Reactor Physics 1

Nuclear Reactor Theory 0302750Nuclear Reactor Theory 0302750

Course webCourse webhttp://nuclear bau edu jo/ju/ju-reactors/http://nuclear.bau.edu.jo/ju/ju reactors/

orhttp://nuclear dababneh com/ju/ju-reactors/http://nuclear.dababneh.com/ju/ju reactors/

Nuclear Reactor Theory, JU, First Semester, 2010-2011(Saed Dababneh).

1

GradingReview Test 10%Mid term Exam 30%Mid-term Exam 30%Projects, quizzes and HWs 20%Final Exam 40%

• Homeworks and small projects are due after one• Homeworks and small projects are due after one week unless otherwise announced.• Remarks or questions marked in red without being• Remarks or questions marked in red without being announced as homeworks should be also seriously considered!considered!• Some tasks can (or should) be sent by email:

saed@dababneh comNuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).2

[email protected]

Review Test

R d

Will do the test

afterwardsRead LamarshChs 1 2

afterwards.

Review relevant

Chs. 1, 2 and 3.Readrelevant

material.Read

Krane Ch. 13.

Nuclear Reactor Theory, JU, First Semester, 2010-2011 (Saed Dababneh).

3

Projects

Topics related to:Topics related to:• Heat removal.• Radiation protection.• Radiation shielding.• Reactor licensing and safety.

N l it• Nuclear security.• Uranium mining or other front-end fuel cycle elements.• Back-endBack end.• Other topics that you would like to suggest…..


4

Projects

• Please do your own thorough research on relevant• Please do your own thorough research on relevant topics you may find appropriate.• Provide your suggestion next week• Provide your suggestion next week.• Final decision on the subject of your project should be taken before mid Octoberbe taken before mid October.• Due date for written version is Monday, December 20th20 .• Presentation date will be decided later.


5

Nuclear Reaction Energetics (revisited)

Conservation Laws• Charge, Baryon number, total energy, linear momentum, angular g y gy gmomentum, parity, (isospin??) …….

bQTT22

θφ

ap X

pb QTTcmcm iffi =−=− 22

φpa XpY Y

+ve Q-value exoergic reaction.-ve Q-value endoergic reaction.

aYb TQTT +=++ve Q-value reaction possible if T 0+ve Q-value reaction possible if Ta 0.-ve Q-value reaction not possible if Ta 0. (Is Ta > |Q| sufficient?).

Conservation of momentum ……


6


• Conservation of momentum.• We usually do not detect Y.

HW HW 11y

Show that:aaYYbYabaaba TmmQmmmTmmTmm

T−+++± ])()[(coscos 2θθ

• The threshold energy (for Ta): (the condition occurs for θ = 0º).bY

aaYYbYabaabab mm

T+

=

gy ( a)

abY

bYTh mmm

mmQT−+

+−=

• +ve Q-value reaction possible if Ta 0.• -ve Q-value reaction possible if Ta > TTh.

abY

a Th• Coulomb and other barriers…….!!!• Neutrons vs. charged particles.


7


Th d bl l d it ti b t T d th

HW HW 11 (continued)(continued)

• The double valued situation occurs between TTh and the upper limit Ta

\.YmQT \

aY

Ya mm

QT−

−=\

• Double-valued in a forward cone.

aaYYbY TmmQmmm ])()[(cos max2 −++

−=θaba Tmmmax


8


Di th hl th 7Li( ) ti

HW HW 11 (continued)(continued)

• Discuss thoroughly the 7Li(p,n) reaction.• During the discussion emphasize on the case

h th i id t t b i 30 k V bwhen the incident proton beam is 30 keV above the threshold.

U ti kill• Use your computing skills.

• Discuss the elasticelastic and inelastic inelastic scatteringscattering of neutronsneutrons using thesescatteringscattering of neutronsneutrons using these relations.


9

Nuclear Reaction Energetics


10



11



12



13


• If the reaction reaches excited states of Y

EQcmEcmcmcmQ =++= 2222 )(

58 61

exbexYaXex EQcmEcmcmcmQ −=−+−+= 0)(

58Ni(α,p)61Cu

l

less proton energy

even less ….

Highest proton energy

See Figures 11.4 i K


14

in Krane



15

Neutron Interactions (revisited)

• Chadwick’s discovery.• Neutrons interact with nuclei, not with atoms. (Exceptions).

• Recall from basic Nuclear Physics:o Inelastic scattering (n,n\). Q = -E* Inelastic gammas.

Threshold?o Elastic scattering (n,n). Q = ?? (Potential and CN).

N t d ti ?Neutron moderation?o Radiative capture (n,γ). Q = ?? Capture gammas.

(n α) (n p) Q = ?? Absorption Reactionso (n,α), (n,p). Q = ?? Absorption Reactions.o (n,2n), (n,3n) Q = ?? Energetic neutrons on heavy water can easily eject the loosely bound neutroncan easily eject the loosely bound neutron.o Fission. (n,f).

HWHW 22 Examples of such exo- and endo-thermic reactions with Q


16

HW HW 22 Examples of such exo and endo thermic reactions with Q calculations.

Neutron Scattering (revisited)

• Elastic or inelastic.• Analogous to diffraction.• Alternating maxima and minima• Alternating maxima and minima.• First maximum at

h=λ R

λθ ≈

(3

1ARR

p

o=

λ

• Minimum not at zero (sharp edge of the nucleus??)• Clear for neutrons• Clear for neutrons.• Protons? High energy, large angles. Why? 222 11)(θσ ⎞

⎜⎛⎞

⎜⎛ zZedg y

• Inelastic Excited states, 2

4sin1

41

4)(

θπεθσ

⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎜⎝

⎛=

Ω ao TzZe

dd


17

energy, X-section and spin-parity.

Reaction Cross Section (revisited)

• Probability.• Projectile a will more probably hit target X if area is larger.• Classically: σ = π(Ra + RX)2.Classical σ = ??? (in b) n + 1H, n + 238U, 238U + 238U

Q t h i ll D2• Quantum mechanically: σ = π D2.

CMXa

Emm

2hh

D =+

=

• Coulomb and centrifugal barriers energy dependence of σ.What about neutrons?What about neutrons?

CMaXaXaaX EEmm µ22

What about neutrons?What about neutrons?• Nature of force:

Strong: 15N(p,α)12C σ ~ 0.5 b at Ep = 2 MeV.Strong: N(p,α) C σ 0.5 b at Ep 2 MeV.Electromagnetic: 3He(α,γ)7Be σ ~ 10-6 b at Eα = 2 MeV.Weak: p(p,e+ν)D σ ~ 10-20 b at Ep = 2 MeV.


18

p• Experimental challenges to measure low X-sections..

Reaction Cross Section (Simple terms)A (A f h ??!!)

XA (Area of what??!!)|v|

Monoenergetic (and Target with N atoms.cm-3 or NAX atomsPosition of a neutron 1 s

b f i iMonoenergetic (and unidirectional) neutrons of speed v (cm.s-1) and

density n (cm-3)

gbefore arriving at target

Volume = vAcontaining nvA neutrons that hit the

NX??y ( ) containing nvA neutrons that hit the

“whole!!” target in 1 s.Beam Intensity I ≡ nvA/A = nv (cm-2s-1)

Number of neutrons interacting with target per secondNumber of neutrons interacting with target per second∝ I, A, X and N= σt I N A XTotal microscopic cross


19

σt I N A XTotal microscopic cross section

Reaction Cross Section (Simple terms)

Number of neutrons interacting with target per second= σt I N A Xt

Total microscopic Total number of

Number of interactions with a single nucleus per second

pcross section nuclei in the

target

Number of interactions with a single nucleus per second = σt I Interpretation and units of σ.

nvA = IA neutrons strike the target per second, of these σtI neutrons interact with any single nucleus. Thus,

measures the probability for a neutron to hit a nucleus (per unit area of target)AAI

I tt σσ=


20

(per unit area of target).AAIEffective cross-sectional area of the nucleus.

Reaction Cross Section (Simple terms)The probability for a neutron to hit a nucleus (per unit area of target):

AAII tt σσ=

AAI

Typical nucleus (R=6 fm): geometrical πR2 ≈ 1 b.Typical σ: <µb to >106 b


21

Typical σ: <µb to >106 b.


Number of neutrons interacting with target per second= σt I N A X t

Volume of the Total

microscopic

Number of interactions per cm3 per second (Collision Density)

targetp

cross section

Number of interactions per cm3 per second (Collision Density)Ft = σt I N = I Σt

Σt = N σtt t

Macroscopic total cross

section

XteIXI = Σ−)( 0section.

Probability per unit path length. t Σ

=1λMean free path


22

g

tΣ

Neutron AttenuationX

I II0 IRecall Σt = N σt XteIXI Σ−= 0)(

Probability per unit path

mfp for scattering λs = 1/Σsmfp for absorption λ = 1/Σ

length.mfp for absorption λa 1/Σa

………….

total mfp λt = 1/Σt

XteXP Σ−=)(i iProbability

total mfp λt 1/Σt

Nuclear Reactor Theory, JU, First Semester,2010-2011 (Saed Dababneh).

23XteXP

eXPΣ−

−

−=1)(

)(

ninteractio

ninteractionoProbability


Homogeneous Mixture

yyxxyx NN σσ +=Σ+Σ=Σ

Molecule xmyn Nx=mN, Ny=nN

yx nm σσσ +=given that events at x and y are independent.

yx


24

Reaction Cross Section

Detector for particle “b”dΩ

θ,φIa\

p

d“b” particles / s

2θ,φ

\\ NIdRd b=σ

cm2

NIa


25


Many different quantities are called“cross section”.

dΩAngular distribution

Krane Table 11.14

),( drdRb πφθ Ω

=Units … !

\\4),(

NIr

dd

πφθσ

=Ω

“Differential” cross sectionσ(θ,φ) or σ(θ )or “cross section” !! 4 NId aπΩor “cross section” …!!

=Ω ddd φθθ2

sin Doubly differential

∫ ∫ ∫ Ω=Ω

Ω=

ddddd

dd σφθθσσ

π π

0

2

0

sin d σ2

dEdσ

0 0 ΩdEddE

σ for all “b” particlesNuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).26

σt for all b particles.


27

nn--TOFTOFCERNCERN


28


29

Neutron Cross Section (Different Features)

1/v Fast neutrons should be

d t dmoderated.

235U thermal cross sectionsσfission ≈ 584 b.σscattering ≈ 9 b.σ ≈ 97 bσradiative capture ≈ 97 b.


