1、Chapter 12 Radiation Heat Transfer12-68Review Problems12-88 The temperature of air in a duct is measured by a thermocouple. The radiation effect on the temperature measurement is to be quantified, and the actual air temperature is to be determined.Assumptions The surfaces are opaque, diffuse, and gr
2、ay. Properties The emissivity of thermocouple is given to be =0.6. Analysis The actual temperature of the air can be determined fromK 1C W/m60) 50()K 8(17.5)(60K 8244484hTTwttthf12-89 The temperature of hot gases in a duct is measured by a thermocouple. The actual temperature of the gas is to be det
3、ermined, and compared with that without a radiation shield.Assumptions The surfaces are opaque, diffuse, and gray. Properties The emissivity of the thermocouple is given to be =0.7. Analysis Assuming the area of the shield to be very close to the sensor of the thermometer, the radiation heat transfe
4、r from the sensor is determined from 2 44428142senor frmad, W/m9.5715.01. )K 380() (K /06.()(TQThen the actual temperature of the gas can be determined from a heat transfer balance to beK 532 ffthTTqq22senor frmcov,senr ocv, /9.57)30C( /10Without the shield the temperature of the gas would be 549.2C
5、 W/m120) 380() 5(67.5)(0K 53 4484hTwttthfAir, Tf Tw = 500 KThermocoupleTth = 850 K = 0.6Air, Tf Tw = 380 KThermocoupleTth = 530 K1 = 0.72 = 0.15Chapter 12 Radiation Heat Transfer12-6912-90E A sealed electronic box is placed in a vacuum chamber. The highest temperature at which the surrounding surfac
6、es must be kept if this box is cooled by radiation alone is to be determined.Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. 4 Heat transfer from the bottom surface of the box is negligible.Properties The emi
7、ssivity of the outer surface of the box is = 0.95. Analysis The total surface area is 2ft 67.3)1(2/8(4sAThen the temperature of the surrounding surfaces is determined to beF43 R 50 )R 590(Btu/h.f 1074.)(m 6.)(9Btu/h )412.30( 442824sr sursursradT TTAQ12-91 A double-walled spherical tank is used to st
8、ore iced water. The air space between the two walls is evacuated. The rate of heat transfer to the iced water and the amount of ice that melts a 24-h period are to be determined.Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. Properties The emissivities
9、of both surfaces are given to be 1 = 2 = 0.15. Analysis (a) Assuming the conduction resistance s of the walls to be negligible, the rate of heat transfer to the iced water in the tank is determined to be m 2AD1220169(.).W107.4 24442821142 0.15.1.0 )K 730() 73)(K /6.5)( 9.(TQ(b) The amount of heat tr
10、ansfer during a 24-hour period iskJ927)s3624)(kJ/s7.(tQThe amount of ice that melts during this period then becomesg.8 kJ/ 7.35ififhmhD2 = 2.04 m T2 = 20C2 = 0.15D1 = 2.01 m T1 = 0C1 = 0.15VacuumIcedwater0CTsurr 8 in100 W = 0.95Ts = 130F12 in12 inChapter 12 Radiation Heat Transfer12-7012-92 Two conc
11、entric spheres which are maintained at uniform temperatures are separated by air at 1 atm pressure. The rate of heat transfer between the two spheres by natural convection and radiation is to be determined.Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray.
