Laboratory Frame Fig-IV v?Center of Mass Frame av?Collision in the Laboratory Frame?? a2v?TheoreticalTask31-Neutronva?vm(T-3) before collision b2?L2-Neutron after collision:Solutions 4of9B.TheModerator13-Moderator Atom before collision ?L3v??v?bm4-?Moderator Atom after collision Consideratom3thetwodimensionalelastic4collision betweenofmassAu.Before?aneutronofmass1uandamoderatorlaboratoryframe(LF).4Let?→collision?allthemoderator?vmatomsareconsideredatrestintherespectivelyintheLF.Let?vv→band→vabethevelocitiesoftheneutronbeforeandaftercollisionmbethevelocityofthecenterofmass(CM)framerelativetoLFandθbetheneutronscatteringangleintheCMframe.Alltheparticlesinvolvedincollisionsaremovingatnon-relativisticspeeds
B1ThecollisioninLFisshownschematicallywithθLasthescatteringangle(Fig-IV).Sketch
thecollisionschematically?inCMframe.Labeltheparticlevelocitiesfor1,termsof?→vb,?→vaandv→2and3in
m
.Indicatethescatteringangleθ.Solution:
B2ObtainvandV,thespeedsoftheneutronandthemoderatoratomintheCMframeafter
thecollision,intermsofAandvb.
Solution:Detailedsolution:BeforethecollisionintheCMframe(vb?vm)andvmwillbemagnitudeofthevelocitiesoftheneutronandmoderatoratomrespectively.
FrommomentumconservationintheCMframe,vb?vm=Avmgivesvm=vbA+1.Afterthecollision,letvandVbemagnitudeofthevelocitiesofneutronandmoderatoratomrespectivelyintheCMframe.Fromconservationlaws,
v=AV
and
1(v212b?vm)+2112Avm=2v2+2
AV2.(→[0.2+0.2]) ?vb1[1.0]
[1.0]
TheoreticalTask3(T-3):Solutions
5of9
Solvinggivesv=AvbandV=vb
A+1A+1.(OR)Fromde?nitionofcenterofmassframevm=vb
.BeforethecollisionintheCMframevb?vm=AvbAthevelocitiesoftheneutronandmoderatoratomAandvmwillbenitude+1mag-of+1respectively.InelasticcollisiontheparticlesarescatteredintheoppositedirectionintheCMframeandso
thespeedsremainsamev=Avb
A+1andV=vbA+1(→[0.2+0.1]).Note:Alternativesolutionsareworkedoutintheendandwillgetappropriateweigh-tage.
B3DeriveanexpressionforG(α,θ)=Ea/Eb,whereEbandEaarethekineticenergiesofthe
neutron,intheLF,beforeandafterthecollisionrespectively,andα≡[(A?1)/(A+1)]2,
Solution:
G(α,θ)=Ea
A2+2Acosθ+1E==1[(1+α)+(1?α)cosθ].b(A+1)22
Detailedsolution:Since?→va=?→v+?v→m,v2a=v2+v2m
+2vvmcosθ(→[0.3]).Substi-tutingthevaluesofvandvm,v2
a=A2v2bv2b2Av22+b(A+1)2+(A+1)(A+1)2cosθ(→[0.2]),so
v2
aEaA2+2Acosv2=E=θ+1.bb(A+1)2
α,θ)=A2G(+1(A+1)2+2A(A+1)2cosθ=1
2
[(1+α)+(1?α)cosθ].
Alternateform
=1?(1?α)(1?cosθ)
2.
Note:Alternativesolutionsareworkedoutintheendandwillgetappropriateweigh-tage.
B4AssumethattheaboveexpressionholdsforD2Omolecule.Calculatethemaximumpos-siblefractionalenergylossfl≡Eb?Ea
Eb
oftheneutronfortheD2O(20u)moderator.Solution:fl=0.181
Detailedsolution:Themaximumenergylosswillbewhenthecollisionisheadonie.,Eawillbeminimumforthescatteringangleθ=π.SoEa=Emin=αEb.
ForD2O,α=0.819andmaximumfractionalloss??
Eb?Emin
??
