4    3 3 4  3  b  1800  Numerical integration of the    0 ε  Ebd  gives 0.42709. Answer: (a) 0.438; (b) 0.427. 2.11 Find the emissivity at 950 K and the solar absorptivity of the diffuse material with the measured spectral emissivity shown in the figure. This will require numerical integration. 2.9 2. Radiative Properties at Interfaces SOLUTION: The emissivity in the various ranges can be expressed as 0    0.70 : ε   0.83    0.70  0.70    1.90 : ε   0.83  0.33    1.20  1.90    2.80 : ε   0.50    2.80  2.80    3.50 : ε   0.50  0.33    0.70    3.50;ε   0.17 The total emissivity is then   ε  T  E T  d  ε (T )   b 0 T 4     0.70   0.83  0.33  1.20   Eb T  d      0   4 4 T T 3.50     2.80   2.80 0.50  0.33    Eb T  d  0.50  Eb T  d   2.80  0.70    1.90   4 4 T T  0.83  0.70 0.17   Eb T  d   3.50 1.90  0.70 Eb T  d  T 4 The blackbody fractions can be used to evaluate the first, third and fifth integrals, but the second and third probably require numerical integration. The results using numerical integration are ε (950 K) = 0.0000+0.0684+0.1185+0.0556+0.0823 = 0.3249; ε (5780 K) = 0.4060+0.3225+0.0215+0.0041+0.0024 = 0.7564. Answer: ε (950 K) = 0.3248; ε (5780 K) =s(5780 K) = 0.7564. 2.10 2. Radiative Properties at Interfaces 2.12 A diffuse surface at 1000 K has a hemispherical spectral emissivity that can be approximated by the solid line shown. (a) What is the hemispherical-total emissive power of the surface? What is the total intensity emitted in a direction 60° from the normal to the surface? (b) What percentage of the total emitted energy is in the wavelength range 5 < < 10 m? How does this compare with the percentage emitted in this wavelength range by a gray body at 1000 K with an emissivity ε = 0.611? 1.0 ε  0.8 0.7 0.55 0.45  0.2 0 0 1 4 5 7  10 , m SOLUTION:   ε  T  E T  d  ; the blackbody fractions for each spectral (a): ε T    0 A b A T band are obtained from Equation 1.37 as given in the table below: A range ε 4 A T range F0T – F0T F0T ε  F0T  0.000032 0.00032 0 0.48086-0.00032 0.48054 0.21624 0.63371-0.48086 0.15285 0.10699 0.80793-0.63371 0.17422 0.13937 0.9134-0.80793 0.10547 0.05801 1 – 0.9134 0.08660 0.017321 ε  ε F0T = 0.53800 = ε T4 = 0.53800x 5.6704×10-8 x 9004 = 30506.9 W/m2 0-1 1-4 4-5 5-7 7 – 10 10 –  0.2 0.45 0.7 0.8 0.55 0.2 0 – 1000 1000-4000 4000-5000 5000-7000 7000-10000 10000 –  Eb Ib =Eb/ = 9710.7 W/m2•sr (b) For the range 5<<10, the percentage is [(0.13937+0.05801)/ 0.53800]x100=36.69%. For a gray surface, the percentage is the same as for a black surface, or 100 (F09000 – F04500) = 100 (0.91339 – 0.63371) = 27.97 %. Answer: (a) 30507 W/m2; 9711 W/m2•sr (b) 36.69 %; 27.97 % 2.11 2. Radiative Properties at Interfaces 2.13 The ε  for a metal at 1000 K is approximated as shown, and it does not vary significantly with the metal temperature. The surface is diffuse. 1.0 ε  0.6 0.35  0.15 0 0 2 , m 4 6  (a) What is  for incident radiation from a gray source at 1200 K with εsource = 0.822? (b) What is for incident radiation from a source at 1200 K made from the same metal as the receiving plate? SOLUTION:      ε   0.822E 1200K  d   (a)     range ε 0-2 2-4 4-  0.6 0.35 0.15  0.822 12004 ε  F Ti range F0Ti – F0Ti b 0 F ε F 0-2400 0.14026- 0 0.14026 0.08415 2400-4800 0.60753-0.14026 0.46727 0.16355 1 – 0.60753 0.39247 0.05887 4800-      =  ε F = 0.30657 (b)  =  (F) / (F) = [ 0.62 x 0.14026 + 0.352 x 0.46727 + 0.152 x 0.39247 ] / 0.30657 = 0.38022 Answer: (a) 0.30657; (b) 0.38022. 2.14 The directional total absorptivity of a gray surface is given by the expression () = 0.450cos2where  is the angle away from the normal to the surface. (a) What is the hemispherical total emissivity of the surface? (b) What is the hemispherical- hemispherical total reflectivity of this surface for diffuse incident radiation (uniform incident intensity)? 