12、3 Air is an ideal gas with constant properties.Properties The emissivities of the surfaces are given to be 1 = 2 = 0.5. The properties of air at 1 atm and the average temperature of (T1+T2)/2 = (350+275)/2 = 312.5 K = 39.5C are (Table A-15) 1-25K 03. .3176.Pr/sm 9CW.08.kAnalysis (a) Noting that Di =
13、 D1 and Do = D2 , the characteristic length ism0.5).5.(2)2iocLThen 5253-1231 104.7)256.0()/s 0697.( ) .(K 73(2)m/s 81.9(Pr)( cTgRaThe effective thermal conductivity is059.m) 25.0() 15.0(m) 2.( 15.0)()( 7/-7/-45/75/74sph oioicDLF CW/. 135)0415.7)(09.(7256.081.C)W/m. 02658.(74(Pr1. 4/4/14/4/ef RaFksph
14、Then the rate of heat transfer between the spheres becomes 2. K)7530()m 05.(2)C/. 3.()(ef oiciTLDkQ(b) The rate of heat transfer by radiation is determined fromW32. 2444822121421 5.019.01) 75() 3(K W/67.5)(m 0.()(75.0DTAD2 = 25 cm T2 = 275 K2 = 0.5D1 = 15 cm T1 = 350 K1 = 0.9Lc =5 cmAIR1 atmChapter
15、12 Radiation Heat Transfer12-7112-93 A solar collector is considered. The absorber plate and the glass cover are maintained at uniform temperatures, and are separated by air. The rate of heat loss from the absorber plate by natural convection and radiation is to be determined.Assumptions 1 Steady op
16、erating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is an ideal gas with constant properties.Properties The emissivities of surfaces are given to be 1 = 0.9 for glass and 2 = 0.8 for the absorber plate. The properties of air at 1 atm and the average temperature of (T1+T2)/2
17、= (80+32)/2 = 56C are (Table A-15) 1-25K 034.)76(1.0Pr/sm 8CW.79.fTkAnalysis For , we have horizontal rectangular enclosure. The characteristic length in this case is the distance between the two glasses Lc = L = 0.03 m Then, 4253-12231 1083.)72.0()/sm 087.( ) .(K 3(34)/s 1.9(Pr)( TgRa2 5.4) (5.WHAs
18、 74.3 118)20cos(03.()2cos(1083.(.in7)20cos(1083.(71 )csRa(Ra).(sia4.Nu 3/446.14 3/16.1 W75 m .C) 5CW/m 9.21LTkNuAQsNeglecting the end effects, the rate of heat transfer by radiation is determined from128919.08)K 32()3)(K/1067.5)(.41)( 4442224radsDiscussion The rates of heat loss by natural convectio
19、n for the horizontal and vertical cases would be as follows (Note that the Ra number remains the same):Horizontal: 812.318)03.(1083.74.118Ra7014.Nu 3/443/ Wm .C)2() 62.)(CW/m 29.(LTkAQsVertical: Solar radiation = 20InsulationAbsorber plateT1 = 80C1 = 0.8 Glass cover,T2 = 32C2 = 0.91.5 mL = 3 cmChapt
20、er 12 Radiation Heat Transfer12-7201.2m 03.)721.0()1083.(42Pr42.0 4/3.012.0/ LHRaNu W54 .Cm 6.)CW/ 79.(TkAQsChapter 12 Radiation Heat Transfer12-7312-94E The circulating pump of a solar collector that consists of a horizontal tube and its glass cover fails. The equilibrium temperature of the tube is
21、 to be determined.Assumptions 1 Steady operating conditions exist. 2 The tube and its cover are isothermal. 3 Air is an ideal gas. 4 The surfaces are opaque, diffuse, and gray for infrared radiation. 5 The glass cover is transparent to solar radiation.Properties The properties of air should be evalu
22、ated at the average temperature. But we do not know the exit temperature of the air in the duct, and thus we cannot determine the bulk fluid and glass cover temperatures at this point, and thus we cannot evaluate the average temperatures. Therefore, we will assume the glass temperature to be 85F, an
23、d use properties at an anticipated average temperature of (75+85)/2 =80F (Table A-15E),sk/ft 10.697/hft 610.FtBu4824-2 R 5401729.PraveTAnalysis We have a horizontal cylindrical enclosure filled with air at 0.5 atm pressure. The problem involves heat transfer from the aluminum tube to the glass cover
24、 and from the outer surface of the glass cover to the surrounding ambient air. When steady operation is reached, these two heat transfer rates must equal the rate of heat gain. That is,(per foot of tube)Btu/h30gainsolr ambient-glslas-tube QQThe heat transfer surface area of the glass cover is(per fo