Eb
=1?α=0.181.[Ac-
ceptableRange(0.170to0.190)]
[1.0]
[0.5]
TheoreticalTask3(T-3):Solutions
6of9
C.TheNuclearReactor
TooperatetheNRatanyconstantneutron?uxΨ(steadystate),theleakageofneutronshastobecompensatedbyanexcessproductiongeometrytheleakagerateisk??1??ofneutronsinthereactor.Forareactorincylindrical2.405R??2+??πH??2??Ψandtheexcessproductionrateisk2Ψ.Theconstantsk1andk2dependonthematerialpropertiesoftheNR.
C1ConsideraNRwithk1=1.021×10?2mandk2=8.787×10?3m?1.Notingthatfora
?xedvolumetheleakagerateistobeminimizedfore?cientfuelutilisationobtainthedimensionsoftheNRinthesteadystate.
Solution:R=3.175m,H=5.866m.
Detailedsolution:ForconstantvolumeV=πR2H,
d????2.405??2??π????dHR+2
H
=0,
d??2.4052πHπ2??
π2dHV+2.4052πH2=V?2H
3=0,
gives
??2.405πR??2
=2
??H??2
.
Forsteadystate,
??1.021×10?2
??2.405
??2
??
R
+??π??2H
Ψ=8.787×10?3Ψ.
HenceH=5.866m[AcceptableRange(5.870to5.890)]R=3.175m[AcceptableRange(3.170to3.180)].
AlternativeNon-CalculusMethodMinimisationoftheexpression??toOptimize2.405??2????
R+π2
H
,fora?xedvolumeV=
πR2H:
SubstitutingforR2
intermsofV,Hweget2.4052πHπ2V+H
2,
whichcanbewrittenas,2.4052πH2.4052πHπ2
2V+2V+H2.
SinceallthetermsarepositiveapplyingAMGMinequalityforthreepositivetermswe
get
2.4052πH2V+2.4052πHπ2????2V+H232.4052πH2.4052πHπ232.4054π43≥2V×2V×H2=4V2
.
[1.5]
TheoreticalTask3(T-3):Solutions
7of9
TheRHSisaconstant.TheLHSisalwaysgreaterorequaltothisconstantim-pliesthatthisistheminimumvaluetheLHScanachieve.Theminimumisachieved
whenall2.4052πH
π2??thethree??positivetermsareequal,whichgivesthecondition
=2??2VπH2?2.405
??2
R=2H.Forsteadystate,
??2
1.021×10?2
??2.405
??R
+??π??2??
3H
Ψ=8.787×10?Ψ.
HenceH=5.866m[AcceptableRange(5.870to5.890)]R=3.175m[AcceptableRange(3.170to3.180)].
Note:PuttingtheconditionintheRHSπ2
??
givestheminimumasH2.Fromthecondi-tionwegetπ32.4052π2π232.4054π
4H3=2V?H2=4V2.
Note:TheradiusandheightoftheTarapur3&4NRinWesternIndiais3.192mand
5.940mrespectively.
C2Thefuelchannelsareinasquarearrangement(Fig-III)withnearestneighbourdistance
0.286m.Thee?ectiveradiusofafuelchannel(ifitweresolid)is3.617×10?2m.EstimatethenumberoffuelchannelsFninthereactorandthemassMofUO2requiredtooperatetheNRinsteadystate.
Solution:Fn=387andM=9.892×104kg.
Detailedsolution:Sincethefuelchannelsareinsquarepitchof0.286m,theef-fectiveareaperchannelis0.2862m2=8.180×10?2m2.
Thecross-sectionalareaofthecoreisπR2=3.142×(3.175)2=31.67m2,sothemaximumnumberoffuelchannelsthatcanbeaccommodatedinthecylinderisthe
integerpartof31.67
0.0818=387.Massofthefuel=387×Volumeoftherod×density
=387×(π×0.036172×5.866)×10600=9.892×104kg.
Fn=387[AcceptableRange(380to394)]
M=9.892×104kg[AcceptableRange(9.000to10.00)]
Note1:(Notpartofgrading)Thetotalvolumeofthefuelis387×(π×0.036172×5.866)=9.332m3.Ifthereactorworksat12.5%e?cienythenusingtheresultofa-(iii)wehavethatthepoweroutputofthereactoris9.332×4.917×108×0.125=
[1.0]
TheoreticalTask3(T-3):Solutions
573MW.
8of9
Note2:TheTarapur3&4NRinWesternIndiahas392channelsandthemassofthefuelinitis10.15×104kg.Itproduces540MWofpower.