2.12 2. Radiative Properties at Interfaces (c) What is the hemispherical-directional total reflectivity for diffuse incident radiation reflected into a direction 75° from the normal? SOLUTION: ε   (a) 1 4    cos  d   2 0  /2 0.9 cos4   0.450 cos  d    0.2250   0 4 0 4 3 (b)  ε 1 – 0.2250= 0.7750. (c) Using Equation 2.43, (r,r) =(,) = 1 – () = 1 – 0.450 cos2 (75°) = 0.9699. Answer: (a) 0.225; (b) 0.775; (c) 0.97. 2.15 Using Figure 3.24, estimate the total absorptivity of typewriter paper for normally incident radiation from a blackbody source at 1200 K. SOLUTION: Use Equations 2.81 and 2.39 for a gray surface and assume that there is no dependence on circumferential angle . From the plot we have symmetry for the values of .    0,TA   1     0,TA   1  2   /2  r 0    0,r ,TA  cos r sinr dr  1  0.1079   0.8921 Evaluation requires numerical integration. Answer: 0.8921. 2.16 A gray surface has a directional emissivity as shown in the figure. The properties are isotropic with respect to circumferential angle . (a) What is the hemispherical emissivity of this surface? (b) If the energy from a blackbody source at 650 K is incident uniformly from all directions, what fraction of the incident energy is absorbed by this surface? (c) If the surface is placed in a very cold environment, at what rate must energy be added per unit area to maintain the surface temperature at 1000 K? 2.13 2. Radiative Properties at Interfaces SOLUTION: (a) From Equation 2.9, ε  2 ε    cos  sin  d  2    /2 1/2 0  sin 0 0.9 sin  d  sin     1 sin 1/2 0.5 sin  d  sin      1  1   1/2 1  0 1   2 2     sin  sin  4   0.5 4   0.600   2 0.9   2 0.9  0.5   2 2 2 2     sin 0 sin 1/2       (b) Because the surface is gray, from Table 2.2,  ε  Hence, = 0.600 = fraction absorbed. (c) Emissive power = ε T4 = 0.600 x 5.67040×10-8 x 10004 = 34,022W/m2. Answer: (a) 0.600; (b) 0.600; (c) 34,022 W/m2 . 2.17 Using Figure 3.44, estimate the ratio of normal total solar absorptivity to hemispherical total emissivity for aluminum at a surface temperature of 650 K with a coating of 0.1-m dendritic lead sulfide crystals. Assume the surface is diffuse. (The solar temperature can be taken as 5780 K.) SOLUTION: From the figure, approximate the normal hemispherical spectral reflectivity as shown below. Using n = 1 – n, the spectral normal absorptivity is approximated as shown below: 2.14 2. Radiative Properties at Interfaces 0.9 0.8  ,n ,n  0.2 0.1 0 0 3 3 m m For n,solar, use Tsolar = 5780 K: then (T)cutoff = 3×5780 = 17340 m•K, and F0(T)cutoff = 0.97880 [Equation 1.37]. Then n,solar = 0.9 F0 (T)cutoff + 0.2 (1 – F0 (T)cutoff) = 0.9 x 0.97880 + 0.2 ( 1 – 0.97880 ) = 0.8852 ε (T = 650 K)  ε n(T = 650 K) = 0.9 F01950 + 0.2 (1 – F01950) = 0.9 x 0.05920 + 0.2 (1 – 0.05920 )= 0.2414 (NOTE: This assumes that the hemispherical ε  ε n.) Thus, n,solar /(T = 650 K) = 0.8852/0.2414 = 3.6662. Answer: 3.666. 2.18 A gray surface has a directional total emissivity that depends on angle of incidence as ε  = 0.788 cos . Uniform radiant energy from a single direction normal to the cylinder axis is incident on a long cylinder of radius R. What fraction of energy striking the cylinder is reflected? What is the result if the body is a sphere rather than a cylinder? Qincident R SOLUTION: () = ε () = 0.788 cos . () = 1 – () = 1 – 0.788 cos . For the cylinder, the intercepted energy on dA per unit length of cylinder is q dA cos  = q (Rd cos , and the reflected energy per unit length is q (Rd cos  ( 1 – 0.788 cos  ). The ratio is  /2 reflected q  0 R cos  1  0.788 cos   d 0.788   1  0.3811  /2 incident 4 q R cos  d  0 2.15 2. Radiative Properties at Interfaces For the sphere, the intercepted energy by a ring element is R sin dA = R d  R q 2R sin Rd cos, and the ratio of reflected to incident energy is  /2 reflected q  0 2 R sin  cos  1  0.788 cos   d   /2 incident q 2 R 2 sin  cos  d 2  0  /2   sin2  0.788 cos3      3  2 0  /2 sin2  2 0 Answer: ; 1 0.788  2 3   0.4747 1 2 . 