Download e = electron charge = 1.6x10-19 C me = electron mass = 9.1x10

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).