Solution Manual For Thermal Radiation Heat Transfer, 6th Edition

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solutions MAnuAlFoRbyThermalradiaTionheaTTransferJohnr.howellrobertsiegelm. Pinarmenguc

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Solutions to HomeworkProblemsTable of ContentsChapter 1Introduction to Radiative Transfer…………..…………….…1.1Chapter 2Radiative Properties at Interfaces………….……....…….….2.1Chapter 3Radiative Properties of Opaque Materials……..…..……….3.1Chapter 4Configuration Factors for Diffuse Surfaces withUniform Radiosity….…...……….……..……………………...4.1Chapter 5Radiation Exchange in Enclosures Composed of Blackand/or Diffuse-GraySurfaces………………….…………….5.1Chapter 6Exchange of Thermal Radiation among NondiffuseNongray Surfaces…..……………………………....…………6.1Chapter 7Radiation Combined with Conduction and Convectionat Boundaries….....…………………...…..…………..……….7.1Chapter 8Inverse Problems in Radiative Heat Transfer…….......….…8.1Chapter 9Properties of Absorbing and Emitting Media…………….….9.1Chapter 10Fundamental Radiative Transfer Relations…..……......….10.1Chapter 11Radiative Transfer in Plane Layers andMultidimensional Geometries………………...……..……...11.1Chapter 12Solution Methods for Radiative Transfer inParticipating Media….………………………………..……. .12.1

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Chapter 13Conjugate Heat Transfer in Participating Media……….….13.1Chapter 14Electromagnetic Wave Theory……………...….………..….14.1Chapter 15Absorption and Scattering by Particlesand Agglomerates…………………………...….………...….15.1Chapter 16Near-Field Thermal Radiation……………………..…..……16.1Chapter 17Radiative Effects in Translucent Solids,Windows, and Coatings………………..……….…………...17.1

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1.Introduction to Radiative Transfer1.1SOLUTIONS- CHAPTER 11.1What are the wave number range in vacuum and the frequency range for thevisible spectrum (0.4 to 0.7m)? What are the wave number and frequency values atthe 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-1Answer: 2.5 x 106 to 1.4286 x 106 m-1; 4.2827 x 1014 to7.4948 x 1014 s-1; 4 x 104 m-1; 1.1992 x 1013 s-1.1.2Radiant energy at a wavelength of 2.0m is traveling through a vacuum. It thenenters 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 cm = 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; 5x105 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

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1.Introduction to Radiative Transfer1.21.3Radiation propagating within a medium is found to have a wavelength within themedium of 1.570m and a speed of 2.500x108m/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 = 2.99792458x108 m/s;m = 1.570x10-6 m; cm = 2.500x108m/s.(a) nm = c0/cm = 1.199(b)0= nmm = 1.882x10-6 m = 1.882m.Answer: (a) 1.199; (b) 1.882m.1.4What range of radiation wavelengths are present within a glass sheet that has awavelength-independent refractive index of 1.33 when the sheet is exposed in vacuumto 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.1.5A material has an index of refractionn(x)that varies with position x within itsthickness. Obtain an expression in terms ofco andn(x)for the transit time for radiationto pass through a thicknessL. Ifn(x)=ni(1 +kx), whereni andkare constants, what isthe relation for transit time? How does wave number (relative to that in a vacuum) varywith position within the medium?SOLUTION: n(x) = co / c(x), so c(x) = co / n(x) = dx/d. Then200000112( )()LLiixxnnkLtn x dxkx dxLcccc =m;= c /m = co /oco / c = n =o /m;m /o = 1/n som/o= n =ni(1 + kx).Answer:nicoL + kL22; ni(1 + kx).