30Fission Barriers

Neutron Induced Reactions22 nXHCCHbY IIIn ++∝ DσX(n,b)Y

Γn(En)Γb(Q+En)2

11vE

∝∝

)(EP )( nln EPvn

∝Probability to ypenetrate the potential barrier

P (E ) 1For thermal neutronsQ >> En

Γb(Q) ≈ constant Po(Ethermal) = 1P>o(Ethermal) = 0

1v

Enn1)( ∝σNon-resonant


31

v

Neutron Induced Reactions


32

Statistical Factor (Introduction)

D

hh bbplL ===

D

Dlb =222

1max, )12( Dπππσ +=−= + lbb lll

)()(7.656)(2

keVEub CMµ

π =DHW HW 33)()( keVEuµ

122 J +Generalization

)1()12)(12(

122max aX

XaaX JJ

J δπσ +++

+= D


33

ω


ExcitedState

ExJπ a + X Y + b Q > 0b + Y X + a Q < 0

Entrance Channela + X

ExitChannel

b Y X a Q 0

Inverse Reactiona X Channelb + YCompound

Nucleus C*

Inverse Reaction

212J +More Generalization

22 )1()12)(12(

12 XaHCCHbYJJ

JIIIaX

XaaXaX +++

+++

= δπσ D

QM StatisticalFactor (ω)

Identicalparticles

• Nature of force(s).• Time-reversal invariance.

212J + 22 )1()12)(12(

12 YbHCCHXaJJ

JIIIbY

YbbYbY +++

+++

= δπσ D

??=aXσHWHW 44Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).34

??=bYσ

HW HW 44

Resonance Reactions

Projectile Projectile

TargetQ l Q-value

TargetQ-value Q-value

Q + ER = Er

Direct ResonantEγ = E + Q - Eex

Direct Capture(all energies)

Resonant Capture(selected energies ( g ) ( gwith large X-section)

2XaHY +∝ γγσ

22XaHEEHE CNrrf +∝ γγσ


35

γγ CNrrf γγ


36

Resonance Reactions

fDamped OscillatorDamped Oscillator Oscillator strength

22

2 )()( δωω +−∝

o

fresponse

Dampingf t

1t

=δ

eigenfrequencyfactor0t

)( ΓΓ∝ baEσ 2

22 )()(

)(Γ+−

∝REE

Eσ

hΓtNuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).37

h=Γ ot

Resonance Reactions

222 )1(12)(

Γ

ΓΓ+

+= ba

aXaXJE δπσ D 2

22 )()(

)()12)(12(

)(Γ+−++ R

aXXa

aX EEJJBreitBreit--Wigner formulaWigner formula Γ+ΓΓBreitBreit Wigner formulaWigner formula

• All quantities in CM systemO l f i l t d

ba Γ+Γ=Γ

• Only for isolated resonances.baR ΓΓ∝σ Reaction Usually Γa >> Γb.

bR

aae

ΓΓΓ∝

σσ Elastic scattering

HWHW 55 When does σ take its maximum value?

y a b.

a

b

e

R

ΓΓ

=σσ HW HW 55 When does σR take its maximum value?


38

Resonance Reactions

Ja + JX + l = JExitChannelExcited

ExJπ

(-1)l π(Ja) π(JX) = π(J) b + YExcitedState

Entrance Ch l(-1)l = π(J) Natural parity.

Compound

Channela + X

Nucleus C*


39

Resonance Reactions

What is the “Resonance Strength” …?What is its significance?What is its significance?In what units is it measured?

ΓΓ+

+ baJ )1(12 δωγΓ

+++

= baaX

Xa JJ)1(

)12)(12(δωγ

ectio

n

Charged particledi ti t ( )

Cro

ss se

E

radiative capture (a,γ)(What about neutrons?)(What about neutrons?)

C ECωγ ∝ Γa ωγ ∝ Γγ

Energy


40

Neutron Resonance Reactions


41

Neutron Activation Analysis

(Z,A) + n (Z, A+1)β-

γ (β-delayed γ-ray)

(Z+1, A+1)

ProjectProject 11Project Project 11NAA and UNAA and UNAA and UNAA and U


42

Neutron Flux and Reaction Rate

Recall Ft = n v σt N = I ΣtSimultaneous beams, different intensities, same energysame energy.Simultaneous beams, different intensities, same energysame energy.

Ft = Σt (IA + IB + IC + …) = Σt (nA + nB + nC + …)vIn a reactorreactor, if neutrons are moving in all directionsall directionsIn a reactorreactor, if neutrons are moving in all directionsall directions

n = nA + nB + nC + …F = Σ nvFt = Σt nv

Not talkingneutron flux φ = nv

Not talking about a beam

anymore.

Reaction Rate Rt ≡ Ft = Σt φ = φ /λt (=nvNσt)Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).43same energysame energy


Different energiesDifferent energiesDensity of neutrons with energy between E and E+dEy gy

n(E)dEReaction rate for those “monoenergetic” neutronsg

dRt = Σt(E) n(E)dE v(E)∞ ∞∞

∫=0

)( dEEnn ∫∫∞∞

==00

)()()( dEEEndEE υφφ

∫∫∞∞

∑=∑= )()()()()( dEEEdEEEnER φυ ∫∫ ∑=∑=00

)()()()()( dEEEdEEEnER ttt φυ

∫∞


44∫∑=0

)()( dEEER ii φ


In general, neutron flux depends on:• Neutron energy, E.gy• Neutron spatial position, r.• Neutron angular direction, Ω.g• Time, t.

Various kinds of neutron fluxes (depending on the ( p gdegree of detail needed).Time-dependent and time-independent angularTime dependent and time independent angular neutron flux. ),,,( tEr Ωφ

),,( ΩErφ),,,(φ


45

),,(φ

Neutron Flux and Reaction RateIn Thermal ReactorsThermal Reactors, the absorptionabsorption rate in a “medium” of thermal (MaxwellianMaxwellian) neutrons

∫∑=Th l

aa dEEvEnER )()()(

Usually 1/v cross section, thus

Thermal

)( 0vEa =∑

∫

)()( 0 EvEa∑

then 000000 )()()()( φEnvEdEEnvER aaThermal

aa ∑=∑=∑= ∫Independent of Independent of n(E)n(E)..

The reference energy is chosen at 0.0253 eV. • Look for Thermal Cross Sections.• Actually, look for evaluated nuclear data.Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).46ENDF

Neutron Moderation

Show that, after one elasticelastic scattering the ratio between the final neutron energy E\ and its initial

HW HW 66gy

energy E is given by:[ ]2222\ sincoscos21 −+++ AAAE CM θθθ 11[ ]

22 )1(sincos

)1(cos21

++

=+

++=

AA

AAA

EE θθθ

⎞⎛⎞⎛ 2\ 1AE

11H ?H ?

For a head-on collision: α≡⎟⎠⎞

⎜⎝⎛

+−

=⎟⎟⎠

⎞⎜⎜⎝

⎛

min 11

AA

EE

After n ss--wavewave collisions:where the average change in lethargy lethargy

ζnEEn −= lnln \

)ln( EEu M=g g gygyis

11ln

2)1(1ln

2

\ +−−

+=⎥⎦⎤

⎢⎣⎡==∆

AA

AA

EEu ζ

)( M

12 +⎦⎣ AAE av

47Nuclear Reactor Theory, JU, First Semester, 2010-2011 (Saed Dababneh).

Average decrease in ln(E) after one collision.

Neutron Moderation HW HW 6 6 (continued)(continued)

• Reproduce the plot.• Discuss the effect of the

On 12C.

thermal motion of the moderator atoms.

First collision.

MostMostMost Most probable probable

and average and average energies?energies?

Second collision.energies?energies?


48

Neutron Moderation HW HW 6 6 (continued)(continued)

Neutron scattering by light nuclei then the average energy loss

EE )1(21\ α+=

EEEE )1(1\ α==∆then the average energy loss and the average fractional energy loss

EEEE )1(2 α−=−=∆

)1(1∆E

• How many collisions are needed to thermalize a 2 MeV

)1(21 α−=

EHow many collisions are needed to thermalize a 2 MeV

neutron if the moderator was:1H 2H 4He graphite 238U ?H H He graphite U ?

• What is special about 1H?• Why we considered elastic scattering?Why we considered elastic scattering?• When does inelastic scattering become important?


Nuclear Fission

Coulomb effectSurface effect

~200 MeV

Coulomb effectSurface effect


Nuclear Fission• B.E. per nucleon for 238U (BEU) and 119Pd (BEPd) ?

2 119 BE 238 BE ?? K E f th• 2x119xBEPd – 238xBEU = ?? K.E. of the fragments ≈ 1011 J/g

B i l 105 J/• Burning coal 105 J/g• Why not spontaneous?

T 119Pd f t j t t hi• Two 119Pd fragments just touchingThe Coulomb “barrier” is:

)46( 2

C d ! Wh t if 79Z d 159S ? L t

MeVMeVfm

fmMeVV 2142502.12

)46(.44.12

>≈=

• Crude …! What if 79Zn and 159Sm? Large neutron excess, released neutrons, sharp potential edge,

h i l U !

f

spherical U…!51Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).

Nuclear Fission

• 238U (t½ = 4.5x109 y) for α-decay.• 238U (t½ ≈ 1016 y) for spontaneous fissionU (t½ ≈ 10 y) for spontaneous fission.• Heavier nuclei??• Energy absorption from a neutron (for example) couldEnergy absorption from a neutron (for example) could form an intermediate state probably above barrier induced fissioninduced fission.• Height of barrier is called activation energy.


Nuclear Fission

Liquid Drop

MeV

)

Shell

Ene

rgy

(va

tion

EA

cti


Nuclear Fission34 Rπ ε+= )1(Ra23 abR =3

24 abπ=

=Rb

abR =

Surface Term B = a A⅔

3abπ ε+1

)1( 22 ++ ε

Volume Term (the same)

Surface Term Bs = - as A⅔

Coulomb Term BC = - aC Z(Z-1) / A⅓

...)1( 52 ++ ε

...)1( 251 +− ε

32

31

52

51 )1( AaAZZa SC >− − fission

47~2

>Z Crude: QM and original shape

could be different from sphericalA could be different from spherical.54Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).

Nuclear Fission

)120( 2

48300

)120(=

300Consistent with activation energy curve for A = 300curve for A = 300.

Extrapolation to 47 10-20 sExtrapolation to 47 ≈ 10 20 s.55Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).

Nuclear Fission

235U + n235U + n

93Rb + 141Cs + 2nN t iNot unique.

Low-energy fissionfission processes.


Nuclear FissionZ1 + Z2 = 92Z 37 Z 55Z1 ≈ 37, Z2 ≈ 55A1 ≈ 95, A2 ≈ 140L tLarge neutron excessMost stable:Z=45 Z=58Z=45 Z=58

Prompt neutronsPrompt neutrons within 10-16 sPrompt neutronsPrompt neutrons within 10 16 s.Number ν depends on nature of fragments and on incident neutron energyon incident neutron energy.The average number is characteristic of the processprocess.


Nuclear Fission

The average number of neutrons isneutrons is different, but the distribution is G iGaussian.


Why only left side of the

mass parabola?


59

Higher than S ?

Delayed neutronsDelayed neutrons

Higher than Sn?

Delayed neutronsDelayed neutrons~ 1 delayed neutron

100 fi i b tper 100 fissions, but essential for control of the reactor

In general, of the reactor.