25、ot of tube)2ft309.1ft)(2/5)(WDAoglasoTo determine the Rayleigh number, we need to know the surface temperature of the glass, which is not available. Therefore, solution will require a trial-and-error approach. Assuming the glass cover temperature to be 85F, the Rayleigh number, the Nusselt number, t
26、he convection heat transfer coefficient, and the rate of natural convection heat transfer from the glass cover to the ambient air are determined to be624323 1092.)7.0()/sft 10675.( )ft 1/5(R8()/)fts .3(PrRa ooDoTg95.14 7290./5.1)(86.Pr/.0a 6Nu 27/81627/8169D FftBu/h 3.0)9.4(ft /5FBuh4.0 DkhoBtu/h 96
27、. )758(91(ft/ 31.0()( 22cnv, TAQoAlso,Btu/h 5.30 R) 3() 4(ft 30.)(Rftu/h 174)(9 42428skyrad,oThen the total rate of heat loss from the glass cover becomesBtu/h5.7.0rad,conv,toal, QD2 =5 inAluminum tubeD1 =2.5 in, T1 1 = 0.9Air space0.5 atmPlastic cover, 2 = 0.9, T2 WaterT = 75FTsky = 60F30 Btu/h.ftC
28、hapter 12 Radiation Heat Transfer12-74which is more than 30 Btu/h. Therefore, the assumed temperature of 85F for the glass cover is high. Repeating the calculations with lower temperatures (including the evaluation of properties), the glass cover temperature corresponding to 30 Btu/h is determined t
29、o be 81.5F.The temperature of the aluminum tube is determined in a similar manner using the natural convection and radiation relations for two horizontal concentric cylinders. The characteristic length in this case is the distance between the two cylinders, which is ft1.25/in.2/)5.(2/)( iocDLAlso,(p
30、er foot of tube)64.0ft) (.WAitubeiWe start the calculations by assuming the tube temperature to be 118.5F, and thus an average temperature of (81.5+118.5)/2 = 100F=640 R. Using properties at 100F,4243223 103.)726.0()/sft 1089.( )ft 1/5.(R.5)/)fts.(Pr)(Ra LTgoiLThe effective thermal conductivity is .
31、ft) 1/5(ft) /5.(ft) (.2/.ln)(/ln 53/-3/-345/3534cy oiiDFFftBu/h 0327. )034.16.0(.720.861)159(86RaPr 4/14/1cy/1ef LkkThen the rate of heat transfer between the cylinders by convection becomesBtu/h 8.10F)5.18(ln(5/2.)FftBuh3.)()/ln(efcov , oioi TDkQAlso, Btu/h 0.25in 5.29.0. R) .4() .7()ft 64(Rft 1074
32、.(18i4rad, oiiiTAThen the total rate of heat loss from the glass cover becomesBtu/h8.350.281rad,conv,tal, iii Qwhich is more than 30 Btu/h. Therefore, the assumed temperature of 118.5F for the tube is high. By trying other values, the tube temperature corresponding to 30 Btu/h is determined to be 11
33、3.2F. Therefore, the tube will reach an equilibrium temperature of 113.2F when the pump fails.Chapter 12 Radiation Heat Transfer12-7512-95 A double-pane window consists of two sheets of glass separated by an air space. The rates of heat transfer through the window by natural convection and radiation
34、 are to be determined.Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is an ideal gas with constant specific heats. 4 Heat transfer through the window is one-dimensional and the edge effects are negligible.Properties The emissivities of glass surfa
35、ces are given to be 1 = 2 = 0.9. The properties of air at 0.3 atm and the average temperature of (T1+T2)/2 = (15+5)/2 = 10C are (Table A-15)1-2551K 034. )270(36.Pr /sm 073././Wm049.atkAnalysis The characteristic length in this case is the distance between the glasses, m 05.Lc4253-12231 198.)76.()/sm
36、 073.4( )05.(K()m/s81.9(Pr)( LTgRa 9.1.).097.0 4/1/4/1 HNu2 6) 3(m2sAThen the rate of heat transfer by natural convection becomes W45.0 m .0C)1( 6)539.1(CW/m0439.( 221LTkNuQsconvThe rate of heat transfer by radiation is determined from W2519.0K 2735 2735)K/m 1067.)( )( 44428214radAsThen the rate of
37、total heat transfer becomes W2754radconvtalQDiscussion Note that heat transfer through the window is mostly by radiation. 5C 15CL = 5 cmH = 2 mQ AirChapter 12 Radiation Heat Transfer12-7612-96 A simple solar collector is built by placing a clear plastic tube around a garden hose. The rate of heat lo
38、ss from the water in the hose by natural convection and radiation is to be determined.Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is an ideal gas with constant specific heats.Properties The emissivities of surfaces are given to be 1 = 2 = 0.9.
