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Exam3DEX1:Physicsofnewenergy 26-1-2016van9:00-12:00 PLEASEREADTHESEINSTRUCTIONSFIRST! Inthisexamwewouldliketosummarizewithyoutheissuesdiscussedinthisclassby4 exercises: 1. Energyingeneral:theenergyproblem,consumptionandstorage. 2. Thermodynamics:howthermodynamicsimposesalimitonthemaximumefficiency toconvertheatintowork. 3. Fusionpower:thebasicprinciplesanditschallengestorealiseafusionpowerplant 4. Solarcells:itsstructureandtheoperatingprinciple AllfourquestionsareposedinEnglish.YoucanchooseyourselftoanswerineitherEnglish orDutch. Foreachofthesubquestionsthenumberofpointsthatcanbescoredisindicated.Thetotal numberofpointsis100.ThefinalresultFiscalculatedaccordingtoF=1.0+0.09x (numberofpointsscore)androundedto1decimal. Wealsoprovideanindicativetimeneededtocompletetheexercise(justourownestimate, maybeithelpsyoutocheckwhetheryourpaceissufficient). Theuseofcalculatorsisallowed,butanyotherbooks,phones,laptops,internetaccess, formularyisstrictlyprohibited. Belowyoufindsomeconstants,whichyoumightneedforsolvingsomeexercises.(Note thatyoudonotnecessarilyneedallofthem,itisjustastandardlist). Constants e= electroncharge = 1.6x10-19 C me= electronmass = 9.1x10-31 kg -27 mp protonmass = 1.67x10 kg c= speedoflight = 2.99x108 m/s -12 ε0= vacuumpermittivity = 8.85x10 F/m μ0= magneticpermeability= 1.26x10-6 Vs/Am -34 h= Planckconstant = 6.63x10 Js kB= Boltzmannconstant = 1.38x10-23 J/K g= gravitationofEarth = 9.81 m/s2 23 NA=Avogadro’snumber = 6.02x10 mol-1 R= Gasconstant = 8.31 J/(molK)(=8.31Pam3/(molK) atm= atmosphere = 1.01x105 Pa ρair= densityofair = 1.3 kg/m3 ρwater=densityofwater = 1000 kg/m3 kw= thermalconductivitywood= 0.1 W/(mK) kr= heatconductivityrubber= 0.15 W/(mK) atomicmass(amu):hydrogen=1,helium=4,carbon=12,oxygen=16 1. Energy–General–15pts–estimatedtime:25minutes a) [5pts]EnergyProblem:Whatarethethreemainfactorsdeterminingtheenergy problemwhichwewillfaceinthenearfuture?Giveonesolutiontothis.Inwhichway cansciencecontributetothis?Provideoneexample. Answer: 1pt:growthofpopulation 1pt:energyuseindevelopingcountriesisincreasing 1pt:fossilefuelsupplyisdecreasing 1pt:CO2 à4points,butmax3forthispart. Solution: 1pt:userenewableenergysources,increasedefficiency,decreaseenergyconsumption ExampleScience:(manypossibilities,ownjudgement)à1pt. (e.q.developnewsources.Likenuclearfusion,newhighefficiencysolarcells,developnew batterieselectricalcars….) b) [5pts]EnergyUse:Thelargestenergyconsumptionbyhouseholdsisfortransport (cars)andheating.Let’shavealookatthelatterone.Whatismostefficientenergyuse forcooking:anelectricalcookingplateoragascooker(gasstove).Why? Calculatethepowerneededtoraisethetemperatureofa5litrepanfilledwithwater from5°Cto100°C(theboilingtemperature)in3minutes.Whatisthepriceforthisifit isdoneelectrically(25ct/kWh,40%efficiency)? 1pt:gasisbestchoice:,efficiencytoproduceelectricityislow(40%),atleastforfossil powerplant:a)chemicalenergyàb)thermalenergyàc)electricalenergyàd)thermal energy:stepcàdveryinefficient 2pt:P=E/t=(c.m.dT)/t=4.18x5x95kJ/180s=11kW 2pt:E=P.t/eff=c.m.dT=4.18x5x95/0.4=5MJ=(5/3.6)kWhàprice=5/3.6*0.25 euro=0.35euro c) [5pts]Energystorage:Inanelectriccartheenergyisstoredinabattery.Typicallythis carcandriveonly100kmonfullelectricenergy,whereasaconventionalcaronpetrol candriveeasily500kmonasingletank.Whatistheessentialdifferenceintheenergy storage? Howisthisdifferentforfuelcellsbasedonhydrogen?Explain 2pt:Batterieshavealowerenergydensityperkgcomparedtopetrol 3pt:Fuel(H)hasevenmuchhigherenergydensityperkgthanpetrol,butthe questionishowhydrogencanbetransported:asafluid(alotofmassneededforkeeping thefuelcool)asagas(heavytankneededtokeepitathighpressure),orashydride(also heavy).Sofinallystillnotahighenergydensityperkg. 