• Waste

g ,β decay favors high • Waste.

• Poison.high

energy.


Nuclear Fission


Nuclear Fission

1/1/vFast neutrons should be

d t dmoderated.

235U thermal cross sectionsσfission ≈ 584 b.σscattering ≈ 9 b.σ ≈ 97 bσradiative capture ≈ 97 b.

Fission Barriers 62Nuclear Reactor Theory, JU, First Semester, 2010-2011 (Saed Dababneh).

Nuclear Fission

• Q for 235U + n 236U is 6.54478 MeV.• Table 13 1 in Krane: Activation energy E for 236U ≈ 6 2 MeV• Table 13.1 in Krane: Activation energy EA for 236U ≈ 6.2 MeV (Liquid drop + shell) 235U can be fissioned with zero-energy neutrons.

238 239• Q for 238U + n 239U is 4.??? MeV.• EA for 239U ≈ 6.6 MeV MeV neutrons are needed.• Pairing term: δ = ??? (Fig 13 11 in Krane)• Pairing term: δ = ??? (Fig. 13.11 in Krane).• What about 232Pa and 231Pa? (odd Z).• Odd-N nuclei have in general much larger thermal fission

63

g gcross sections than even-N nuclei (Table 13.1 in Krane).Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).

Nuclear FissionWhy not use it?Why not use it?

σf,Th 584 2.7x10-6 700 0.019 b


64

Nuclear Fission• Thermal neutron fission of 235U forms compound nucleus that splits up in more than 40 different ways, yielding over 80 primary fission fragments (products).235

92U + 10n 9037Rb + 144

55Cs + 210n

23592U + 10n 87

35Br + 14657La + 31

0n

23592U + 10n 72

30Zn + 16062Sm + 41

0n • The fission yield is defined as the proportion• The fission yield is defined as the proportion (percentage) of the total nuclear fissions that form products of a given mass number.

65Nuclear Reactors, BAU, 1st Semester, 2007-2008 (Saed Dababneh).

p g


Nuclear Fission

• Remember neutron excess.(A Z) (A Z+1) or (A 1 Z)• (A,Z) ⇒ (A,Z+1) or (A-1,Z).

Only leftOnly left side of the

mass parabola


66

parabola.

Nuclear Fission• 235U + n 93Rb + 141Cs + 2n ≈ 165 MeV

average kinetic• Q = ????• What if other fragments?

Diff t b f t

average kinetic energy carried

by fission• Different number of neutrons.• Take 200 MeV as a representative value.

by fission fragments per

fission.

66 MeV 98 MeV

Heavy LightHeavyfragments

g tfragments

67

miscalibratedNuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).

Nuclear Fission• ν neutrons

itt demitted per fission.• ν depends• ν depends on fissioning nuclide and

India?

on neutron energy i d iinducing fission.


68

Nuclear Fission• Mean neutron energy ≈ 2 MeVMeV.• ≈ 2.5 neutrons per fission (average) ≈ 5 MeV(average) 5 MeV average kinetic energy carried by prompt neutrons per fission.

• Show that the average momentum carried by a neutron is only ≈1.5 % that carried by a fragment. • Thus neglecting neutron momenta, show that the ratio between kinetic energies of the two fragments is the inverse of the ratio of their masses E

69

their masses.

1

2

2

1

mm

EE

≈14095

9866

≈Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).

Nuclear Fission

EeE E 292sinh4530)( 036.1−=χHW HW 77 EeE 29.2sinh453.0)( =χ

The experimental pspectrum of prompt neutrons is fitted by ythe above equation. Calculate the mean and the most probable neutron penergies.


70

Nuclear Fission• The fission gamma radiation

Prompt with average energy of 0.9 MeV.β delayed gammas.

HW HW 88•• Investigate how prompt Investigate how prompt gammas interact with gammas interact with water, uranium and lead.water, uranium and lead.



Fission Products• β and γemissions fromemissions from radioactive fission products carry partproducts carry part of the fission energy, even after shut down. • On approaching end of the chain, the decay energy decreases and half-life increases Long-lived isotopes constitute the mainand half-life increases. Long-lived isotopes constitute the main hazard.• Can interfere with fission process in the fuel. Example?Example? (poisoning).(poisoning).p pp (p g)(p g)• Important for research.• β-decay favors high energy ~20 MeV compared to ~6 MeV for γ.


• Only ~ 8 MeV from β-decay appears as heat. Why?Why?

Nuclear FissionSegrè Distribution of fission energy

aLost … ! b

c

•• How much is recoverable?How much is recoverable?How much is recoverable?How much is recoverable?•• What about capture gammas? What about capture gammas? (produced by (produced by νν--1 1 neutronsneutrons))•• Note again that c < (Note again that c < (a+ba+b).).

73

Note again that c < (Note again that c < (a+ba+b).).Nuclear Reactors, BAU, 1st Semester, 2007-2008

(Saed Dababneh).Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).

Nuclear Fission

Distribution of fission energyEnge Distribution of fission energyg

Krane sums

them up as β Lost … !

decays.


Fission Productsβ-f

A, Zi

A, Z-1k

A-1, Zj

(n,γ)j

β-(n,γ)

A, Z+1A+1, Z

dNi/dt = Formation Rate - Destruction rate - Decay Rate

i NNNNNdN λφλφφγ ++ iiiikkjjffii NNNNN

dtλφσλφσφσγ −−++=

Ni saturates and is higher with higher neutron flux, larger “fission yield” and


Ni saturates and is higher with higher neutron flux, larger fission yield and longer half-live.


Fission Products

HWHW 99HW HW 99Investigate the activity, decay and gamma

i f fi i d t f ti f tienergies of fission products as a function of time. Comment on consequences (e.g. rod cooling).

iikki NN

dtdN λλ −=• Shutdown

HWHW 1010Investigate both and

dt

iikk NN λλ > iikk NN λλ <HW HW 1010

giving full description for the buildup and decay of fission fragment i.

iikk



Fission Products[ ] sMeVTttxtP /)(101.4)( 2.02.011 −− +−= [ ])()(

per watt of original operating power.T = time of operation.

Fission product activity after

reactor shutdown?



Nuclear Fission

It is necessary to evaluate the potential hazards• It is necessary to evaluate the potential hazards associated with an accidental release of fission products into the environment.products into the environment.

• It is required to determine a proper cooling time of the spent fuel (before it becomes ready for reprocessing) p ( y p g)that depends on the decay times of fission products.

• It is necessary to estimate the rate at which the heat is released as a result of radioactive decay of the fission products after the shut down of a reactor.

• The poisoning is needed to be calculated (the parasitic capture of neutrons by fission products that accumulate during the reactor operation)during the reactor operation).



Nuclear Fission• Recoverable energy release ≈ 200 MeV per 235U fission.• Fission rate = 2 7x1021 P fissions per day P in MW• Fission rate = 2.7x1021 P fissions per day. P in MW.• 3.12x1016 fissions per second per MW, or 1.2x10-5 gram of 235U per second per MW (thermal).

B tB t 1 05 P /d P i MW• Burnup rateBurnup rate: 1.05 P g/day. P in MW. • The fissioning of 1.05 g of 235U yields 1 MWd of energy. • Specific Burnup Specific Burnup = 1 MWd / 1.05 g ≈ 950000 MWd/t (pure (pure 235235U !!!!!!!!!).U !!!!!!!!!).• Fractional Burnup Fractional Burnup = ???• Thermal reactor loaded with 98 metric tons of UO2, 3% enriched, operates at 3300 MWt for 750 days.

Actually much less (all heavy material).Actually much less (all heavy material).

y• ≈ 86.4 t U. Specific burnup ≈ 28650 MWd/t. • Fast fission of 238U.• 238U converted to plutonium ⇒ more fission Not all fissions from 235UU converted to plutonium ⇒ more fission. Not all fissions from U.


Nuclear Fission

)(EσγC t t fi i ti)()(

)(E

Efσ

σα γ=• Capture-to-fission ratio:

1 05(1 ) /d• Consumption rateConsumption rate: 1.05(1+α) P g/day.

•• Read all relevant material in Lamarsh Read all relevant material in Lamarsh ChCh 44 We will come back to this laterWe will come back to this later

•Two neutrinos are expected immediately from the

Ch. Ch. 44. We will come back to this later.. We will come back to this later.

•Two neutrinos are expected immediately from the decay of the two fission products, what is the minimumflux of neutrinos expected at 1 km from the reactor


80

flux of neutrinos expected at 1 km from the reactor.4.8x1012 m-2s-1

Nuclear Fission• 3.1x1010 fissions per second per W.

I th l t j it f fi i i• In thermal reactor, majority of fissions occur in thermal energy region, φ and Σ are maximum.

T t l fi i t i th l t f l• Total fission rate in a thermal reactor of volume V

φVΣ• Thermal reactor powerThermal reactor power (quick calculation)(quick calculation)

φfVΣ

VP fφΣ

10101.3 xP f

th =



Controlled Fission• 235U + n X + Y + ν n

Fast second generation neutrons

• Moderation of second generation neutrons Chain reaction.• Water, D2O or graphite moderator.

R ti f b f “ t ” (fi i ) i ti t

g

• Ratio of number of “neutrons” (fissions) in one generation to the preceding ≡ k∞ (neutron reproduction or multiplication factor)factor).

• k ≥ 1 Chain reaction.Infinite medium (ignoring leakage at the surface).

Chain reacting pileChain reacting pile• k < 1 subcritical.• k = 1 critical system.

k 1 iti l

Chain reacting pileChain reacting pile

• k > 1 supercritical.For steady release of energy (steady-state operation) we need k =1

82

state operation) we need k =1.


Controlled Fission• Average number of allall neutrons released per fission

ν (for thermal neutrons 0 0253 eV)ν (for thermal neutrons, 0.0253 eV).• 233U : 2.492• 235U : 2.418• 239Pu : 2.871• 241Pu : 2.927

• Reactor is critical (k = 1): rate of neutrons produced by fission = rate• Reactor is critical (keff = 1): rate of neutrons produced by fission = rate of neutrons absorbed absorbed + leaked.

Size and composition of the reactor. 83Nuclear Reactor Theory, JU, First Semester,2010-2011 (Saed Dababneh).

Controlled FissionProbability for a thermal neutron to

fi i 235 i235U thermal cross sections

cause fission on 235U isU t e a c oss sect o s

σfission ≈ 584 b.σscattering ≈ 9 b. σ

=≈1fCheck Check

numbers!numbers!σradiative capture ≈ 97 b. ασσ γ ++ 1f

If each fission produces an average of ν neutrons, then the mean number of fastfast fission neutrons produced per thermal neutron = ηnumber of fastfast fission neutrons produced per thermal neutron η

νσσ ff

αν

σσσ

νσσ

νηγ +=

+==

1f

f

a

f η <ν


84

Controlled Fission235U• Assume Assume natural natural uranium:uranium:

99 2745% 238U 0 7200% 235U99.2745% 238U, 0.7200% 235U.

