Q T-3 The Design of a Nuclear Reactor1
Page 1 of 2 (Total Marks: 10)
Uranium occurs in nature as UO2 with only 0.720% of the uranium atoms being 235U. Neutron induced fission occurs readily in 235U with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other 235U nuclei. This forms the basis of the power generating nuclear reactor (NR).
A typical NR consists of a cylindrical tank of height H and radius R filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural UO2 in solid form of height H, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry.
Fig-I Fig-II Fig-III Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown).
A Fuel Pin 1. Molecular weight Mw = 0.270 kg mol-1 Data for UO2 3. Melting point Tm = 3.138×103 K 2. Density ρ = 1.060×104 kg m-3 4. Thermal conductivity λ = 3.280 W m-1 K-1 A1 A2 A3 A4 Consider the following fission reaction of a stationary 235U after it absorbs a neutron of negligible kinetic energy. 235U + 1n → 94Zr + 140Ce + 2 1n + Estimate (in MeV) the total fission energy released. The nuclear masses are: m(235U) = 235.044 u; m(94Zr) = 93.9063 u; m(140Ce) = 139.905 u; m(1n) = 1.00867 u and 1 u = 931.502 MeV c-2. Ignore charge imbalance. Estimate N the number of 235U atoms per unit volume in natural UO2. Assume that the neutron flux density, φ = 2.000×1018 m-2 s-1 on the fuel is uniform. The fission cross-section (effective area of the target nucleus) of a 235U nucleus is σf = 5.400×10-26 m2. If 80.00% of the fission energy is available as heat, estimate Q (in W m-3), the rate of heat production in the pin per unit volume. 1MeV = 1.602×10-13 J The steady-state temperature difference between the center (Tc) and the surface (Ts) of the pin can be expressed as Tc?Ts = k F(Q,a,λ), where k = 1 ∕ 4 is a dimensionless constant and a is the radius of the pin. Obtain F(Q,a,λ) by dimensional analysis. Note that λ is the thermal conductivity of UO2. 0.8 0.5 1.2 0.5 1
Joseph Amal Nathan (BARC) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the
principal authors of this problem. The contributions of the Academic Committee, Academic Development Group and the International Board are gratefully acknowledged.
Q T-3 Page 2 of 2 The desired temperature of the coolant is 5.770×102 K. Estimate the upper limit au on the radius a of the A5 1.0 pin. B The Moderator
Consider the two dimensional elastic collision between a neutron of mass 1 u and a moderator atom of mass A u. Before collision all the moderator atoms are considered at rest in the laboratory frame (LF). Let ? ??? and ???? ???? be the velocity be the velocities of the neutron before and after collision respectively in the LF. Let ?
of the center of mass (CM) frame relative to LF and θ be the neutron scattering angle in the CM frame. All the particles involved in collisions are moving at nonrelativistic speeds. The collision in LF is shown schematically, where θL is the scattering angle (Fig-IV). Sketch the collision schematically in CM frame. Label the particle velocities for 1, 2 and 3 in terms of ? ??? ??? ???? . Indicate , ? and ? the scattering angle θ. ???? Fig-IV 2
B1 ???? 1θL 34Collision in the Laboratory Frame 1-Neutron before collision 2-Neutron after collision 3-Moderator Atom before collision 4-Moderator Atom after collision 1.0 Obtain v and V, the speeds of the neutron and moderator atom in the CM frame after collision, in terms of A 1.0 and . Derive an expression for G(α, θ) = Ea ∕ Eb , where Eb and Ea are the kinetic energies of the neutron, in the B3 1.0 LF, before and after the collision respectively and . Assume that the above expression holds for D2O molecule. Calculate the maximum possible fractional B4 energy loss of the neutron for the DO (20 u) moderator. 0.5 2 B2 C The Nuclear Reactor
