e = electron charge = 1.6x10-19 c me = electron mass = 9.1x10-31

Exam3DEX1:Physicsofnewenergy26-1-2016van9:00-12:00

PLEASEREADTHESEINSTRUCTIONSFIRST!Inthisexamwewouldliketosummarizewithyoutheissuesdiscussedinthisclassby4exercises:

1. Energyingeneral:theenergyproblem,consumptionandstorage.2. Thermodynamics:howthermodynamicsimposesalimitonthemaximumefficiency

toconvertheatintowork.3. Fusionpower:thebasicprinciplesanditschallengestorealiseafusionpowerplant4. Solarcells:itsstructureandtheoperatingprinciple

AllfourquestionsareposedinEnglish.YoucanchooseyourselftoanswerineitherEnglishorDutch.Foreachofthesubquestionsthenumberofpointsthatcanbescoredisindicated.Thetotalnumberofpointsis100.ThefinalresultFiscalculatedaccordingtoF=1.0+0.09x(numberofpointsscore)androundedto1decimal.Wealsoprovideanindicativetimeneededtocompletetheexercise(justourownestimate,maybeithelpsyoutocheckwhetheryourpaceissufficient). Theuseofcalculatorsisallowed,butanyotherbooks,phones,laptops,internetaccess,formularyisstrictlyprohibited.Belowyoufindsomeconstants,whichyoumightneedforsolvingsomeexercises.(Notethatyoudonotnecessarilyneedallofthem,itisjustastandardlist).Constants e= electroncharge = 1.6x10-19 Cme= electronmass = 9.1x10-31 kgmp protonmass = 1.67x10-27 kgc= speedoflight = 2.99x108 m/sε0= vacuumpermittivity = 8.85x10-12 F/mμ0= magneticpermeability= 1.26x10-6 Vs/Amh= Planckconstant = 6.63x10-34 JskB= Boltzmannconstant = 1.38x10-23 J/Kg= gravitationofEarth = 9.81 m/s2

NA=Avogadro’snumber = 6.02x1023 mol-1R= Gasconstant = 8.31 J/(molK)(=8.31Pam3/(molK)atm= atmosphere = 1.01x105 Paρair= densityofair = 1.3 kg/m3

ρwater=densityofwater = 1000 kg/m3kw= 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:25minutesa) [5pts]EnergyProblem:Whatarethethreemainfactorsdeterminingtheenergy

problemwhichwewillfaceinthenearfuture?Giveonesolutiontothis.Inwhichwaycansciencecontributetothis?Provideoneexample.

Answer:1pt:growthofpopulation1pt:energyuseindevelopingcountriesisincreasing1pt:fossilefuelsupplyisdecreasing1pt:CO2à4points,butmax3forthispart.Solution:1pt:userenewableenergysources,increasedefficiency,decreaseenergyconsumptionExampleScience:(manypossibilities,ownjudgement)à1pt.(e.q.developnewsources.Likenuclearfusion,newhighefficiencysolarcells,developnewbatterieselectricalcars….)b) [5pts]EnergyUse:Thelargestenergyconsumptionbyhouseholdsisfortransport

(cars)andheating.Let’shavealookatthelatterone.Whatismostefficientenergyuseforcooking:anelectricalcookingplateoragascooker(gasstove).Why?Calculatethepowerneededtoraisethetemperatureofa5litrepanfilledwithwaterfrom5°Cto100°C(theboilingtemperature)in3minutes.Whatisthepriceforthisifitisdoneelectrically(25ct/kWh,40%efficiency)?

1pt:gasisbestchoice:,efficiencytoproduceelectricityislow(40%),atleastforfossilpowerplant:a)chemicalenergyàb)thermalenergyàc)electricalenergyàd)thermalenergy:stepcàdveryinefficient 2pt:P=E/t=(c.m.dT)/t=4.18x5x95kJ/180s=11kW2pt:E=P.t/eff=c.m.dT=4.18x5x95/0.4=5MJ=(5/3.6)kWhàprice=5/3.6*0.25euro=0.35euroc) [5pts]Energystorage:Inanelectriccartheenergyisstoredinabattery.Typicallythis

carcandriveonly100kmonfullelectricenergy,whereasaconventionalcaronpetrolcandriveeasily500kmonasingletank.Whatistheessentialdifferenceintheenergystorage?Howisthisdifferentforfuelcellsbasedonhydrogen?Explain