2.19 A flat metal plate 0.1 m wide by 1.0 m long has a temperature that varies only along the long direction. The temperature is 900 K at one end, and decreases linearly over the one meter length to 350 K. The hemispherical spectral emissivity of the plate does not change significantly with temperature but is a function of wavelength. The wavelength dependence is approximated by a linear function decreasing from ε  = 0.85 at  = 0 to ε  = 0.02 at  = 10 m. What is the rate of radiative energy loss from one side of the plate? The surroundings are at a very low temperature. SOLUTION: The rate of energy loss is given by the total emissive power integrated over the plate length, or Q  0.1(m2 )  0.1 1 1  10 x 0  0 10  ε  Eb T  d dx 2 C1 0.83    0.85  10  d  dx  416.6W .     C   2  3 exp    1     350  550 x    where the final result is obtained by numerical integration of the double integral. Answer: 416.6 W. x 0 0 2.16 2. Radiative Properties at Interfaces 2.20 A thin ceramic plate, insulated on one side, is radiating energy from its exposed side into a vacuum at very low temperature. The plate is initially at 1200 K, and is to cool to 300 K. At any instant, the plate is assumed to be at uniform temperature across its thickness and over its exposed area. The plate is 0.25 cm thick, and the surface hemispherical-spectral emissivity is shown in the figure and is independent of temperature. What is the cooling time? The density of the ceramic is 3200 kg/m3, and its specific heat is 710 J/(kg•K).  1.0 0.85 ε   0 1.0 10 m SOLUTION: For an area A, the energy equation is  2C1 dT  Vc  qA  A  ε  d  0 dt  C2   5  exp    1  T    0.85    1  0 9  A 10  2C1 2C1 d  A  0.85 d 10  C2    C2   5 5  exp   exp    1   1  T    T     Integrating gives  cV 1200 t A  T 300 10 0.85    1 1 9  dT  2 C1 2 C1 d   0.85 5 d  5   10  exp C2 / T   1  exp C2 / T   1 Numerical integration is required, and results int = 1796 s = 29.9 min. (Note: V/A = 0.25×10-2 m.) Answer: t= 29.9 min. 2.17 3. Radiative Properties of Opaque Materials SOLUTIONS-CHAPTER 3 3.1 An electrical insulator has a refractive index of n = 1.332 and has a smooth surface radiating into air. What is the directional emissivity for the direction normal to the surface? What is it for the direction 60o away from the normal? SOLUTION: From Equation 3.12 with n1 = 1, n2 = 1.332, εn = 1 – [(n2 – 1)/(n2+1)]2 = 1 – [(1.332- 1)/( 1.332+ 1)] 2 = 0.9797. Find  from Equation 3.3 using n1 = 1, n2 = 1.332,  = 60°. sin  = (n1/n2 ) sin  = (1/1.332) sin 60° = 0.6502 Then = sin -1(0.6502) = 40.55°. ε(60°) = 1 – (60°); using Eq. 3.7a ε(70°) = 1 – (1/2)[sin2()/sin2()] {1 + [cos2()/cos2()]} = 1 – (1/2)[sin2(19.45°)/ sin2(100.55°)] {1+[cos2(100.55°)/cos2(19.45°)]} = 1 – (1/2)(0.1108/0.9665)2 [1 + (-0.1831/0.8891)2] = 0.9405. Answer: 0.9797; 0.9405. 3.2 A smooth hot ceramic dielectric sphere with an index of refraction n = 1.43 is photographed with an infrared camera. Calculate how bright the image is at locations B and C relative to that at A. (Camera is distant from sphere.) Camera A 1.5 cm 1.9 cm B C 2 cm 3.1 3. Radiative Properties of Opaque Materials SOLUTION: A blackbody has the same intensity for all directions. Hence, to compare intensities (brightness), compare the directional emissivities.  