To operate the NR at any constant neutron flux ψ (steady state), the leakage of neutrons has to be
compensated by an excess production of neutrons in the reactor. For a reactor in cylindrical geometry the leakage rate is k1 [(2.405 ∕ R)2 + (π ∕ H)2] ψ and the excess production rate is k2 ψ. The constants k1 and k2 depend on the material properties of the NR. Consider a NR with k1 = 1.021×10-2 m and k2 = 8.787×10-3 m-1. Noting that for a fixed volume the leakage C1 1.5 rate is to be minimized for efficient fuel utilization, obtain the dimensions of the NR in the steady state. The fuel channels are in a square arrangement (Fig-III) with the nearest neighbour distance 0.286 m. The C2 effective radius of a fuel channel (if it were solid) is 3.617×10-2 m. Estimate the number of fuel channels Fn 1.0 in the reactor and the mass M of UO2 required to operate the NR in steady state. Fig-I Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a TheoreticalTask3(T-3)fuel channel (1-:Solutions1of9Fuel Pins) Fig-II: A view of the NR TheDesignofaNuclearReactor1) (2-Fuel ChannelsFig-III: Top view of NR (3-Square Arrangement of UraniumoccursinnatureasUO2withonly0.720%oftheuraniumatomsbeing235U.NeutronFuel and havinghighinduced?ssionoccursreadilyin235Uwiththeemissionof2-3Channels?ssionneutrons4-Typical Neutron Paths). kineticenergy.This?ssionprobabilitywillincreaseiftheneutronsinducing?ssionhavelowkineticenergy.Sobyreducingthekineticenergyofthe?ssionneutrons,onecaninduceachainOnly components relevant of?ssionsinother235Unuclei.Thisformsthebasisofthepowergeneratingnuclearreactorto the problem are shown (NR). (e.g. control rods and coolant are not shown). AtypicalNRconsistsofacylindricaltankofheightHandradiusR?lledwithamaterialcalledmoderator.Cylindricaltubes,calledfuelchannels,eachcontainingaclusterofcylindricalfuelpinsofnaturalUO2insolidformofheightH,arekeptaxiallyinasquarearray.Fissionneutrons,comingoutwardfromafuelchannel,collidewiththemoderator,losingenergy,andreachthesurroundingfuelchannelswithlowenoughenergytocause?ssion(FigsI-III).Heatgeneratedfrom?ssioninthepinistransmittedtoacoolant?uid?owingalongitslength.Inthecurrentproblemweshallstudysomeofthephysicsbehindthe(A)FuelPin,(B)Moderatorand(C)NRofcylindricalgeometry.
Fig-II Fig-III A.FuelPin
1.MolecularweightMw=0.270kgmol3.MeltingpointTm=3.138×103K
?1
DataforUO2
2.Densityρ=1.060×104kgm?3
4.Thermalconductivityλ=3.280Wm?1K?1
235
A1Considerthefollowing?ssionreactionofastationary
negligiblekineticenergy.
235
1
Uafteritabsorbsaneutronof
U+1n?→94Zr+140Ce+21n+?E
JosephAmalNathan(BARC)andVijayA.Singh(ex-NationalCoordinator,ScienceOlympiads)weretheprincipalauthorsofthisproblem.ThecontributionsoftheAcademicCommittee,AcademicDevelopmentGroupandtheInternationalBoardaregratefullyacknowledged.
TheoreticalTask3(T-3):Solutions
2of9
Estimate?E(inMeV)thetotal?ssionenergyreleased.Thenuclearmassesare:m(235U)=235.044u;m(94Zr)=93.9063u;m(140Ce)=139.905u;m(1n)=1.00867uand1u=931.502MeVc?2.Ignorechargeimbalance.Solution:?E=208.684MeV
Detailedsolution:Theenergyreleasedduringthetransformationis
?E=[m(235U)+m(1n)?m(94Zr)?m(140Ce)?2m(1n)]c2
Sincethedataissuppliedintermsofuni?edatomicmasses(u),wehave
?E=[m(235U)?m(94Zr)?m(140Ce)?m(1n)]c2
=208.684MeV[AcceptableRange(208.000to209.000)]
fromthegivendata.A2EstimateNthenumberof
235
UatomsperunitvolumeinnaturalUO2.