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1.Introduction to Radiative Transfer1.31.6Derive Equation1.26by analytically finding the maximum of theEb/T5vs.Trelation (Equation1.20).SOLUTION:Take the derivative ofEb/T5with respect toT. To simplify notation, let.T551125226522122212622(/)22/ exp(/)1()()()exp(/)1/ ()exp(/)2[ 5] / exp(/)1exp(/)12/[ 5exp(/)11exp(/)bd ETCddCCdddCCCCCCCCCCSetting the result = 0 to find the maximum,22maxmax2max2max//51exp(/)1exp/CTCCCTClearly, the product (T)maxat the maximum of the Planck curve is equal to a constant,whichmustbe found by iteration.Using iteration or a root-finding program gives3max2897.8TCm K.1.7A 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 blacksurface at= 3.75m?(b) What is the spectral intensity emitted at= 45° with respect to the normal ofthe 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, andwhat 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,21,/521b nCTCIe= 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,

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1.Introduction to Radiative Transfer1.42max1,max/5max21bCTCIe= 12499.08 W/m2•m•sr(e) From Equation 1.32 and Table A-4, Eb =T4= 5.67040x10-8 W/m2•K4 x (1250)4 K4 = 138,437.5 W/m2Answers: (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 .1.8Plot the hemispherical spectral emissive powerEb for a blackbody in air[W/(m2·m)] as a function of wavelength (m) for surface temperatures of 2000 and6250K.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 .

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1.Introduction to Radiative Transfer1.51.9For 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 = 4.09570x10-12 x 22505 = 236.17 kW /m•m2•sr(b) Eb =T4 = 5.67040x10-8(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 = 0.56429F2T1T =0.41336Eb(12)=0.3482T4=600.730kW/m2(d) Using21,/521b nCTCIe,[ Ib(=2) ] / [Ib(=8) ] = 1.5861x105/ 2.9695x103 = 53.4139Answers: (a) 236.17 kW/m2•m•sr; (b)1453.26 kW/m2;(c) 600.730 kW/m2; (d) 53.4139

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1.Introduction to Radiative Transfer1.61.10Determine the fractions of blackbody energy that lie below and above the peak ofthe blackbody curve.SOLUTION:Atthepeak,maxT=2893mK.UsingEquation1.37with214388()4.9652897.8()Cm KTm Kgives02897.80.25005F. Therefore about 25 percent ofthe blackbody energy is at wavelengths below the peak, and 75 percent is at longerwavelengths.1.11A 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 wavelengthsof= 2.0m and= 5m?(c) At what wavelength does the peak energy in the radiated spectrum occur forthis 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.)Using21521/bCTCIe,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=0.59656x 5.67040x10-8(W/m2•K4) x (1250)4(K4)=82587.2W/m2Answer:(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 areoccurring. Various methods can be used to estimate the resulting effective radiatingtemperature of the sun, such as determining the best fit of a blackbody spectrum to theobserved solar spectrum. Use two other methods (below), and compare the results tothe oft-quoted value ofTsolar= 5780 K.a)Using Wien's Law and taking the peak of the solar spectrum as 0.50m,estimate the solar radiating temperature.

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1.Introduction to Radiative Transfer1.7b) Given the measured solar constant in Earth orbit of 1368 W/m2, and using the"inversesquarelaw"forthedropinheatfluxwithdistance,estimatethesolartemperature. The mean radius of the Earth's orbit around the sun as 149x106km andthe diameter of the sun as 1.392x106km.SOLUTION:a)3solarmax2898 m KTC2898 m K;T5796K0.50 mb)1/ 42224SunSunsolarsolarSunSunSun222Earth orbitEarth orbitSun2Earth orbit1/ 42252824262RRqq1353(W / m )qT:TRRRR1368(W / m )5766K1.39210/ 2km5.670410(W / m K )14910km  Answers:a) 5796K; b) 5766 K.1.13The surface of the sun has an effective blackbody radiating temperature of 5780K.(a) What percentage of the solar radiant emission lies in the visible range= 0.4to 0.7m?(b) What percentage is in the ultraviolet?(c) At what wavelength and frequency is the maximum energy per unitwavelength emitted?(d) What is the maximum value of the solar hemispherical spectral emissivepower?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, or36.7 %(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 %