2pt:Batterieshavealowerenergydensityperkgcomparedtopetrol 3pt:Fuel(H)hasevenmuchhigherenergydensityperkgthanpetrol,butthequestionishowhydrogencanbetransported:asafluid(alotofmassneededforkeepingthefuelcool)asagas(heavytankneededtokeepitathighpressure),orashydride(alsoheavy).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 athermodynamiccycleaccordingtothepathDgAgBgCgD.ThestepsDAenBCareisochoric.ThestepsABandCDareisothermal.InstateD,thepressureis2⋅105Paandthetemperatureis360K.InstateB,thevolumeisVB=3VDandthepressureisPB=2PC.

a) [2pts]DrawthePVdiagramforthecycleDABCD.b) [8pts]Calculatetheworkandtheheatforeachstepofthecycle.c) [5pts]Calculatetheefficiencyofthecyclicprocess.

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3. Fusion–35pts–estimatedtime50minutesFusionEnergyisaverypromisingnewenergysource:

• Thefuelisabundantlypresentforthousandsofyears• NoCO2emission• Inherentsafe• Largescale

Nevertheless,westilldonothaveaworkingfusionreactoryet.Beforethiswillberealisedmanychallengeshavetobeovercome.Let’shaveacloserlookatafewofthose.

a) (6pts)Challenge1:Thefuel:theeasiestfusionreactionistheonebetween

deuteriumandtritiumnuclei.However,inaplasmamixofdeuteriumandtritiumalsootherfusionreactionsarepossible.Giveatleasttwootherreactionsthatwilloccurinthisplasmaandusethepicturebelowtoestimatetheamountofenergyreleasedinthisreaction.

Differentpossibilities(energiesonlyapproximate,countvaluecorrectifdifferenceislessthen1MeV)D+DàT+p+4MeVD+Dà3He+n+3.3MeVT+Tà4He+2n+11MeV3He+Dà4He+p+18MeV3He+Tà4He+D(orp+n)+12-14MeV3He+3Heà4He+p+13MeV

b) (3pts)Challenge2:TheBurnCondition.Tohavenetfusionpowerweneedtofulfilthefollowingcondition:Morepowershouldbeproducedbyfusionreactionsthanweneedtoprovidetoheattheplasma.Thisleadstothefusiontripleproduct.Whicharethethreeparameterswhichdeterminethiscondition?

nxTxτE>constantn=densityT=temperatureτE=energyconfinementtime

c) (7pts)Challenge3:Magneticconfinement.Weneedtoconfinethehotplasmawith

amagneticfieldBof5T.• (2pt)calculatetheaveragespeedofadeuteriumionina15keVplasma• (2pt)calculatetheradiusatwhichthisiongyratesaroundthemagneticfield

line.• (3pt)Sketchthemotionofthisioninthefollowingthreesituations:theionis

movingparalleltothemagneticfield,perpendiculartothemagneticfieldandoblique(i.e.underanangle)tothemagneticfield.Indicateclearlythedirectionsofmagneticfieldandvelocity.

Averagevelocity:0.5*md*v2=kTàv=√(2kT/md)

=√(2x15000*1.6.10-19/2x1.6x10-27)=√1011=3.3x105m/s

(Note:kTisenergyunit.Thisisgivenhereas15keV,toconvertthisbacktoJoule,youhavetomultiplybye=1.6x10-19J/eV)Radius:rlarmor=mxv/(qxB)=2x1.6x10-27x3.3x105/(1.6x10-19x5)=1.3x10-3mMotion(1pteach):- parallel:noforce,ioncontinuestomovestraighton- perpendicular:circulatingthemagneticfield.Checkalsodirection(-0.5ptifthisis

notindicatedorincorrect! - oblique:combinationofboth:spiral(alsocheckdirection)

d) (4pts)Challenge4:Temperature.FusionoftheD-Treactioniseasiestat15keV

(equaltoapproximately165MillionK).Thiscanbedonebyeitherinjectingparticlesorinjectingelectromagneticwaves

• (2pts)Whichparticlesarebesttoinjecttobesttoinjecttoheattheplasma.Explainthisheatingprinciplein2-3sentences.

