喜欢热与传质基本原理答案分章节版

PROBLEMKNOWN:Irradiationofsa litefromearthandsun.TwoemissivitiesassociatedwiththeFIND:(a)Steady-statesa litetemperaturewhensa liteisonbrightsideofearthforE/s>1andE/s<1,(b)Steady-statesa litetemperaturewhensa liteisondarksideofearthforE/s>1andE/s<1,(c)Schemetominimizetemperaturevariationsofthe E G=340E 1=s2= G=1353sASSUMPTIONS:(1)Steady-stateconditions,(2)Opaque,diffusegrayYSIS:Performinganenergybalanceonthe lite,itfollowsthatEinEoutEGE(D2/4)sGs(D2/4)ET4(D2/2)sT4(D2/2) GG1/TsatE ss2(Es)BrightSideofEarth(Gs=1353ForE=E=2=0.3,s=s=1=0.60.3340W/m20.61353W/ Tsat

2(0.30.6)5.67108W/m2K4ForE=E=1=0.6,s=s=2=0.30.6340W/m20.31353W/ Tsat

2(0.60.3)5.67108W/m2K4DarkSideofEarth(Gs=0ForE=E=1=0.6,s=s=2=…PROBLEM12.134 Tsat

2(0.60.3)5.67108W/m2K4ForE=E=2=0.3,s=s=1=0.6 Tsat

2(0.30.6)5.67108W/m2K4Tominimizethetemperaturevariationsofthesa lite,wewouldhavethehighemissivitycoatingalwaysfacingearth.COMMENTS:Iftheentiresa litewerecoveredwitheithercoating,thetemperaturesonthebrightanddarksidesofearthwouldbeTs=294Kand197K,respectively.Useofthetwoemissivitycoatingsreducestemperaturevariationsfrom294K–197K=97Kto278K–211K=67问题查找：(a)αE/αs>1且αE/αs<1星位于地球亮面时的稳态卫星温度，(b)αE/αs1αE/αs1，(c卫星GE340αEαsα1=0.6α2=GS1353假设：(1)(2)E=Ein22αG(πD/4)+αG(πD/4.εσT(πD/2).εσT(πD/2)0EEssEsatssat或1/EE+αsGsTsata)地球的明亮面（Gs1353W/m2）αEεEα20.3,αsεsα1，嚯=..=308K<2(0.3+0.6)×5.67×10W/m.KαE=εEα10.6,αsεsα20.34T=....=278K坐20.6+0.3)5.6710W/m.Kb)Gs0W/m2αEεEα10.6，αsεsα20.3，继续...问题12.134（续）2嚯=..=211K<2×(0.6+0.3)×5.67×1