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1.Introduction to Radiative Transfer1.8(c) The maximum energy is atmax, where, from Table A.4, C3 = 2897.8mK, somax = 2897.8 / 5780 = 0.50134m.The corresponding frequency ismax = co /= 2.9979x108 (m/s) / 0.50134 x 10-6 (m) = 5.9798x1014 Hz(d) Atmax T = C3 = 2897.8m K, using 2max1max bC /T5max2 CEe1gives Emax,b) =8.301x107 W/(m2·m).Answer: (a) 36.7%;(b) 12.2%;(c) 0.5013m, 5.98x1014 Hz;(d) 8.301x107 W/(m2·m).1.14A blackbody radiates such that the wavelength at its maximum emissive poweris 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, soT = 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: 0.821.1.15A 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 maximumemissive 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.1.16A radiometer is sensitive to radiation only in the interval 3.68.5m. Theradiometer is used to calibrate a blackbody source at 1000K. The radiometer recordsthat the emitted energy is 4600 W/m2. What percentage of the blackbody radiatedenergy in the prescribed wavelength range is the source actually emitting?

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1.Introduction to Radiative Transfer1.9SOLUTION:mT,m KF 0-T3.636000.40368.585000.87417F1T2T = 0.87417 - 0.4036 = 0.47058Eb,= F1T2TT4=0.47058x 5.67040x10-8x10004=26,684W/m2Percentage sensed = (4600/26684) x 100 = 17.239%Answer: 17.24 %.1.17What temperature must a blackbody have for 25 percent of its emitted energy tobe in the visible wavelength region?SOLUTION: There are two solutions to this problem- one for the temperature at whichthe bulk of the energy is in the IR, and one for a higher temperature where the majorportion is in the UV. We can assume a temperature and, by trial and error or by using aroot solver, find the fraction in the visible (0.40.7), and continue until the fraction is0.25.For example, for the lower temperature, try T = 4600 K. Then1T = 4600 K x0.4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 lowertemperature and proceed to convergence.Using a root solver in a computational software package, find T such that0.704()0.250bET dT

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1.Introduction to Radiative Transfer1.10This gives the two results T = 4,343 and 12,460 K.Answer: 4,343 K, 12,460 K. (Note, two solutions are possible!)1.18Show that the blackbody spectral intensity Ib increases with T at any fixedvalue ofSOLUTION: From Equation 1.15, Ib = 2C1/{5 [ e C2/T - 1 ] }. The change in Ibwith T for a fixedis:222221212226522211CCTTbCCTTCC eIC CeTTeTe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 thatIb itself increases as T increases by noting that the denominator of Planck's spectraldistribution of intensity decreases with increasing T at any fixed wavelength.

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1.Introduction to Radiative Transfer1.111.19Blackbody radiation is leaving a small hole in a furnace at 1200 K (see thefigure.) What fraction of the radiation is intercepted by the annular disk? What fractionpasses through the hole in the disk?SOLUTION: From Table 1.2, the fraction of radiation to an annular disk is:12221222sinsin2sinsin2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) = 0.1000Answer: 0.1000; 0.1000.1.20A sheet of silica glass transmits 85% of the radiation that is incident in thewavelength range between 0.38 and 2.7m, and is essentially opaque to radiationhaving longer and shorter wavelengths. Estimate the percent of solar radiation that theglass 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, whatpercent of this radiation will be transmitted through the glass?1.5 cmHole infurnace wall1 cm3 cm211.51.53

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Solution Manual For Thermal Radiation Heat Transfer, 6th Edition - Page 16 preview image

1.Introduction to Radiative Transfer1.12SOLUTION: Part 1;1T =0.38m x 5780 K =2196.4m K;2T =2.7m x 5780 K=15606m K. Using Equation1.37,F = F02T-F01T= 0.96951-0.10022 =0.86929.The percent transmitted is then0.86929 x 0.85x 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%1.21Derive Wien's displacement law in terms of wave number by differentiation ofPlanck'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,222231122612/1CCTTbCTTCeCCTeEeSetting this equal to zero givesTCmax2TCmax2max2eTC1e3for which the solution is (T/max ) = constant. Substitution of(T/max) = 5099.4m K shows this solution to be valid.Garden at40CSolarradiationGlass

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