1.5  B 2  B = sin-1(1.5/2) = 48.590°; B = sin-1(1.5/2n) = 31.633° C = sin-1(1.9/2) = 71.805°; C = sin-1(1.9/2n) = 41.631° ε(=0) = A = 1 – [(n-1) / (n+1)]2 = 1 – (0.43/2.43)2 = 0.968. ε  1 2 2 1 sin      cos      1    2 sin2      cos2      εB  1 2 2 1 sin 16.957   cos  80.223   1     0.9548 2 sin2  80.223   cos2 16.957   2 2 1 sin  30.174   cos 113.436   εC  1  1    0.8182 2 sin2 113.436   cos2  30.174   Thus, εB/εA = 0.9548/0.968 = 0.986; εC/εA = 0.8182/0.968 = 0.845. Answer: 0.986; 0.845. 3.3 A particular dielectric material has a refractive index of n = 1.350. For a smooth radiating surface, estimate: (a) the hemispherical emissivity of the material for emission into air. (b) the directional emissivity at  = 60° into air. (c) the directional hemispherical reflectivity in air for both components of polarized reflectivity. Plot both components for n=1.350 on a graph similar to Figure 3.5. Let  be the angle of incidence. SOLUTION: (a) From Equation 3.11, ε= (1/2) – [ (3 x 1.350 + 1) x (1.350 -1)]/[6 x (1.350 +1 )2] – {1.3502x(1.350 2 1)2/(1.350 2 + 1)3} ln[(1.350 – 1)/( 1.350 + 1)] + 2 x 1.350 x (1.350 2 + 2×1.350 -1)/[ (1.350 2 + 1) x (1.350 4 – 1)] 3.2 3. Radiative Properties of Opaque Materials – {8 x 1.350 4x (1.350 4 + 1)/[( 1.350 2 + 1) x (1.350 4 – 1)2]} ln (1.350) = 0.9309 The result may also be found from Figure 3.3. (b) From Figure 3.2, ε(= 60°) = 0.85. Could also use Equation 3.6 to obtain 39.9°, substitute  into Equation 3.9 for(  )=0.0629, and use ε = 1 – . This gives ε(= 60°) = 0.9371. (For this particular set of parameters, the hemispherical emissivity and the directional emissivity at  = 60° are nearly the same.) Answer: (a) 0.9309, (b) 0.9371. 3.4 A smooth dielectric material has a normal spectral emissivity of ε,n = 0.725 at a wavelength in air of 6 m. Find or estimate values for: (a) the hemispherical spectral emissivity εat the same wavelength. (b) the perpendicular component of the directional hemispherical spectral reflectivity   , () at the same wavelength and for incidence at  = 40°. SOLUTION: From Equation 3.12 εn = 4n/(n+1)2 ; n2 + 2n + 1 = 4n/0.725 ; so n = 3.2053 (or 0.3120, but n cannot be < 1). a) From Figure 3.3: ε/εn = 0.965; so  = 0.965×0.725 = 0.70. Equation 3.8 gives 0.7034 b) From Equation 3.8,   , () = {[(3.2053 2-0.4132)1/2 – 0.7660]/[( 3.2053 2-0.4132]1/2 + 0.7660]}2 = 0.3694. Answer: (a) 0.703; (b) 0.369. 3.5 An inventor wants to use a light source and some Polaroid glasses to determine when the wax finish is worn from her favorite bowling alley. She reasons that the wax will reflect as a dielectric with n = 1.40, and that the parallel component of light from the source will be preferentially absorbed and the perpendicular component strongly reflected by the wax. When the wax is worn away, the wood will reflect diffusely. At what height should the light source be placed to maximize the ratio of perpendicular to parallel polarization from the wax as seen by the viewer? h=? 2m 20 m 3.3 3. Radiative Properties of Opaque Materials SOLUTION: To make   / II a maximum, use Brewster's angle (Equation 3.7 et seq.): = tan-1(n2/n1) = tan-1(1.40) r 2m h=?  20 – x x 20 m Since reflection is specular, r =  Thus, x/2 = tan = 1.40 = (20-x)/h, so x = 2.80 m, h = 12.286 m (a high ceiling!!) Answer: 12.3 m. 3.6 A smooth ceramic dielectric has an index of refraction n = 1.58, that is independent of wavelength. If a flat ceramic disk is at 1100 K, how much emitted energy per unit time is received by the detector when it is placed at  = 0° or at  = 60°? Use relations from electromagnetic theory. 