• (2pts)toheatthedeuteriumnucleiintheplasma,whichfrequencyofelectromagneticwaveisbesttoinjectinareactorwithamagneticfieldof5T?

Besttoinjectfuelparticles(0.5pt)DeuteriumortritiumTheseshouldbeneutral(0.5pt)withanenergymuchhigherthantheplasmatemperature(0.5pt).Neutralparticlesareinjectedintotheplasma,particlesgetionizedandthenfollowthemagneticfieldlines.Thentheycollidewiththeplasmaparticlesandtransfertheirkineticenergytotheplasma(0.5pt)Frequencytoheatdeuterium:f=eB/(2pi.Md)=1.6x10-19x5/(2x3.14x2x1.67x10-27)Hz=38MHz(note:ifangularfrequencyisgiven,ie..withoutfactor2pithisisalsocorrectiftherightunitsisgiven)

e) (7pts)Challenge5:Wallpowerload.Theaimofafusionpowerplantistoproduceelectricity.Assumewehaveafusionpowerreactorofthetypetokamak,producing4GWoffusionpower(fromtheD-Treaction).Themajorradiusis6meter,theminorradius=2meter.Assumethetorushasacircularcrosssection.

• (2pt)Describehowthisfusionpowerisconvertedtoelectricityandgiveacoarseestimateofthetheelectricoutputpowerofthisreactor.

• (3pt)Thefusionpowerisdistributedbetweentheneutronsandthealphaparticles.Let’sconcentrateontheneutrons.Whatisthepoweroftheseneutrons?Whereisitdeposited?Besidesthepowerproduction,whatotherusedotheneutronshave?

• (2pt)Calculatethepowerwallload(inMW/m2)asaresultoftheneutrons.Fusionpoweriskineticenergyofneutrons(80%)andalphaparticles(20%).Thiswillbedepositedinthereactorwall(neutrons)orinthedivertor(plasmalossesincludingalphaparticlepower).Coolingpipesinreactorwallandivertorwillheatupthewater,generatingsteam,rotatingturbine,producingelectricityinconventionalmanner.The4GWoffusionpowerwillleadwithaconversionefficiencyofabout25%to1GWofelectricpower.PowerofNeutrons:80%of4GW=3.2GWWheredeposited:inreactorwallOtherreaction:produceTritium:n+6,7LiàHe+T+(n)Powerwallload:Totalarea=4π2Rxa=474m2àpowerload=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.ThisDopplershiftisameasureoftheelectronvelocity.Fullspectrumrepresentsvelocitydistribution.Itwidthdeterminesthetemperature.

b) Components:laser,lensestodetect,spectrometer.(AlternativelyECEcanbeexplained:microwaves,blackbodyradiation,intensityproportionaltotemperature)g) (4pts)Challenge7:Fusionisonlycompetitivewithalternativeenergysourcesifthe

costs/kWharecomparable.Afusionreactorisextremelyexpensive(about10BillionEuro’s)becauseoftheinfrastructure.

• (2pt)Giveoneargumentwhythecosts/kWhcanstillbereasonable• (2pt)Alsogiveanargumentwhythecosts/kWhareatriskofrisingtoohigh

(assumethatwewillbeabletoreachtherequiredfusioncriteriontoproducenetenergy).

Investmentscostareverylarge,butrunningcostverylow,soeffectivelythecost/kwHcanstillbereasonable.Systemiscomplex.Ifthereactorisdown(lowavailability)thenthecosts/kwHrise.

4. SolarCells–35pts–estimatedtime50minutes

Analyseandcarefullydescribeinyourownwordsthestructureofacrystallinesiliconsolarcellbyaddressingthefollowingquestions:a) (4pts)Makeacompletesketchofac-Sisolarcellandindicateitscomponents.

b) (3 pts) The p-n junction in a solar cell is essential because it takes care of the

separationof theelectronsandholes,once theyhavebeengenerated bysunlightabsorption in the semiconductor. Make an accurate sketch of the energy banddiagram of a p-n junction in a solar cell and indicate in which directions theelectrons andholeswill be transportedwhen they experience the electric fieldofthejunction.

c) (6 pts) Explainwhy it is not possible to convert all the photocurrent (i.e. all thephoto-generatedcharges)inusefulelectricalcurrentfromasolarcell.Makeuseof

theelectricalcircuitofasolarcell.