Solution:N=1.702×1026m?3
Detailedsolution:ThenumberofUO2moleculesperm3ofthefuelN1isgiveninthetermsofitsdensityρ,theAvogadronumberNAandtheaveragemolecularweightMwas
N=
ρNA
1M=
10600w
×6.022×1023
0.270=2.364×1028m?3EachmoleculeofUO2containsoneuraniumatom.Sinceonly0.72%oftheseare235
U,
N=0.0072×N1
=1.702×1026m?3[AcceptableRange(1.650to1.750)]A3Assumethattheneutron?uxφ=2.000×1018m?2s?1onthefuelisuniform.The?ssion
cross-section(e?ectiveareaofthetargetnucleus)ofa235Unucleusisσf=5.400×10?26m2.If80.00%ofthe?ssionenergyisavailableasheat,estimateQ(inWm?3)therateofheatproductioninthepinperunitvolume.1MeV=1.602×10?13J.
Solution:Q=4.917×108W/m3
Detailedsolution:Itisgiventhat80%ofthe?ssionenergyisavailableasheatthustheheatenergyavailableper?ssionEfisfroma-(i)Ef=0.8×208.7MeV
=166.96MeV=2.675×10?11J
Thetotalcross-sectionperunitvolumeisN×σf.Thustheheatproducedperunit
[0.8]
[0.5]
[1.2]
TheoreticalTask3(T-3):Solutions
3of9
volumeperunittimeQisQ=N×σf×φ×Ef
=(1.702×1026)×(5.4×10?26)×(2×1018)×(2.675×10?11)W/m3=4.917×108W/m3[AcceptableRange(4.800to5.000)]
A4Thesteady-statetemperaturedi?erencebetweenthecenter(Tc)andthesurface(Ts)ofthe
pincanbeexpressedasTc?Ts=kF(Q,a,λ)wherek=1/4isadimensionlessconstantandaistheradiusofthepin.ObtainF(Q,a,λ)bydimensionalanalysis.
Solution:TQa2
c?Ts=4λ
.
Detailedsolution:ThedimensionsofTc?Tsistemperature.WewritethisasTc?Ts=[K].OncecansimilarlywritedownthedimensionsofQ,aandλ.EquatingthetemperaturetopowersofQ,aandλ,onecouldstatethefollowingdimensionalequation:K=Qαaβλγ
=[ML?1T?3]α[L]β[ML1T?3K?1]γ
Thisyieldsthefollowingalgebraicequationsγ=-1equatingpowersoftemperature
α+γ=0equatingpowersofmassortime.Fromthepreviousequationwegetα=1Next?α+β+γ=0equating2
powersoflength.Thisyieldsβ=2.
ThusweobtainTQac?Ts=
whereweinsertthedimensionlessfactor1/4assug-gestedintheproblem.Nopenalty4λ
ifthefactor1/4isnotwritten.Note:Samecreditforalternatewaysofobtainingα,β,γ.
A5Thedesiredtemperatureofthecoolantis5.770×102K.Estimatetheupperlimitauon
theradiusaofthepin.
Solution:au=8.267×10?3m.
Detailedsolution:ThemeltingpointofUO2is3138Kandthemaximumtempera-tureofthecoolantis577K.Thissetsalimitonthemaximumpermissibletemperature(Tc?Ts)tobelessthan(3138-577=2561K)toavoid“meltdown”.Thusonemaytakeamaximumof(Tc?Ts)=2561K.Notingthatλ=3.28W/m-K,wehave
a22561×4×3.28
u=
4.917×108
WherewehaveusedthevalueofQfromA2.Thisyieldsau??8.267×10?3m.Soau=8.267×10?3mconstitutesanupperlimitontheradiusofthefuelpin.Note:TheTarapur3&4NRinWesternIndiahasafuelpinradiusof6.090×10?3m.
[0.5]
[1.0]