Thermal σf = 0 b 584 bTh l 2 75 b 97 b

24 RπWhy?

Thermal σγ = 2.75 b 97 b

NN yyxxyx σσ +=Σ+Σ=Σ

4 Rπ

238UΣ / N (0 992745)(0)

Nyyxx )( σγσγ +=

• Σf / N = (0.992745)(0) + (0.0072)(584)

= 4 20 b24 Rπ

Doppler effect?Doppler effect?

4.20 b.• Σγ / N = (0.992745)(2.75) +

(0.0072)(97)

85

= 3.43 b.Nuclear Reactor Theory, JU, First Semester, 2010-

2011 (Saed Dababneh).

Using the experimental elastic scattering data the radius of the nucleus can be estimated.

Controlled Fission• Probability for a thermal neutron to cause fission in natural natural uraniumuranium

55.0433204

20.4=

+=

Compare to pure 235U and to 3% enriched fuel.

• If each fission produces an average of ν = 2.4 neutrons, then the mean number of fast fission neutrons produced per thermal

43.320.4 + to 3% e c ed ue

mean number of fast fission neutrons produced per thermal neutron = η = 2.4 x 0.55 ≈ 1.3• This is close to 1. If neutrons are still to be lost, there is a danger of losing criticality. (Heavy water?).• For enriched uraniumenriched uranium (235U = 3%) η = ????? (> 1.3). (Light water?)water?).• In this case η is further from 1 and allowing for more neutrons to be lost while maintaining criticality.

86

be lost while maintaining criticality.


Moderation (to compare x-section)

1H2H 1H(n,n)(n,n)

2H

(n,γ) (n,γ)(n,γ) (n,γ)

• Resonances?Resonances?• 3H production.


Controlled FissionHW HW 1111

∑ Σ= f ii )()(1 νη• Verify ∑Σ if

a

ii )()(νηy

• Comment on the calculation for thermal neutrons and a mixture of fissile and non-fissile materials, ,giving an example.• Comment for fast neutrons and a mixture of fissionable materials, giving an example.


88

Conversion and Breeding

Converters: Convert non thermally fissionable materialConverters: Convert non-thermally-fissionable material to a thermally-fissionable material.

_239min23239238 νβ ++⎯⎯ →⎯→+ −NpUnU

_2393.2 νβ ++⎯⎯→⎯ −Pud

σf,th = 742 bf,th

_233min22233232 νβ ++⎯⎯ →⎯→+ −PaThnTh

_23327 νβ

β

++⎯⎯→⎯ −UdIndia


βσf,th = 530 b


Conversion and Breeding

If 2 C i d fi i ibl• If η = 2 Conversion and fission possible.• If η > 2 Breeder reactor.• 239Pu: Thermal neutrons (η ~ 2 1) hard for breeding• Pu: Thermal neutrons (η ~ 2.1) hard for breeding.

Fast neutrons (η ~ 3) breeding fast breeder reactors.

• After sufficient time of breeding, fissile material can be easily (chemically) separated from fertile material(chemically) separated from fertile material.Compare to separating 235U from 238U.• Reprocessing.

90

p g

Nuclear Reactors, BAU, 1st Semester, 2007-2008 (Saed Dababneh).


Controlled Fission• Note that η is greater than 2

t th l i dat thermal energies andalmost 3 at high energies.

Th “ t ” t

Variations in Variations in ηη• These “extra” neutrons are Used to convert fertile into fi il f lfissile fuel.• Plutonium economy.

I di d th i• India and thorium.• Efficiency of this process is d t i d b tdetermined by neutronenergy spectrum.


91

Controlled Fission

• Conversion ratio Conversion ratio CR is defined as the average rate of fissile atom production to the average rate of fissile atom consumption.• For LWR's CR ≅ 0.6.• CR is called BR for values > 1 (fast breeder).• They are called “fast” because primary fissions inducing neutrons are fast not thermal, thus η > 2.5 but σf is only a few barns.• Moderator??


92

Controlled Fission• N thermalthermal neutrons in one generation have produced have produced so far so far ηηNN fastfast neutronsneutronsηηNN fastfast neutrons.neutrons.• Some of these fastfast neutrons can cause 238U fission more fast neutrons fast fission factorfast fission factor = ε (= 1.03 for natural uranium).neutrons fast fission factor fast fission factor ε ( 1.03 for natural uranium).• Now we have Now we have εηεηNN fastfast neutrons.neutrons.• We need to moderate these fast neutrons use graphite as an example for 2 MeV neutrons we need ??? collisions. How many for 1 MeV neutrons?• The neutron will pass through the 10 100 eV region during the• The neutron will pass through the 10 - 100 eV region during the moderation process. This energy region has many strongstrong 238Ucapture resonances (up to ????? b) Can not mix uranium and captu e eso a ces (up to b) Ca ot u a u a dmoderator.• In graphite, an average distance of 19 cm is needed for

93

thermalization the resonance escape probability resonance escape probability p (≈ 0.9).Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).Reactor design.

Controlled Fission• Now we have Now we have ppεηεηN N thermalthermal neutrons.neutrons.• Moderator must not be too large to capture thermal neutrons;Moderator must not be too large to capture thermal neutrons; when thermalized, neutrons should have reached the fuel.• Graphite thermal cross section = 0.0034 b, but there is a lot of it present.• Capture can also occur in the material encapsulating the fuel l t ( l d)elements (clad).

• The thermal utilization factor thermal utilization factor f (≈ 0.9) gives the fraction of thermal neutrons that are actually available for the fuelthermal neutrons that are actually available for the fuel.• Now we have Now we have ffppεηεηNN thermalthermal neutronsneutrons, could be > or < Nthus determining the criticality of the reactor.

k fk∞ = fpεη The fourThe four--factor formula.factor formula.

k = k ff = fpεη(1-lf )(1-l h l)94

k keff fpεη(1 lfast)(1 lthermal)Fractions lost at surfaceNuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).

Controlled Fission

k∞ = fpεη, leaknoneff Pfk −= ρεη

• Fast from thermal, as defined in HW 11.

• Fast from fast ε∑ Σ

Σ=

if

a

ii )()(1 νη• Fast from fast, ε.• Thermal from fast, p.• Thermal available for fuel

a

d

∑= idtl df l

fuelaf• Thermal available for fuel

Thinking QUIZThinking QUIZ

..mod +∑+∑+∑+∑+∑ poisona

rodsa

eratora

clada

fuela

f

Thinking QUIZThinking QUIZ• For each thermal neutron absorbed, how many fast neutrons are produced? Will need this when discuss two group diffusion


95

neutrons are produced? Will need this when discuss two-group diffusion.

Neutron reproduction

Neutron Life Cycle

x x 00..99

factork eff = 1.000

Thermal Thermal utilization utilization factor “f”factor “f”

x η

x x 00..99Resonance Resonance

escape escape i i ” ”i i ” ”

x x 11..0303Fast fissionFast fission

probability ”p”probability ”p”What is:• Migration length? Fast fission Fast fission

factor “factor “εε””

g g• Critical size?How does the

t ff t thgeometry affect the reproduction factor?


Neutron Life Cycley

Why should we b t th ?worry about these?

How?f

How?


Controlled Fission

k = fpεη(1-lf )(1-l h l) Not fixed !k fpεη(1 lfast)(1 lthermal)• Thermal utilization factor f can be changed, as an

Not fixed…!

g ,example, by adding absorber to coolant (PWR)(chemical shim, boric acid), or( , ),by inserting movable control rods in & out. Poison.• Reactors can also be controlled by altering neutron y gleakages using movable neutron reflectors.• f and p factors change as fuel is burned.p g• f, p, η change as fertile material is converted to fissilematerial.


98

Controlled Fission• Attention should be paid also to the fact that reactor power changes occur due to changes in resonance escape probability p. If Fuel T↑, p↓ due to Doppler broadening ofresonance peaks.

U d d tiUnder-moderation and

over-moderationover moderation.


99

Controlled FissionTime scale for neutron multiplicationTime scale for neutron multiplication• Time constant τ includes moderation time (~10-6 s) and diffusion time of thermal neutrons (~10-3 s).

Time Average number of thermal neutronsTime Average number of thermal neutronst n

t + τ knt + τ knt + 2τ k2n

nkndn• For a short time dtτ

nkndtdn −

=

•• Show thatShow that τtkentn )1(0)( −=

100

0)(Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).

Controlled Fissionτtkentn )1()( −=

• k = 1 n is constant (Desired).• k < 1 n decays exponentially

entn 0)( =Reactivity.

• k < 1 n decays exponentially.• k > 1 n grows exponentially with time constant τ / (k-1).• k = 1.01 (slightly supercritical..!) e(0.01/0.001)t = e10 = 22026 in in 11s. s. ( g y p )• Design the reactor to be slightly subcritical for prompt neutrons.• The “few” “delayed” neutrons will be used to achieve criticality, allowing enough time toallowing enough time tomanipulate the controlrods (or use shim or …).

101

( )Cd control rodsCd control rods


Fission ReactorsEssential elements:Essential elements:• Fuel [fissile (or fissionable) material].• Moderator (not in reactors using fast neutrons). Core

• Reflector (to reduce leakage and critical size).• Containment vessel (to prevent leakage of waste).• Shielding (for neutrons and γ’s).• Coolant.• Control system.• Emergency systems (to prevent runaway during failure).

Chapter 4 in Lamarsh


p

Fission ReactorsTypes of reactors:Types of reactors:Used for what?Used for what?• Power reactors: extract kinetic energy of fragments as heat boil water steam drives turbine electricity.• Research reactors: low power (1-10 MW) to generate neutrons (~1013 n.cm-2.s-1 or higher) for research.• Converters and breeders: Convert non-thermally-fissionable material (non-fissile) to a thermally-fissionable material (fissile).• ADS.• Fusion. What are neutron generators?What are neutron generators?


What are neutron generators?What are neutron generators?

Fission ReactorsWhat neutron energy?What neutron energy?• Thermal, fast reactors.• Large, smaller but more fuel.What fuel?What fuel?• Natural uranium, enriched uranium, 233U, 239Pu,, , , ,Mixtures.

From converter or breeder reactorHow??? breeder reactor.How???


Fission ReactorsWhat assembly?What assembly?

H t d t d f l l d• Heterogeneous: moderator and fuel are lumped. • Homogeneous: moderator and fuel are mixed together.

I h t it i i t l l t d f• In homogeneous systems, it is easier to calculate p and ffor example, but a homogeneous natural uranium-graphite mixture (for example) can not go critical Why?graphite mixture (for example) can not go critical. Why?

What coolant?What coolant?• Coolant prevents meltdown of the core.• It transfers heat in power reactors.• Why pressurized-water reactors.• Why liquid sodium?