Thephotocurrent(Iph)anddiodecurrent(Id)areinoppositedirectionssincethefirstrelateswiththecurrentofthechargesgeneratedbysunlightabsorptionandbeingsweptacrossthejunctionviadriftmechanism,whilethediodecurrentisarecombinationcurrentduetothediffusionofelectronsfromntopandholesfrompton.Thisrecombinationdiodecurrentisalwayspresentbecausetheseparationofchargesinducesaforwardbias(p-typesiliconbecomesp-polarizedandn-typesiliconbecomesn-polarized)whichisthenresponsibleforthedevelopmentofthediffusion(diode)recombinationcurrent.ThisisthereasonwhyapartofIphwillbealwayslostintheformofdiodecurrentandtheusefulcurrentIwillbealwayslessthanIph.

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,thisisindiscrepancywiththevalueofthepenetrationdepthofthesunlightinto c-Si,which suggests thatwewouldneedat least a thicknessof1000µmto takeadvantageofallphotonswithenergyabovethec-Siband-gap(1.1eV).Canyouexplainthisdiscrepancybytakingintoconsiderationthestructureofthesolarcell?

Thediscrepancybetweenthetworesultsarisesfromthefactthatthefirstansweristheresultoftheoptimizationinthedesignofthesolarcell.Specifically,wecanuse“only”200-250µmofc-Si because the optimized structure of the cell includes ametal (preferably of Au, Al or Cu)back contact which serves also as back-reflector and reflects back into the c-Si absorber allthosephotonswhichhavenotbeenabsorbedyetfromc-Si.

e) (5pts)Theefficiencyofthesunlight-to-electricityconversionprocess is limitedbythe2nd law of thermodynamics. Why? Furthermore, provide an estimate of thethermodynamic limit values, according to the twomodelswhichwedescribedduringclasses.Whydothetwomodelsprovidedifferentthermodynamicefficiencyvalues?

Thethermodynamiclimitisduetothefactthatwecannotconvertwith100%efficiencytheheat(photons)fromthesunintousefulelectricalwork,otherwisewewouldgoagainstthe2ndlawofthermodynamics.AccordingtotheShockley-Queissermodel,thislimitisequalto44%.TheDetailedBalancemodelmodelmakesamoreaccuratecalculation:31%.TheDetailedBalancemodelcorrectstheShockleyQueissermodelby:1)consideringthatthecurrentisalwayslowerbecausefundamentalchargerecombinationoccurs;2)themaximumvoltagewecangainfromasolarcellwillbeneverequaltothebandgapvalueofthesemiconductor,butequaltotheVoc(Voc<Bg).

f) (8pts)Providealistofalllossmechanismsoccurringinacommercialsolarcellandexplainthemindetailbymakinguseofsketches.

Spectralmismatchisduetothemismatchbetweenthesunlightspectrumandthebandgapofthesemiconductor.PhotonswithlowerenergythanBgwillnotbeabsorbedwhilephotonsabovetheBgwillbeabsorbedbuttherestoftheenergywillbelostasheat.Shadowingandreflectionlossesarerelatedtotheopticalpathofthelightwhenhittingacell.Lightreflectingonthemetalcontactswillnotbeabsorbedandlightcanbereflectedalsoattheglasssurfacetoo.Chargescanrecombinebecauseelectronscanfallbackintothevalencebands.Furthermorethepresenceofdefectsatsurfacesandinterfacestrapchargestoo.Afewsketchesareherepresented:

g) (5pts)Whichisthemajorlossmechanism?Canyouprovideandexplainindetailone

approachtoaddressthisspecificlossmechanism?Makealsouseofasketchtopresent

yourapproach.Spectralmismatchisthemajorlossinsolarcells.Oneapproachistotakeadvantageofsolarcellsmadeoutofmultiplejunctions(tandem,triple)wheretwoorthreesemiconductorswithdifferentbandgapvaluesareusedtobetterexploitthesolarspectrum.Anotherapproachistoconvertphotonsnotusefulforthespecificband-gaptoenergyrangeswhichhaveabettermatchwiththebandgapvalue:wetalkthenaboutconvertors(upconvertersanddownconverters).