太阳辐射计算.pdf

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1、RenewableEnergyTechnologiesI–HS09ExerciseB.1:Solarradiationandangleformulation1)Solarangles(18p)Inthefiguresbelowlabeltheangleswhichdescribe:a.thepositionofthesunrelativetoahorizontalsurface(Figure1).b.thepositionofthesunrelativetoatiltedsurface,e.g.acollector(Figure2).c.thepositiono

2、fthetiltedsurfacerelativetoahorizontalsurface(Figure3).Figure1Figure2Figure32)Solartime(52p)Forthe“Greencommunityvillage”nearDubai(latitudeangle=25°N,locallongitudeangle=55°12’E,standardtimezone=UTC+4,nodaylightsavingtime)onFebruary3at14.00determine:a.theapparentsolartime.b.hourangle

3、andsolardeclination.c.solaraltitudeandsolarazimuthangles.d.CompareyourresultswiththecorrespondingSunPathDiagraminthelecturenotes.3)Radiationontiltedsurface(30p)Aflatplatecollectorfacingsouthandtiltedup25°isinstalledonahousein“Greencommunityvillage”.a.Determinetheangleofincidenceatloc

4、alsolarnoon(February3)?b.Assumingthecollectorcantrackthesunthecollectorcantrackthesun,determinetheoptimalslopeandsurfaceslopeandsurfaceazimuthangleofyourcollector(atlocalsolarnoon,February3)?RenewableEnergyTechnologiesI–HS09SolutionB.1:Solarradiationandangleformulation1)Solaranglesan

5、gleofincidencezenithangleθzsolaraltitudeangleαsθγssolarazimuthangleFigure1Figure2βslopeγsurfaceazimuthangleFigure32)Solartimea.equations:4minlocalsolartime:AST=LST+ET+l-l(stlocal)⋅°equationoftime:ET=9.87sin2B-7.53cosB-1.5sinBET=9.87sin2B-7.53cosB-1.5sinB⋅(⋅)⋅()⋅()360n81°(−)withB=364r

6、esults:3603481°(−)February3→n=34→B==−46.48°3642RenewableEnergyTechnologiesI–HS09ET=9.87sin2⋅(⋅−(46.48°))-7.53cos⋅(−46.48-1.5sin°)⋅(−46.48°)=−13.95minllocal=55°12’E=−55.2°(conversionofanglein°,negativesincelocationisinEast!)FromStandardTimeMeridiantableinlecturenotes:UTC+4→lst=60°E=−6

7、0°(negativesincelocationisinEast!)Nodaylightsavingtime(i.e.summer/wintertime)→LST=14:004minAST=14:00+(−13.95min+)(−60°−−(55.2°⋅))=13:2713:27°b.equations:hourangle:ω=15°hoursfromlocalsolarnoon⋅()=15°AST⋅(−12)360°(284+n)⋅solardeclination:δ=23.45°sin⋅s365results:AST=13:27=13.45h(c

8、onversionoft