2. Thermodynamics–15pts–estimatedtime:25minutes Two moles of an ideal mono-atomic gas (cV= 12.47 J/(mol⋅K), cp=20.78 J/(mol⋅K)) follow a thermodynamiccycleaccordingtothepathDgAgBgCgD.ThestepsDAenBCareisochoric. ThestepsABandCDareisothermal.InstateD,thepressureis2⋅105Paandthetemperatureis 360K.InstateB,thevolumeisVB=3VDandthepressureisPB=2PC. a) [2pts]DrawthePVdiagramforthecycleDABCD. b) [8pts]Calculatetheworkandtheheatforeachstepofthecycle. c) [5pts]Calculatetheefficiencyofthecyclicprocess. VD = nRTD = 0.03m3 PD VD = VA TD = TC = 360 K VC = VB = 3VD = 0.09 m3 PC = nRTC = 0.66 ⋅ 105 Pa VC PB = 2 PC = 1.33 ⋅ 105 Pa TB = TA = PA = PBVB = 720K nR nRTA = 4 ⋅ 105 Pa VA ΔU = Q − W 3 ⎧ ⎪ΔU = 2 nR (TD − TA ) = 8.98kJ ⎪ stapDA⎨W = 0 ⎪Q = 8.98kJ ⎪ ⎩ ⎧ΔU = 0 J ⎪ V ⎪ staAB⎨W = nRT ln B = 13.15kJ VA ⎪ ⎪⎩Q = 13.15kJ 3 ⎧ ⎪ΔU = 2 nR (TC − TB ) = −8.98kJ ⎪ stapBC ⎨W = 0 J ⎪Q = −8.98kJ ⎪ ⎩ ⎧ΔU = 0 J ⎪ V ⎪ stapCD ⎨W = nRT ln D = −6.6kJ VC ⎪ ⎪⎩Q = −6.6kJ η= Wtot 13.15 − 6.6 = = 0.29 QH 13.15 + 8.98 3. Fusion–35pts–estimatedtime50minutes FusionEnergyisaverypromisingnewenergysource: • Thefuelisabundantlypresentforthousandsofyears • NoCO2emission • Inherentsafe • Largescale Nevertheless,westilldonothaveaworkingfusionreactoryet.Beforethiswillbe realisedmanychallengeshavetobeovercome.Let’shaveacloserlookatafewofthose. a) (6pts)Challenge1:Thefuel:theeasiestfusionreactionistheonebetween deuteriumandtritiumnuclei.However,inaplasmamixofdeuteriumandtritium alsootherfusionreactionsarepossible.Giveatleasttwootherreactionsthatwill occurinthisplasmaandusethepicturebelowtoestimatetheamountofenergy releasedinthisreaction. Differentpossibilities(energiesonlyapproximate,countvaluecorrectifdifference islessthen1MeV) D+DàT+p+4MeV D+Dà3He+n+3.3MeV T+Tà4He+2n+11MeV 3He+Dà4He+p+18MeV 3He+Tà4He+D(orp+n)+12-14MeV 3He+3Heà4He+p+13MeV b) (3pts)Challenge2:TheBurnCondition.Tohavenetfusionpowerweneedtofulfil thefollowingcondition:Morepowershouldbeproducedbyfusionreactionsthan weneedtoprovidetoheattheplasma.Thisleadstothefusiontripleproduct. Whicharethethreeparameterswhichdeterminethiscondition? nxTxτE>constant n=density T=temperature τE=energyconfinementtime c) (7pts)Challenge3:Magneticconfinement.Weneedtoconfinethehotplasmawith amagneticfieldBof5T. • (2pt)calculatetheaveragespeedofadeuteriumionina15keVplasma • (2pt)calculatetheradiusatwhichthisiongyratesaroundthemagneticfield line. • (3pt)Sketchthemotionofthisioninthefollowingthreesituations:theionis movingparalleltothemagneticfield,perpendiculartothemagneticfieldand oblique(i.e.underanangle)tothemagneticfield.Indicateclearlythedirections ofmagneticfieldandvelocity. Averagevelocity:0.5*md*v2=kTàv=√(2kT/md) =√(2x15000*1.6.10-19/2x1.6x10-27) =√1011=3.3x105m/s (Note:kTisenergyunit.Thisisgivenhereas15keV,toconvertthisbacktoJoule,you havetomultiplybye=1.6x10-19J/eV) Radius:rlarmor=mxv/(qxB)=2x1.6x10-27x3.3x105/(1.6x10-19x5)=1.3x10-3m Motion(1pteach): - parallel:noforce,ioncontinuestomovestraighton - perpendicular:circulatingthemagneticfield.Checkalsodirection(-0.5ptifthisis - notindicatedorincorrect! oblique:combinationofboth:spiral(alsocheckdirection) d) (4pts)Challenge4:Temperature.FusionoftheD-Treactioniseasiestat15keV (equaltoapproximately165MillionK).Thiscanbedonebyeitherinjecting particlesorinjectingelectromagneticwaves • (2pts)Whichparticlesarebesttoinjecttobesttoinjecttoheattheplasma. Explainthisheatingprinciplein2-3sentences. • (2pts)toheatthedeuteriumnucleiintheplasma,whichfrequencyof electromagneticwaveisbesttoinjectinareactorwithamagneticfieldof 5T? Besttoinjectfuelparticles(0.5pt)Deuteriumortritium