0.6 cm Detector 0.6 cm 0.3 m n Detector  0.3 m Disk 0.5 cm SOLUTION: Neglect reflected energy from the surroundings. Then the energy reaching the detector is I cos  d dA, where dA = ( x 0.0052/4); d = ( x 0.0062/4) (1/0.32). In the normal direction, εn = 4n/(n+1)2 = 4×1.58/(2.58)2 = 0.9495. Then the energy for ( = 0) = [ 11004/] ( x 0.0062/4) (1/0.32) ( x 0.0052/4) x 0.9495 = 1.5477×10-4 W. At  = 60°,  = sin-1 (sin 60°/1.58) = 33.24°, so 3.4 3. Radiative Properties of Opaque Materials    75   2 2 1 sin      cos      1     0.1021 2 sin2      cos2      ε() = 1 – () = 0.8979. The energy rate at  =60° is then =[11004/] cos(60°) (x 0.0062/4) (1/0.32) (x 0.0052/4) x 0.8979= 7.3183×10-5 W. Answer: 15.48×10-5 W at = 0°; 7.318×10-5 W at  = 60°. 3.7 At a temperature of 300 K these metals have the resistivities (Table 3.2): Copper Gold Aluminum 1.72 x 10-6 Ohm-cm 2.27 x 10-6 Ohm -cm 2.73 x 10-6 Ohm -cm What are the theoretical normal total emissivities and hemispherical total emissivities of these metals, and how do they compare with tabulated values for clean, unoxidized, polished surfaces? SOLUTION: Using Equation 3.40 for the normal total emissivity : εn= 0.578 ( reT)1/2 – 0.178 reT + 0.0584 (reT)3/2 and Eq. 3.45 for the total hemispherical emissivity: ε= 0.776 (reT)1/2 – [0.309 – 0.0889 ln(reT)]reT – 0.0175 (reT)3/2 (reT)1/2 Material reT Copper 516×10-6 22.7×10-3 681×10-6 26.1×10-3 819×10-6 28.6×10-3 Gold Aluminum εn ε 11.72×10-6 17.7×10-6 0.0130 0.0171 0.02 0.0150 0.0196 0.018 23.4 x10-6 0.0164 0.0214 0.04 (reT) 3/2 εn (Table B.1) Alternatively, using Equation 3.39 gives for the normal value εcopper = 0.0131; εgold = 0.0150; εaluminum = 0.0164. Answer: εn: 0.0131; 0.0157; 0.0168; ε: 0.0170. 0.0202, 0.0217. 3.8 A highly-polished metal disk is found to have a measured normal spectral emissivity of 0.095 at a wavelength of 12 m. What is: (a) the electrical resistivity of the metal (Ohm -cm) ? (b) the normal spectral emissivity of the metal at  = 10 m? (c) the refractive index n of the metal at  = 10 m? (Note any assumptions that you make in obtaining your answers.) SOLUTION: (a) From Equation 3.35 (assumes n =  for large wavelength.), εn = 0.095 = 36.5 (re/o)1/2 – 464 (re/0); 0 = 12 m Solve for re = 8.7161 x 10-5 Ohmcm. 3.5 3. Radiative Properties of Opaque Materials (b) Substitute the re from part (a) into Equation 3.35 with 0 = 10 m to obtain εn ( = 10 m) = 0.1037. (c) Can solve using Equation 3.33 or Figure 3.6: Using the figure with n = , n = about 18.2. Using Equation 3.33 with εn(= 10 m) = 0.1037 and solving for n gives n = 18.26. Answer: (a) 8.7161 x 10-5 Ohm -cm; (b) 0.104; (c) 18.3. 3.9 Show using Equation 3.7 that the parallel component of reflectivity becomes zero n  when   tan 1  2  .  n1  SOLUTION: 2 1/2 2   n  2     n  2 cos     2  sin2      i i   n1   n1       i    1/2  2  n  2     n2  2    cos  i    2   sin  i    n1      n1  To make     0, the numerator must be = 0, so must prove that 1/2 2  n  2   n2  2 2   cos    sin        n n    1  1  n   0 . Substituting  2   tan  gives  n1  1/2 2  n  2   n2  2 2   cos    sin         n1    n1  1/2  sin2   sin2  2    sin   cos    cos2     tan2  cos     tan2   sin2   1/2 1/2   sin2  1   sin    1  2 cos    cos    1/2 1/2       1  cos2     1    sin   tan     1   sin   tan      2 2  cos      cos       1/2   sin2       sin   tan       sin   tan   tan    0 2  cos      (Must choose negative sign, or reflectivity can be < 0.) 3.10 A clean metal surface has a normal spectral emissivity of ε,n = 0.06 at a wavelength of 12 m. Find the value of the electrical resistivity of the metal. SOLUTION: From Equation 3.23b 3.6