AlternativeSolutionstosub-partsB2andB3:LetσbethescatteringangleoftheModeratoratomintheLF,takenclockwisewithrespecttotheinitialdirectionoftheneutronbeforecollision.LetUbethespeedoftheModeratoratom,intheLF,aftercollision.FrommomentumandkineticconservationinLFwehave
vb=vacosθL+AUcosσ,0=vasinθL?AUsinσ,
11212
.vb=AU2+va
222
Squaringandaddingeq(1)and(2)toeliminateσandfromeq(3)weget
22A2U2=va+vb?2vavbcosθL,
22
A2U2=Avb?Ava,
(1)
(2)(3)
(4)(5)
whichgives
22
2vavbcosθL=(A+1)va?(A?1)vb.
(ii)LetvbethespeedoftheneutronaftercollisionintheCOMF.Fromde?nitionofcentervb
ofmassframevm=.
A+1vasinθLandvacosθLaretheperpendicularandparallelcomponentsofva,intheLF,resolvedalongtheinitialdirectionoftheneutronbeforecollision.TransformingthesetotheCOMFgivesvasinθLandvacosθL?vmastheperpendicularcomponentsofv.Substitut-??andparallel22sinθ+v2cos2θ+v2?2vvcosθingforvmandfor2vavbcosθLfromeq(5)inv=vaLLamLam
Avb2
=.SquaringthecomponentsofvtoeliminateθLgivesvaandsimplifyinggivesv=
A+1
2
v2+vm+2vvmcosθ.Substitutingforvandvmandsimplifyinggives,
2
vaEaA2+2Acosθ+1
.==2vbEb(A+1)2
Ea2A1A2+1
G(α,θ)=+cosθ==[(1+α)+(1?α)cosθ].
Eb(A+1)2(A+1)22
(OR)
vb
.Afterthecollision,letvandVA+1
bemagnitudeofthevelocitiesofneutronandmoderatoratomrespectivelyintheCOMF.FromconservationlawsintheCOMF,
(iii)Fromde?nitionofcenterofmassframevm=
v=AV
and
11211(vb?vm)2+Avm=v2+AV2.2222
AvbvbSolvinggivesv=AandV=.Wealsohavevcosθ=vacosθL?vm,substitutingforvm
+1A+1
andforvacosθLfromeq(5)andsimplifyinggives
2vaEaA2+2Acosθ+1==.2vbEb(A+1)2
TheoreticalTask3(T-3):Solutions
A2+1Ea2A1
G(α,θ)==+cosθ=[(1+α)+(1?α)cosθ].
Eb(A+1)2(A+1)22
(OR)
(iv)Fromde?nitionofcenterofmassframevm=
9of9
vb
.Afterthecollision,letvandVA+1
bemagnitudeofthevelocitiesofneutronandmoderatoratomrespectivelyintheCMframe.FromconservationlawsintheCMframe,
v=AV
and
11211(vb?vm)2+Avm=v2+AV2.2222
Avbb
Solvinggivesv=AandV=Av.UsinσandUcosσaretheperpendicularandparallel+1+1componentsofU,intheLF,resolvedalongtheinitialdirectionoftheneutronbeforecollision.TransformingthesetotheCOMFgivesUsinσand?Ucosσ+vmastheperpendicularand
2
?2Vvmcosθ.SinceV=vmparallelcomponentsofV.SowegetU2=V2sin2θ+V2cos2θ+vm
2
(1?cosθ).SubstitutingforUfromeq(4)andsimplifyinggiveswegetU2=2vm
2
vaEaA2+2Acosθ+1==.2vbEb(A+1)2
EaA2+12A1
G(α,θ)==+cosθ=[(1+α)+(1?α)cosθ].
Eb(A+1)2(A+1)22
√
A2+2Acosθ+1Note:Wehaveva=vb.Substitutingforva,v,vminvcosθ=vacosθL?vm
A+1
givestherelationbetweenθLandθ,
cosθL=√Acosθ+1
.
A2+2Acosθ+1Treatingtheaboveequationasquadraticincosθgives,
??
2
?sinθL±cosθLA2?sin2θL
cosθ=.
A
ForθL=0?therootwiththenegativesigngivesθ=180?whichisnotcorrectso,
??
cosθLA2?sin2θL?sin2θL
cosθ=.
A
2va
Substitutingtheaboveexpressionforcosθintheexpressionfor2givesanexpressioninterms
vb
ofcosθL??
22
vaEaA+2cosθLA2?sin2θL+cos2θL
==.22vbEb(A+1)