More on ModeratorsWhat moderator?What moderator?1. Cheap and abundant.2. Chemically stable.3. Low mass (high ζ logarithmic energy decrement).4. High density.5. High Σs and very low Σa.• Graphite (1,2,4,5) increase amount to compensate 3.• Water (1,2,3,4) but n + p → d + γ enriched uranium.• D2O (heavy water) (1!) but has low capture cross

section natural uranium, but if capture occurs, produces tritium (more than a LWR).

• ….. 106Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).

More on Moderators

Moderating ratio ≡ s∑ζg

Calculate both a∑HW HW 1212 α+→→+ LiBnB 7*1110

moderating power and ratio for water, heavy

B-101010BBywater, graphite, polyethylene and boronboron.

1/v region

p y yTabulate your results and comment.


More on Moderators

HW HW 12 12 (continued)(continued)

Calculate the moderating power and ratio for pure D O ll f D O t i t d ith ) 0 25%D2O as well as for D2O contaminated with a) 0.25% and b) 1% H2O.C t th ltComment on the results.In CANDU systems there is a need for heavy water

dupgradors.


108

More on Moderators

ζnEE −= lnln \ )/ln( \nEEn =Recall ζnEEn −= lnln

ζn =Recall

After n collisionsAfter n collisions

)/ln( thf EE After one collision

11ln

2)1(1ln

2

\ +−−

+=⎥⎦⎤

⎢⎣⎡==∆

AA

AA

EEu

av

ζζ)( thfn =

ζ

Total mean free path = n λf

Total mean free path = n λsIs it random walk or there is a

f d di ti ???Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).109

preferred direction??? th

More on Moderators


110

More on Moderators

A ⎞⎛21 After one collision.

Recall (head-on). Then the maximum energy loss is (1-α)E, or αE ≤ E\ ≤ E.

EEAAE α≡⎟

⎠⎞

⎜⎝⎛

+−

=min\

11

For an ss--wavewave collision:EEPdEEEP

E 1)(1)( \\\ =→∴=→∫Assumptions:Assumptions:1.1. Elastic scattering.Elastic scattering. E↓Flat-top probability

EEEPdEEEP

E )1()(1)(

αα −→∴→∫

1.1. Elastic scattering. Elastic scattering. E↓2.2. Target nucleus at rest. Target nucleus at rest.

E↑3.3. Spherical symmetry in Spherical symmetry in

CMCMEE )1(2

1\ α+=CM.CM.

Obviously

2

⎪

⎪⎨⎧ ≤≤

−=≡→EEE

EE

dEdEE

ss

s )1()(

)(|

|\ α

ασσσ


111

⎪⎩ otherwisedE 0

More on ModeratorsHW HW 13 13 (or (or 66\\)) •• Scattering Kernel.Scattering Kernel.

•• Slowing down density.Slowing down density.

[ ]2\ 1cos21++ AAE CMθ

(Re)-verifyg yg y

•• Migration length.Migration length.•• Fermi age and continuous fermi model.Fermi age and continuous fermi model.

[ ]

[ ]22 cos)1()1(

21

)1(cos21

−++=+

++=

AAA

EE CMθααθ

[ ]2

222

)1(sincos

+−+

=A

A θθ)(

For doing so, you need to verify and useCMA θ1

CM

CM

AAA

θθθcos21

cos1cos2 ++

+=


112

More on ModeratorsHW HW 13 13 (or (or 66\\) ) continued…continued…

• Forward scattering is preferred for “practical”preferred for practical moderators (small A).• If isotropic neutronIf isotropic neutron scattering (spherically symmetric) in thesymmetric) in the laboratorylaboratory frame average cosine of the Show that

2)(cos == θµaverage cosine of the scattering angle is zero.

Show that A3

)(cos θµ


113

More on Moderators

1dHW HW 13 13 (or (or 66\\) ) continued…continued…

Spherically symmetric in CM )(41)( E

dd

sCM

sCMs σ

πθσσ

==Ω

CM 2312

Show thatCM

CMs

s AAAE

θθ

πσθσ

cos1)1cos2(

4)()( 1

2312

−

−−

+++

=

• Neutron scatteringscattering is isotropic in the laboratory system?! valid for neutron scattering with heavysystem?! valid for neutron scattering with heavy nuclei, which is not true for usual thermal reactor moderators (corrections are applied).moderators (corrections are applied).Distinguish fromDistinguish from

A l t di t ib tidi t ib tiNuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).114

• Angular neutron distribution.distribution.

More on ModeratorsModeratorModerator--toto--fuel ratio fuel ratio ≡≡ Nm/Nu. Self regulation.• Ratio ↑ p ↑ Σa of the moderator ↑ f ↓ (leakage ↓).• Ratio ↓ p ↓ f ↑ (leakage ↑).• T ↑ ratio ↓ (why).• Other factors alsoOther factors also change.• Temperature coefficient of reactivity.• Moderator temperaturetemperature coefficient of reactivity.


115

One-Speed Interactions• Particular general.Recall:Recall:• Neutrons don’t have a chance to interact with each other (BAU 2007 review test!) Simultaneousother (BAU 2007 review test!) Simultaneous beams, different intensities, same energy:

Ft = Σt (IA + IB + IC + …) = Σt (nA + nB + nC + …)vFt Σt (IA IB IC …) Σt (nA nB nC …)v• In a reactor, if neutrons are moving in all directions

n = nA + nB + nC + …n nA nB nC …

Rt = Σt nv = Σt φRt Σt nv Σt φ


116

One-Speed Interactions

Ωdrn ),( ωrr

ωr≡ Neutrons per cm3 at )( ωr whose velocity

vector lies within dΩΩdabout ω.

∫ Ω= ω),()( drnrn rrr

rr∫π4

• Same argument as before ΩddI )()( rrrrΩ= vdrnrdI ),(),( ωω

rrrr

),()(),( rdIrrdF trrrrr ωω ∑=

)()()()(),()(),()()(

),()(),(

4

rrrvnrdrnvrrdFrFrR

rdIrrdF

ttt

trrrrrrrrrrr φωω

ωω

πω

∑=∑=Ω∑===

∑

∫∫Scalar


117

4πω

Ω= ∫ drnvrπ

ωφ4

),()( rrrwhere

Multiple Energy Interactions

ΩdEdErn )( ωrr

≡ Neutrons per cm3 at r with energy• Generalize to include energy

ΩdEdErn ),,( ω ≡ Neutrons per cm3 at r with energy interval (E, E+dE) whose velocity vector lies within dΩ about ωvector lies within dΩ about ω.

∫ Ω= ω),,(),( dEdErndEErnrrr ∫ ∫

∞

Ω= ),,()( ω dEdErnrnrrr

∫π4

∫ ∫0 4π

dEErErdEEvErnErdEErR ttt ),(),()(),(),(),( rrrrr φ∑=∑=

∫∞

∑= ),(),()( dEErErrR ttrrr φ

Scalar0Thus knowing the material properties Σt and the neutron flux φ, both as functions of space and energy, we can calculate the interaction rate

Scalar


118

p gy,throughout the reactor.

Neutron Current• Similarly and so on …∫

∞

∑=0

),(),()( dEErErrR SSrrr φ

• Redefine as

0Scalar

ΩddI )()( rrrr ΩdvrnrId rrrrrr )()( ωω• Redefine as Ω= vdrnrdI ),(),( ωω Ω= dvrnrId ),(),( ωω

Ω= ∫ drnvr )()( ωφrrr

Ω= ∫ drnvJ )( ωrrrrOne group!

Ω= ∫ drnvr ),()(4

ωφπ

Ω= ∫ drnvJ ),(4

ωπ

Neutron current densityNeutron current densityNeutron current densityNeutron current density

Jr• From larger flux to smaller flux!

Neutrons are not pushed! J• Neutrons are not pushed!• More scattering in one direction than in the other

119

than in the other.xJxJ =• ˆ

r


Equation of ContinuityNNetet flow of neutrons per second per unit area normal to the x direction:

∫ Ω==•π

θω4

cos),(ˆ dvrnJxJ xxrrr

π4In general: nJnJ =• ˆ

r

Equation of ContinuityEquation of ContinuityEquation of ContinuityEquation of Continuity

∫∫ ∫∫ •−∀∑−∀=∀∂∂

a dAntrJdtrrdtrSdtrn ˆ),(),()(),(),( rrrrrr φ±

∫∫ ∫∫∂ ∀ ∀∀ Aat

)()()()()( φ

R f h i P d i Ab i “L kRate of change in number of neutrons

Production rate

Absorption rate

“Leakage in/out” rate

So rce S rface Normal to A

Nuclear Reactor Theory, JU, First Semester,2010-2011 (Saed Dababneh).

120

Volume Source distribution

function

Surface area

bounding ∀

Normal to A (outwards)

Equation of ContinuityUsing Gauss’ Divergence Theorem ∫∫ •∇=•

VS

rdBAdB 3rrrr

Recall:

∫ Q

∀•∇=• ∫∫∀

dtrJdAntrJA

),(ˆ),( rrrrr

0

0

ερ

ε

=⋅∇

=⋅∫E

AdES

Q

∀A0ε

∫∫ ∫∫ •−∀∑−∀=∀∂∂

∀ ∀∀ Aa dAntrJdtrrdtrSdtrn

tˆ),(),()(),(),( rrrrrr φ

∂ ∀ ∀∀ AtBoth flux and current!!

Convert current to flux?One group!

),(),()(),(),(1 trJtrrtrStrtv a

rrrrrrr•∇−∑−=

∂∂ φφtv ∂

E ti f C ti itE ti f C ti itNuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).121

Equation of ContinuityEquation of Continuity

Equation of Continuity

Steady statey

0)()()()( •∇∑ rJrrrS rrrrrr φ 0)()()()( =•∇−∑− rJrrrS a φ

Non-spacial dependence

)()()( ttStnt a φ∑−=∂∂t∂

Delayed sources? Will do it later.Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).122

Delayed sources? Will do it later.

Fick’s Law

• The exact interpretation of neutron transport in heterogeneous domains is so complexheterogeneous domains is so complex.• Assumptions and approximations.• Simplified approachesSimplified approaches.• Simplified but accurate enough to give an estimateestimate of the average characteristicsaverage characteristics of neutron populationneutron populationthe average characteristics average characteristics of neutron populationneutron population.• Numerical solutions.• Monte Carlo techniquesMonte Carlo techniques.


123

Fick’s LawAssumptions:1. The medium is infinite.1. The medium is infinite.2. The medium is uniform 3. There are no neutron sources in the medium.

).(rnot r∑∑

3. There are no neutron sources in the medium.4. Scattering is isotropic in the lab coordinate system.5. The neutron flux is a slowly varying function of5. The neutron flux is a slowly varying function of

position.6. The neutron flux is not a function of time.6. The neutron flux is not a function of time.

htt // i /EJ/ ti l /0143 0807/26/5/023/ j 5 5 023 df


124

http://www.iop.org/EJ/article/0143-0807/26/5/023/ejp5_5_023.pdf

Fick’s Law

Lamarsh puts it more bluntly:Lamarsh puts it more bluntly:“Fick’s Law is invalid: a) in a medium that strongly absorbs neutrons;a) in a medium that strongly absorbs neutrons; b) within three mean free paths of either a neutron source or the surface of a material; andsource or the surface of a material; and c) when neutron scattering is strongly anisotropic.”