39、The properties of air are at 1 atm and the film temperature of (Ts+T)/2 = (40+25)/2 = 32.5C are (Table A-15)1-25K 037. )7.3(1.0Pr/m 6CW.7.kAnalysis Under steady conditions, the heat transfer rate from the water in the hose equals to the rate of heat loss from the clear plastic tube to the surroundin
40、gs by natural convection and radiation. The characteristic length in this case is the diameter of the plastic tube, .m062DLplasticc 5253-1232 10842.)75.0()/sm 063.( ) 6.(K4(7)/s 81.9(Pr)( TgRs.71./9.1)8(.r/5.01Ra 76.Nu 27/869527/8169D 22 m 85.0) (16. C.W/43C. LADkhplasticThen the rate of heat transf
41、er from the outer surface by natural convection becomes 12.7 )540)(1.)(C/475.()( 222ThQsconvThe rate of heat transfer by radiation from the outer surface is determined from W26. K) 27315() 27340)(KW/m 1067.5)( 180)(9 442824radskyAFinally, 9.382.71, lostaQDiscussion Note that heat transfer is mostly
42、by radiation. D2 =6 cmGarden hoseD1 =2 cm, T1 1 = 0.9Air spacePlastic cover, 2 = 0.9, T2 =40C WaterT = 25CTsky = 15CChapter 12 Radiation Heat Transfer12-7712-97 A solar collector consists of a horizontal copper tube enclosed in a concentric thin glass tube. The annular space between the copper and t
43、he glass tubes is filled with air at 1 atm. The rate of heat loss from the collector by natural convection and radiation is to be determined.Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is an ideal gas with constant specific heats.Properties The
44、 emissivities of surfaces are given to be 1 = 0.85 for the tube surface and 2 = 0.9 for glass cover. The properties of air at 1 atm and the average temperature of (T1+T2)/2 = (60+40)/2 = 50C are (Table A-15) 1-25K 0396. )70(18.Pr/sm 9CW.073.kAnalysis The characteristic length in this case is m 0.2=)
45、.5-.(2)(21DLc 850,1)72.0()/s 1798.( )m .(K406(3/s 81.9Pr 253-23 cLTgRaThe effective thermal conductivity is130.) 05.(m) 09.() 2.(/ln)(/ln 3/-3-345/3534cyl oiiDLF CW/m. 05321.)8,10(3.728.061.)CW/m. 02735.(86RaPr1. 4/4/14/1cyl4/ef FkkThen the rate of heat transfer between the cylinders becomes(Eq. 1).
46、C)46()5./9ln(m3.)()/ln(efcov oioTDkQThe rate of heat transfer by radiation is determined from W13.4 95.018. )K 27340() 2736)(K/m 067.5)( .0(1)( 4422142radAFinally,(per m length) .24., lostaQD2 =9 cmCopper tubeD1 =5 cm, T1 = 60C1 = 0.85Air spacePlastic cover, T2 = 40C2 = 0.9WaterChapter 12 Radiation
47、Heat Transfer12-7812-98 A cylindrical furnace with specified top and bottom surface temperatures and specified heat transfer rate at the bottom surface is considered. The temperature of the side surface and the net rates of heat transfer between the top and the bottom surfaces, and between the bottom and the side surfaces are to be determined. Assumptions 1 Steady