Theseshouldbeneutral(0.5pt)withanenergymuchhigherthantheplasma temperature(0.5pt).Neutralparticlesareinjectedintotheplasma,particlesget ionizedandthenfollowthemagneticfieldlines.Thentheycollidewiththe plasmaparticlesandtransfertheirkineticenergytotheplasma(0.5pt) Frequencytoheatdeuterium: f=eB/(2pi.Md)=1.6x10-19x5/(2x3.14x2x1.67x10-27)Hz=38MHz (note:ifangularfrequencyisgiven,ie..withoutfactor2pithisisalsocorrectif therightunitsisgiven) e) (7pts)Challenge5:Wallpowerload.Theaimofafusionpowerplantistoproduce electricity.Assumewehaveafusionpowerreactorofthetypetokamak,producing 4GWoffusionpower(fromtheD-Treaction).Themajorradiusis6meter,the minorradius=2meter.Assumethetorushasacircularcrosssection. • (2pt)Describehowthisfusionpowerisconvertedtoelectricityandgivea coarseestimateofthetheelectricoutputpowerofthisreactor. • (3pt)Thefusionpowerisdistributedbetweentheneutronsandthealpha particles.Let’sconcentrateontheneutrons.Whatisthepowerofthese neutrons?Whereisitdeposited?Besidesthepowerproduction,whatother usedotheneutronshave? • (2pt)Calculatethepowerwallload(inMW/m2)asaresultoftheneutrons. Fusionpoweriskineticenergyofneutrons(80%)andalphaparticles(20%).This willbedepositedinthereactorwall(neutrons)orinthedivertor(plasmalosses includingalphaparticlepower).Coolingpipesinreactorwallandivertorwillheat upthewater,generatingsteam,rotatingturbine,producingelectricityin conventionalmanner.The4GWoffusionpowerwillleadwithaconversion efficiencyofabout25%to1GWofelectricpower. PowerofNeutrons:80%of4GW=3.2GW Wheredeposited:inreactorwall Otherreaction:produceTritium:n+6,7LiàHe+T+(n) Powerwallload:Totalarea=4π2Rxa=474m2 àpowerload=3.2GW/474m2=6.8MW/m2 f) (4pts)Challenge6:control.Tocontroltheplasmaweneedtomeasureinrealtime someplasmaparameters,liketheplasmatemperature. • (2pt)Describeameasurementtechnique(physicsprinciple)todothis. • (2pt)Useasketchtoillustratethisandindicatethemainhardware component. Temperature: a) ThomsonScattering:injectionoflaser.Laserphotonsscatteron electrons.PhotonsaredetectedataDopplershiftedfrequency.This Dopplershiftisameasureoftheelectronvelocity.Fullspectrum representsvelocitydistribution.Itwidthdeterminesthetemperature. b) Components:laser,lensestodetect,spectrometer. (AlternativelyECEcanbeexplained:microwaves,blackbodyradiation,intensity proportionaltotemperature) g) (4pts)Challenge7:Fusionisonlycompetitivewithalternativeenergysourcesifthe costs/kWharecomparable.Afusionreactorisextremelyexpensive(about10 BillionEuro’s)becauseoftheinfrastructure. • (2pt)Giveoneargumentwhythecosts/kWhcanstillbereasonable • (2pt)Alsogiveanargumentwhythecosts/kWhareatriskofrisingtoohigh (assumethatwewillbeabletoreachtherequiredfusioncriterionto producenetenergy). Investmentscostareverylarge,butrunningcostverylow,soeffectivelythe cost/kwHcanstillbereasonable. Systemiscomplex.Ifthereactorisdown(lowavailability)thenthecosts/kwHrise. 4. SolarCells–35pts–estimatedtime50minutes Analyseandcarefullydescribeinyourownwordsthestructureofacrystallinesilicon solarcellbyaddressingthefollowingquestions: a) (4pts)Makeacompletesketchofac-Sisolarcellandindicateitscomponents. b) (3 pts) The p-n junction in a solar cell is essential because it takes care of the separation of the electrons and holes, once they have been generated by sunlight absorption in the semiconductor. Make an accurate sketch of the energy band diagram of a p-n junction in a solar cell and indicate in which directions the electrons and holes will be transported when they experience the electric field of thejunction. c) (6 pts) Explain why it is not possible to convert all the photocurrent (i.e. all the photo-generatedcharges)inusefulelectricalcurrentfromasolarcell.Makeuseof