125

Fick’s Law


126

Fick’s Lawφ(x) Negative Flux Gradient

Current Jx• Diffusion: random walk of an ensemble of particles

High flux

pfrom region of high “concentration” to region g f

More collisions

gof small “concentration”.• Flow is proportional to

Low flux

p pthe negative gradient of the “concentration”.

x

Less collisions

• From larger flux to smaller flux!Recall:

x

DJ x ∂∂

−=φ

From larger flux to smaller flux!• Neutrons are not pushed!• More scattering in one direction


127

xx ∂than in the other.

Fick’s Lawz Number of neutrons scatteredscattered per

second from d∀ at rr and going d∀

θ ∀∑ Σ− dedAr rz tcos)( θφ r

g gthrough dAz

d∀

rdAz

θ ∀∑ der

rs 24)(

πφ

R dy

z

φ ∑∑

Removed en route

(assuming no

xφ

Slowly varying)(rnot ssr

∑∑ (assuming no buildup)

Isotropic


128

Fick’s Law


129

Fick’s Law

∞Σ π π2 2/dAHW HW 1414

[ ]∫ ∫ ∫= = =

Σ−− Σ=

φ θ

φθθθφπ 0 0 0

sincos)(4 r

rzszz ddrderdAdAJ t

r

= = =φ θ0 0 0r

⎞⎛⎞⎜⎛ ∑ ∂φ

?=+zz dAJ

023

⎟⎠⎞

⎜⎝⎛⎟⎠

⎞⎜⎜⎝

⎛Σ∑

−=−= −+

zJJJ

t

szzz ∂

∂φand show that

D ≈1

and generalize2sDDJ ∑

=∇−= φr Diffusion Diffusion

coefficientcoefficient

sΣ3

g23 tΣ

φ coefficientcoefficient

The current density is proportional to the negative of the gradient Total removal


130

y p p g gof the neutron flux.

Fick’s LawValidity:1. The medium is infinite. Integration over all space.1. The medium is infinite. Integration over all space.

after few mean free paths 0 corrections at the surface are still required.

rte ∑−

corrections at the surface are still required.2. The medium is uniform.

φ and Σ are functions of space re-)(rnot ssr

∑∑)(rsr

∑ φ and Σ are functions of space rederivation of Fick’s law? locally larger Σs extra JJ cancelled by iff ???

)(s

rr sat ee )( ∑+∑−∑− = HWHW 1515JJ cancelled by iff ???Note: assumption 5 is also violated!

3. There are no neutron sources in the medium.

ee HW HW 1515

3. There are no neutron sources in the medium.Again, sources are few mean free paths away and corrections otherwise.


131

co ect o s ot e se

Fick’s Law4. Scattering is isotropic in the lab. coordinate system!

If reevaluate D.02)(cos ≠== θµ HWHW 1616If reevaluate D.3

)(A

µ

11 trD λ===

HW HW 1616

33)(3 trst

Dµ ∑∑−∑

λ

Weekly absorbing Σt = Σs.

For “practical” moderators:µ

λλ−

≅1

str

5. The flux is a slowly varying function of position.Σa ↑ variation in φ ↑.Σa ↑ variation in φ ↑.

)(2

2

rr

r

∂∂ φ ??


132

r∂

Fick’s Law

Estimate the diffusion coefficient of graphite at 1 eV.HW HW 1717

g pThe scattering cross section of carbon at 1 eV is 4.8 b.

Scattering OthScattering Other materials?

Absorption


133

Fick’s Law6. The neutron flux is not a function of time.

Time needed for a thermal neutron to traverse 3 mean free paths ∼ 1 x 10-3 s (How?).If flux changes by 10% per second!!!!!! g y p

43 101101.011

/ −− ==∆

=∆ xxmsφφφφ

Very small fractional change during the time 1 1smsφ

y g gneeded for the neutron to travel this “significant” distance.

φ∇−≅r

DJNuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).134

φ∇≅ DJ

Back to the Continuity Equation

1 ∂ ),(),()(),(),(1 trJtrrtrStrtv a

rrrrrrr•∇−∑−=

∂∂ φφ

),(),()(),(),(1 trDtrrtrStr rrrrrrr φφφ ∇•∇+∑−=∂ ),(),()(),(),( trDtrrtrStrtv a φφφ ∇∇+∑∂


135

The Diffusion Equation

),(),()(),(),(1 trDtrrtrStrtv a

rrrrrrr φφφ ∇•∇+∑−=∂∂tv ∂

If D is independent of r (uniform medium)Laplacian

)()()()()(1 2 trDtrrtrStr rrrrr φφφ ∇+∑∂

Laplacian

),(),()(),(),( trDtrrtrStrtv a φφφ ∇+∑−=∂

l H l h lt ti

)()()()(0 2 rDrrrS arrrr φφ ∇+∑−=

or scalar Helmholtz equation.

)()()()( a φφ

)()()(0 2 rDrr rrr φφ ∇+∑Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).136

)()()(0 rDrra φφ ∇+∑−= Buckling equation.

Steady State Diffusion Equation)()()()(0 2 rDrrrS arrrr φφ ∇+∑−=

DDL∑

=2Define L ≡ Diffusion LengthL2 ≡ Diffusion Area a∑ us o eaModeration Length

S1DS

L−=−∇ φφ 2

2 1

B d C ditiB d C diti01

22 =−∇ φφ

L

Boundary ConditionsBoundary Conditions• Solve DE get φ.

Solution must satisfy BC’s2 φφL • Solution must satisfy BC’s.

• Solution should be real and non-negative.


137

non negative.

Steady State Diffusion EquationOneOne--speed neutron diffusion in infinite mediumspeed neutron diffusion in infinite medium

Point sourcePoint source

0)(1)( 22 =−∇ r

Lr rr φφ 2L

2HWHW 1818

0)(1)(2)( 22

2

=−+ rL

rdrd

rr

drd φφφ

HW HW 1818

LdrrdrLrLr //−General solution

reC

reA

LrLr //

+=φ


138

A, C determined from BC’s.

Steady State Diffusion EquationBC BC r → ∞ φ → 0 C = 0.

e Lr /−HW HW 18 18 (continued)(continued)

reA=φ

S eS Lr /−

Show that D

SAπ4

=r

eD

S4

=π

φa

DL∑

=2

neutrons per second absorbed in the ring.φπ adrr ∑24 neutrons per second absorbed in the ring.

rdr22

φa

Show that22 6Lr =

r


139

Steady State Diffusion Equation

HWHW 1919HW HW 1919Study example 5.3 and solve problem 5.8 in Lamarsh.Study example 5.3 and solve problem 5.8 in Lamarsh.

Multiple Point Sources?


140

Steady State Diffusion EquationOneOne--speed neutron diffusion in a finite mediumspeed neutron diffusion in a finite medium

At th i t f A B

BA φφ =• At the interface

BA φφdDdDJJ BA

AAφφ

−=−⇒=

What if A or B is a ac m?

dxD

dxDJJ BABA =⇒=

x• What if A or B is a vacuum?• Linear extrapolation distance.


141

More realistic multiplying mediumOneOne--speed neutron diffusion in a multiplying mediumspeed neutron diffusion in a multiplying mediumThe reactor core is a finite multiplying mediumThe reactor core is a finite multiplying medium.• Neutron flux?• Reaction rates?• Reaction rates?• Power distribution in the reactor core?Recall:Recall:• Critical (or steady-state):Number of neutrons produced by fission = numberNumber of neutrons produced by fission = number of neutrons lost by:absorption rate productionneutron (S)k =absorptionandleakage

)( rate absorptionneutron Ak =∞

)( rate productionneutron Sk =


142

leakage)( rate leakageneutron )( rate absorptionneutron LEA

keff +=

More realistic multiplying medium

yprobabilit leakage-nonleaknoneff P

LEAA

kk

−=+

=Things to be used later…!

LEAk∞ +SALE area surface∝ For a critical reactor:

Recall:

aSALEVolumeVS

12

∝For a critical reactor:

Keff = 1K > 1

aaa

VSA

SLE 1

3 =∝∝

St d t t h tSt d t t h t

K∞ > 1

)()()(0 2 rDrrk aarrr φφφ ∇+∑−∑= ∞

Steady state homogeneous reactorSteady state homogeneous reactor)()()( aa φφφ∞

2222 10)()( kBrBr −≡=+∇ ∞rr φφ


143

2)()(L

φφ

Material buckling

More realistic multiplying medium0)()( 22 =+∇ rBr rr φφ

)(2 rφ)(

)(22

rrB r

r

φφ∇

−=)(rφ

• The buckling is a measure of extent to which the flux curves or “buckles ”curves or buckles.• For a slab reactor, the buckling goes to zero as “a”goes to infinity There would be no buckling or curvaturegoes to infinity. There would be no buckling or curvature in a reactor of infinite width. • Buckling can be used to infer leakage The greater the• Buckling can be used to infer leakage. The greater the curvature, the more leakage would be expected.


144

More on One-Speed DiffusionHW HW 2020

Show that for a critical homogeneous reactorcritical homogeneous reactorShow that for a critical homogeneous reactorcritical homogeneous reactorφφP aa1 ∑

=∑

==φφφφ DBDLB

Paa

leaknon 2222 1 +∑=

∇−∑=

+=−

Infinite Bare Slab ReactorInfinite Bare Slab Reactor (( d diff i )d diff i )Infinite Bare Slab Reactor Infinite Bare Slab Reactor (one(one--speed diffusion)speed diffusion)φ• Vacuum beyond.

R t t 0

z

xa/2

• Return current = 0.d = linear extrapolation distance

0 71 λ (for plane s rfaces)

Reactor

aa/2

a0/2= 0.71 λtr (for plane surfaces)= 2.13 D.


145

d d


022

=+ φφ BdFor the infinite slab Show that the02 =+ φBdx

For the infinite slab . Show that the general solution

BxCBxAx sincos)( +=φwith BC’s

BxCBxAx sincos)( +=φ0)

2( 0 =±

aφFlux is symmetric about

0)(2

0

=xdx

xdφFlux is symmetric about

the origin.0=xdx

0cos)( φφ == ABxAx 0)( φφ

,...5,3,)(0)(cos)( 000 πππφ =±⇒=±=±aBaBAa


146

,...2

,2

,2

)2

(0)2

(cos)2

(φ ±⇒±± BBA

More on One-Speed Diffusion

,...2

5,2

3,2

)2

( 0 πππ=±

aBHW HW 21 21 (continued)(continued)

222253a πππ

= ,...,,0 BBBa =

Fundamental mode the only mode significant inFundamental mode, the only mode significant in critical reactors. 2

⎞⎜⎛ ππ BucklinglGeometricacos)(

0

2

00 ≡⎟

⎠

⎞⎜⎜⎝

⎛==

aBx

ax ππφφ

For a critical reactor the geometrical buckling is equalFor a critical reactor, the geometrical buckling is equal to the material buckling.To achieve criticality

21k −⎞

⎜⎛ π


147

To achieve criticality 2

0

1L

ka

−=⎟

⎠

⎞⎜⎜⎝

⎛ ∞π

More on One-Speed Diffusion???0φ

2⎞⎛

• To achieve criticality 2

2

0

1L

ka

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ∞π

• But criticality at what power level??• φ can not be determined by the geometry alone

0 La ⎠⎝

• φ0 can not be determined by the geometry alone.

π xa

Px0

0 cos,..,..)()( πφφ =


148

More on One-Speed DiffusionSpherical Bare Reactor (oneSpherical Bare Reactor (one--speed diffusion)speed diffusion)

22 46 aa π3

343

46aa

aa

ππ

>CubeCube SphereSphere

Minimum leakage minimum fuel to achieve criticality.φ2 2

2 φφ ddHWHW 2222 02 22 =++ φφφ B

drd

rdrdHW HW 2222

CA BrrCBr

rA sincos +=φ

r

r0Br

rr

rC ππφ == 0

0

,sin Continue!Reactor


149

r0Brr 0


Infinite planer source in an infinite Infinite planer source in an infinite didi

φφmedium.medium.

LxeSLx /)( −=φxSxd )(1)(2 δφφ

φφ

eD

x2

)( =φDLdx

)()(22 φφ

−=−

HW HW 2424Infinite planer source in a finiteInfinite planer source in a finite

xa/2( )[ ]2/2sinh LxaSL

Infinite planer source in a finite Infinite planer source in a finite medium.medium.

aa/2

a0/2( )[ ]

)2/cosh(2/2sinh

2 0

0

LaLxa

DSL −

=φ


150Source

More on One-Speed DiffusionInfinite planer source in a multiInfinite planer source in a multi--region medium.region medium.

aa ±=± )2/()2/( φφ

FiniteInfinite InfinitedDdD

aa

=

±=±

22

11

21 )2/()2/(φφ

φφ

φφFiniteInfinite Infinite

BCmoredxdx axax

+±=±= 2/

22/

1

P j tP j t 22Project Project 22


151

Back to Multiplication Factor

k∞ = fpεη, leaknoneff P

kk

−= leaknoneff Pfk −= ρεηThings to be used later…!Recall:

k∞leaknoneff

• Fast from thermal,F t f f t

∑ ΣΣ

=i

f ii )()(1 νη• Fast from fast, ε.• Thermal from fast, p.

Th l il bl f fi i

Σ ia

fuelaf ∑• Thermal available for fission

Thi ki QUIZThi ki QUIZ

poisona

eratora

clada

fuela

af∑+∑+∑+∑

∑= mod

Recall:Thinking QUIZThinking QUIZ• For each thermal neutron absorbed, how many fast

t d d?neutrons are produced?


152

Two-Group Neutron Diffusion•• Introductory to multiIntroductory to multi--group (Hence crude).group (Hence crude).• All neutrons are either in a fast or in a thermal energy group.• Boundary between two groups is set to ~1 eV.• Thermal neutrons diffuse in a medium and cause fission, are captured, or leak out from the system.• Source for thermal neutrons is provided by the slowing down of fast neutrons (born in fission).• Fast neutrons are lost by slowing down due to elastic scattering in the medium, or leak out from the system, or due to fission or capture.• Source for fast neutrons is thermal and fast neutron


153

fission.

Two-Group Neutron Diffusion

FastdErErMeV

∫=10

1 ),()( rr φφ

ThermaldErEreV

eV

∫

∫1

1

)()( rr φφ ThermaldErEr ∫=0

2 ),()( φφ

φνφν ∑+∑

221122

212

1

222111

φφφφφνφν ff

eff DDk

∑+∑+∇−∇−

∑+∑=


154

22112211 φφφφ aaDD ∑+∑+∇∇

Two-Group Neutron Diffusion)()()(0 1

21111 rDrrS a

rrr φφ ∇+∑−=

RemovalRemoval cross section Depends on th l d f t

Fast diffusion coefficient

= fission + capture + scattering to group 2

thermal and fast fluxes.

)()()()(0 12

1112211 rDrrr affrrrr φφφνφν ∇+∑−∑+∑=

koror

)()()(0 12

11122 rDrrkaa

rrr φφφρ

∇+∑−∑= ∞


155

ρ

Two-Group Neutron Diffusion)()()(0 2

22222 rDrrS a

rrr φφ ∇+∑−=

Thermal diffusionThermal absorption D d f t Thermal diffusion

coefficientcross section = fission + capture.

Depends on fast flux.

)()()(0 22

222121 rDrr asrrr φφφ ∇+∑−∑= →

)()()(0 2D rrr φφφ ∇∑∑oror

)()()(0 22

22211 rDrr aarrr φφφρ ∇+∑−∑=


156


)()()(0 12

11122 rDrrkaa

rrr φφφρ

∇+∑−∑= ∞

ρ)()()(0 2

222211 rDrr aa

rrr φφφρ ∇+∑−∑= )()()( 222211 aa φφφρ ∑∑

• A coupled system of equations; both depend on p y q pboth fluxes.• Recall also, for a steady state system:y y

0)()( 12

12 =+∇ rBr rr φφ

0)()(

)()(

22

22

11

=+∇ rBr rr φφ

φφ


157

)()( 22 φφGeometrical


2 ∑∑ ∞kBD

Homogeneous system Determinant of coefficients matrix = 0Review Cramer’s

02

221

22

11 =−∑−∑

∑−∑− ∞

BD

BD

aa

aa

ρρ

Review Cramer s rule!

Do we need it here?

0))(( 122

222

11 =∑∑−−∑−−∑− ∞kBDBD aaaa ρ

here?

0))(( 122

222

11

122211

=∑∑−+∑+∑ ∞kBDBD

aaaa ρ

011)1)(1(

0))((

222

22

2

122211

=−++

∑∑+∑+∑ ∞

kBB

kBDBD aaaa

0)1)(1(

0))((

2222

2222

=++

++ ∞

kLBLB

LLkB

LB

L ThermalFastThermalFast


158

0)1)(1( =−++ ∞kLBLB ThermalFast

Two-Group Neutron Diffusion0)1)(1( 2222 =−++ ∞ThermalFast

kkLBLB

1)1)(1( 2222 =

++∞

ThermalFast LBLBk

11k1

11

12222 ++

== −−∞ FastThermal

Thermalleaknon

Fastleaknon

eff

LBLBPP

kk

∞ FastThermal

1kkFor large reactors

22

222

11)(1 M

kBLLB

k

ThermalFast

−=⇒=

++∞∞


159

Two-Group Neutron Diffusion222FastThermal LLM += If any ↑

leakageFastThermal

D 1λ

leakage ↑.

traa

tr

aThermal

DL∑∑

=∑

=∑

=3

13

2 λ

FastnL∑∑

=≡3

ageFermi 2

trsFast ∑∑3

g

• Slowing down density.g y• Fermi model.


160

Reactor Model: One-Group• Before considering multi-group.• So far we did 1-D.So far we did 1 D.• Back to oneone--groupgroup but extend to 33--DD.

z

F h h i fi iHW HW 2121||

φ

ReactorFor the homogeneous infinite slab reactor, extend the

iti lit diti th t

xa/2Reactor

criticality condition that you found in HW 21.

ad d

a0/2

DLkB

aB af

mg

∑−∑=

−==⎟

⎠

⎞⎜⎜⎝

⎛= ∞

νπ2

22

2 1 11--DD


161

DLa ⎠⎝ 0

Reactor Model: One-Group

)( 22 xd φ

• In 3-D

02222 ∂∂∂ φφφφ B

Daf ∑−∑ν

0)()( 22 =+ xB

dxxd φφ 02

222 =+∂∂

+∂∂

+∂∂ φφφφ B

zyx

Bxcos0φφ = zByBxB zyx coscoscos0φφ =k f ∑−∑−⎞

⎜⎛ νπ 2

22 1

kBBBBB af ∑−∑−⎞⎜⎜⎛

+⎞

⎜⎜⎛

+⎞

⎜⎜⎛

++ ∞νπππ 2

2222222 1DL

kBa

B afmg

∑∑===⎟

⎠

⎞⎜⎜⎝

⎛= ∞

νπ2

2

0

2 1

DLB

cbaBBBB f

mzyxg ===⎠

⎜⎜⎝

+⎠

⎜⎜⎝

+⎠

⎜⎜⎝

=++= ∞2

2

000

2222

Critical dimensions (size), for the given material properties, predicted by the model.


162

Reactor Model: One-Group• Transient case.

t !

),()(),()(),(),(1 trrDtrrtrStrtv a

rrrrrrrr φφφ ∇•∇+∑−=∂∂tv ∂

Moderator structure

fuelfuelf

fuela γ∑+∑=∑t !

• Delayed neutrons!!

Moderator, structure, coolant, fuel, …

• Reflectors!!• For homogeneous 1-D:

2

),(),(),(),(12

2

txx

DtxtxStxtv a φφφ

∂∂

+∑−=∂∂


163

xtv ∂∂),( txf φν ∑


),(),(),(),(12

2

txDtxtxtx af φφφνφ∂∂

+∑−∑=∂∂ ),(),(),(),( 2xtv af φφφφ

∂∂HW HW 2525Separation of variables: )()(),( tTxtx ψφ =

21 T ∂∂ ψ2

1x

DTTTtT

v af ∂∂

+∑−∑=∂∂ ψψψνψ

constant)(12

2

=−≡⎥⎦

⎤⎢⎣

⎡∑−∑+

∂∂

=∂∂ λψνψ

afDvTT

)(2 ⎥⎦

⎢⎣ ∂∂

ψψ afxtT

)0()( teTtT λ−= )( 2DBv ∑+∑= νλShow that

= 0 for steady state.


164

,)0()( eTtT = )( fa DBv ∑−+∑= νλShow that

Reactor Model: One-GroupHW HW 25 25 (continued)(continued)

0)( 0±aφ xBx cos)( =ψ

22 ⎞

⎜⎛ nB πt0)

2( 0 =±φ xBx nn cos)( =ψ

0

2

⎠⎜⎜⎝

=a

Bntry

2 )( 2fnan DBv ∑−+∑= νλeigenvalues

⎞⎛ π?Solution ∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

oddn

tn a

xneAtx n

0

cos),( πφ λ?

?

Initial condition ∑ ⎟⎠

⎞⎜⎜⎝

⎛= n a

xnAx cos)0,( πφ

?