theelectricalcircuitofasolarcell. Thephotocurrent(Iph)anddiodecurrent(Id)areinoppositedirectionssincethefirst relateswiththecurrentofthechargesgeneratedbysunlightabsorptionandbeing sweptacrossthejunctionviadriftmechanism,whilethediodecurrentisa recombinationcurrentduetothediffusionofelectronsfromntopandholesfrompto n.Thisrecombinationdiodecurrentisalwayspresentbecausetheseparationof chargesinducesaforwardbias(p-typesiliconbecomesp-polarizedandn-typesilicon becomesn-polarized)whichisthenresponsibleforthedevelopmentofthediffusion (diode)recombinationcurrent.ThisisthereasonwhyapartofIphwillbealwayslostin theformofdiodecurrentandtheusefulcurrentIwillbealwayslessthanIph. d) (4 pts) According to the Matlab simulation, a crystalline silicon (c-Si) wafer with a thickness of just 200-250 µm is sufficient to quantitatively absorb the solar light. However,thisisindiscrepancywiththevalueofthepenetrationdepthofthesunlight into c-Si, which suggests that we would need at least a thickness of 1000 µm to take advantageofallphotonswithenergyabovethec-Siband-gap(1.1eV).Canyouexplain thisdiscrepancybytakingintoconsiderationthestructureofthesolarcell? Thediscrepancybetweenthetworesultsarisesfromthefactthatthefirstansweristheresult oftheoptimizationinthedesignofthesolarcell.Specifically,wecanuse“only”200-250µmof c-Si because the optimized structure of the cell includes a metal (preferably of Au, Al or Cu) back contact which serves also as back-reflector and reflects back into the c-Si absorber all thosephotonswhichhavenotbeenabsorbedyetfromc-Si. e) (5pts)Theefficiencyofthesunlight-to-electricityconversionprocessislimitedbythe 2nd law of thermodynamics. Why? Furthermore, provide an estimate of the thermodynamic limit values, according to the two models which we described during classes.Whydothetwomodelsprovidedifferentthermodynamicefficiencyvalues? Thethermodynamiclimitisduetothefactthatwecannotconvertwith100%efficiencythe heat(photons)fromthesunintousefulelectricalwork,otherwisewewouldgoagainstthe2nd lawofthermodynamics.AccordingtotheShockley-Queissermodel,thislimitisequalto44%. TheDetailedBalancemodelmodelmakesamoreaccuratecalculation:31%.TheDetailed BalancemodelcorrectstheShockleyQueissermodelby:1)consideringthatthecurrentis alwayslowerbecausefundamentalchargerecombinationoccurs;2)themaximumvoltagewe cangainfromasolarcellwillbeneverequaltothebandgapvalueofthesemiconductor,but equaltotheVoc(Voc<Bg). f) (8pts)Providealistofalllossmechanismsoccurringinacommercialsolarcelland explainthemindetailbymakinguseofsketches. Spectralmismatchisduetothemismatchbetweenthesunlightspectrumandtheband gapofthesemiconductor.PhotonswithlowerenergythanBgwillnotbeabsorbed whilephotonsabovetheBgwillbeabsorbedbuttherestoftheenergywillbelostas heat.Shadowingandreflectionlossesarerelatedtotheopticalpathofthelightwhen hittingacell.Lightreflectingonthemetalcontactswillnotbeabsorbedandlightcanbe reflectedalsoattheglasssurfacetoo.Chargescanrecombinebecauseelectronscanfall backintothevalencebands.Furthermorethepresenceofdefectsatsurfacesand interfacestrapchargestoo.Afewsketchesareherepresented: g) (5pts)Whichisthemajorlossmechanism?Canyouprovideandexplainindetailone approachtoaddressthisspecificlossmechanism?Makealsouseofasketchtopresent yourapproach. Spectralmismatchisthemajorlossinsolarcells.Oneapproachistotakeadvantageofsolar cellsmadeoutofmultiplejunctions(tandem,triple)wheretwoorthreesemiconductorswith differentbandgapvaluesareusedtobetterexploitthesolarspectrum.Anotherapproachisto convertphotonsnotusefulforthespecificband-gaptoenergyrangeswhichhaveabetter matchwiththebandgapvalue:wetalkthenaboutconvertors(upconvertersand downconverters).