Show that ⎠

⎜⎝oddn a0

∫+

⎟⎠

⎞⎜⎜⎝

⎛= 2

0

0cos)0,(2 a

a dxxnxAnπφ


165

∫− ⎠⎜⎝2

000 aa

Reactor Model: One-Group2

2 ⎟⎠

⎞⎜⎜⎝

⎛=

nBnπ

...25

23

21 <<< BBB

0 ⎠⎜⎝ a

)( 2fnan DBv ∑−+∑= νλ

531

...25

23

21 <<< λλλ)( fnan 531

)( 2DBv ∑−+∑= νλ Slowest decaying eigenvalue)( 11 fa DBv ∑+∑= νλ Slowest decaying eigenvalue.

xBeAxeAtx tt coscos)( 11 λλ πφ −− =⎞

⎜⎜⎛

≅ xBeAa

eAtx 110

1 coscos),(φ =⎠

⎜⎜⎝

≅


166

Reactor Model: One-Group0)( 2

11 =∑−+∑= fa DBv νλFor steady state

222 af BBB ≡∑−∑ν

CriticalityCriticality 0=λ1 mf

g BD

BB ≡==CriticalityCriticality 01 =λ

22mg BB < 01 <λSuper criticalitySuper criticality LE ↓

22mg BB > 01 >λSub criticalitySub criticality LE ↑g 1


167

Reactor Model: One-Group• That was for the bare slab reactor.• What about more general bare reactor models?g

),()(),()(),(),(1 trrDtrrtrStrt a

rrrrrrrr φφφ ∇•∇+∑−=∂∂

• For steady state, homogeneous model:

tv ∂

y g

0),(1),(),(),( 222 =

−+∇=

∑−∑+∇ ∞ tr

Lktrtr

Dtr af rrrr φφφ

νφ

• BC: φ(extrapolated boundary) = 0.

2LDφ( p y)


168

Reactor Model: One-Group• R0, H0 are the extrapolated dimensions.

1 22∂⎞

⎜⎛ ∂∂ φφ 01 2

2 =+∂

+⎠⎞

⎜⎝⎛ ∂

∂∂ φφφ B

dzdrr

rrR

• BC’s:0),( 0 =zRφ H

z

0),(

),(

2

0

0 =± Hrφ

φ Hx

yrθ

• Let)()(),( zrzr Ζℜ=φ

B lReactor

θ

• Solve the problem and discuss criticality condition.Solve the problem and discuss criticality condition.cosBesselHW HW 2626


169


R• Briefly, we go through HW 26.

1 2∂⎞⎛ ∂∂ φφ 01 22 =+

∂+⎠⎞

⎜⎝⎛ ∂

∂∂ φφφ B

dzdrr

rr)()()( Ζℜφ

z

H)()(),( zrzr Ζℜ=φ2Ζ πzd

y

Reactor

coscos00

22

⎞⎛ ℜ

==Ζ⇒=Ζ+Ζ πλλ

ddH

zzdzd x rθ

Reactor02 =ℜ+⎟⎠⎞

⎜⎝⎛ ℜ α

drdr

drd

)()( 00 rCYrAJ αα +=ℜNuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).170


0)(0 =⇒−∞→⇒→ CxYx n

00 4048.20)4048.2( RJ α=⇒=Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).171

Reactor Model: One-GroupR

0 cos)4048.2(,...)(H

zR

rJPA πφ =

H

z00 HR

Hx

yrθ

φφ

Reactor

rθ


172


R0


173H0/2

Reflected Slab: One-GroupzReflected Slab ReactorReflected Slab Reactor

or or1 ∂

x

Core

Ref

lect

o

Ref

lect

o

),()(),()(

),()(),(1

trrDtrr

trrtrtv

a

f

rrrrrr

rrr

φφ

φνφ

∇•∇+∑−

∑=∂∂

xaa/2 bb

),()(),()(a φφ

For steady-state, homogeneous, 1-D)(2 xd Cφ

)(

0)()()(

2

2 =∑−∑+

xd

xdx

xdD

R

CCa

Cf

CC

φ

φνφ

C ≡ Core1

Recall:

0)()(2 =∑− x

dxxdD RR

aR φφ

R ≡ Reflector0)(1)( 2

2 =−∇ rL

r rr φφ

)2

()2

(),2

()2

(,0)2

(BCs aJaJaaba CRCRR ===+ φφφ


174

Reflected Slab: One-Group

C

Ca

CfC

mCm

CC

DBxBAx

∑−∑== 2)()cos()(ν

φ

RRRR DL

xba

A =⎥⎥⎤

⎢⎢⎡ −+

= 2)(2sinhφ Ra

R LL

A∑

=⎥⎥

⎦⎢⎢

⎣

= )(sinhφ

⎤⎡C bB

BC

⎤⎡

⎥⎦⎤

⎢⎣⎡=

RC

RR

CmC

bDBLbAaBA sinh)

2cos(

⎥⎦⎤

⎢⎣⎡= RR

RCmC

mC

Lb

LDaBBD coth)

2tan(

⎥⎦⎤

⎢⎣⎡= R

RR

RCmCC

mC

LbA

LDaBABD cosh)

2sin(

⎥⎦⎢⎣LL2


175

Reflected Slab: One-Group

⎥⎦⎤

⎢⎣⎡= RR

RCmC

mC

Lb

LDaBBD coth)

2tan(Criticality condition.Criticality condition.

⎦⎣LL2

For bare slab CC• For bare slab CC was π / 2.

Smaller core for• Smaller core for reflected reactor ( ith )(even with a0 > a).• Save fuel.


176

Criticality “Calculation”• Can we solve “real” reactor problems analytically?• The previous discussion provides understanding of the concepts b t l i di t th d f t ti l t h it ti l t h ibut also indicates the need for computational techniques.computational techniques.

),()(),()(),()(),(1 trrDtrrtrrtr frrrrrrrrr φφφνφ ∇•∇+∑−∑=

∂

• Assume:

),()(),()(),()(),( trrDtrrtrrtrtv af φφφνφ ∇∇+∑∑

∂

)(),( retr t rr ψφ λ= )(),( ψφ

)()()()()()()( rrDrrrrr rrrrrrrrr ψψψνψλ ∇•∇+∑∑=

• Adjust parameters so that λ = 0 (Steady-state)

)()()()()()()( rrDrrrrrv af ψψψνψ ∇•∇+∑−∑=

Adjust parameters so that λ 0 (Steady state).• What parameters and how to adjust them?


177

Criticality “Calculation”)()()()()()()( rrDrrrrr

v afrrrrrrrrr ψψψνψλ ∇•∇+∑−∑=

• Fixed design and geometry one free variable is krr ν

∑∑ )()()()()()( rrk

rrrrD ffudge

arrrrrr ψνψψ ∑=∑+∇•∇−

operators are ,1 FMFK

Mfudge

ψψ =

•• As we did earlier (be guided by HW As we did earlier (be guided by HW 2020):):∑∑∑ νψν

fudge

221 LBDk aff

fudge +

∑∑=

∑+∇•∇−

∑=

νψψ

ψνrr


178

1 LBD a +∑+∇•∇− ψψ

Criticality “Calculation”

22k affudge

∑∑=ν

ψψ FK

M 1=

221 LBfudge + K fudge

• Build an algorithm.g• “Guess” (reasonably) initial kfudge and ψ (or φ) for the zeroth iteration.• Calculate the initial source term.• Iterate: .andGuess 00 kφ

get 1 10

00

01

kSF

kM ⇒== φφφ

φ

converges.flux until .....on soand0

101

11

SSkk

FS

=

= φ


179

g0S

Criticality “Calculation”• Or:

==Fk sourcesfission φ

∫∫ ++ ii dVSdVF

Mk

sinks11φ

φ

∫

∫

∫ ++

+ == volume

i

volume

i

i

dVSdVFk

11

1

φ

∫∫ + ii

volume

idVS

kdVM

k 11φ

• If for example k > 1, take action to reduce source or

volumevolume k

increase absorption.•• How?How?


180

How to Adjust CriticalityReactor Kinetics Reactor Kinetics Reactor kinetics refers to the manipulation of parameters that affect k and to the subsequent direct response of the reactor system Examples are:the subsequent direct response of the reactor system. Examples are:

• Absorber rods or shim movements to compensate for fuel burnup. • Safety scram rods to rapidly shutdown the chain reaction• Safety scram rods to rapidly shutdown the chain reaction. • Control rods to provide real-time control to keep k = 1 or to maneuver up and down in power.• …..

Reactor Dynamics Reactor Dynamics Reactor dynamics refers to the more indirect feedback mechanisms due to power level effects and other overall system effects such as:

• Temperature feedback. • VoidVoid feedback. • Pump speed control (affects water density and temperature).


181

Pump speed control (affects water density and temperature). • …

How to Adjust Criticality

B f llB f llBefore all:Before all:

Core Design Core Design The transient response of the reactor to the above direct and indirect changes in basic parameters is highly dependent on the design details of the reactor. Sample issues are: p

• Where should the control rods be placed for maximum effectiveness? • Will the power go up or down if a void is introduced into the reactor? Will the power go up or down if a void is introduced into the reactor? • Will the power go up or down if core temperature goes up? • How often should the reactor be refueled? • and so on• and so on...


182

Multi-group Model• Wide neutron spectrum.• One-group, two-group? Should be generalized.g p g p g

GG

Fraction of an eV

),()(),()(),(1

11 \\\

\\\\ Strrtrrtr

tvextg

G

gggsg

G

ggfgggg

grr

rrrrr φφνχφ +∑+∑=∂∂ ∑∑

==

Flux averaged ),()(),()(),()( trrDtrrtrr gggsggagrrrrrrrr φφφ ∇•∇+∑−∑−

Identify the terms NOW∫−

≡1

)()(gE

dEtErtr rr φφ

Flux-averaged quantities.


183

Identify the terms, NOW.∫≡ ),,(),(gE

g dEtErtr φφ

Multi-group Model

Total fissionactio

n

Scattering in er s

ourc

es

)()()()()(1 S extGG rrrrr φφφ ∑∑

∂ ∑∑

Total fission

Fra Scattering in

Oth

e

)()()()()()(

),()(),()(),(1

11 \\\

\\\\

trrDtrrtrr

Strrtrrtrtv

extg

gggsg

ggfgggg

grrrrrrrr

rrrrr

φφφ

φφνχφ

∇•∇+∑−∑−

+∑+∑=∂∂ ∑∑

==

),()(),()(),()( trrDtrrtrr gggsggag φφφ ∇•∇+∑∑

LeakageScattering outAbsorption LeakagegAbsorption

Fraction of


184

an eV

Multi-group ModelMaxwellian

Fission1/E

ss o


185

55--group example.group example.

Multi-group Model

5

Total fission

[ ]

5

1\\\\ ),()( φνχ

ggfggg trr =∑∑

=

rr

[ ]111222333444555 φνφνφνφνφνχ fffffg ∑+∑+∑+∑+∑

Th l fi i F t fi iThermal fission (~ 97%)

Fast fission (~ 3%)


186

Multi-group Model

5

Scattering in

1\\\ ),()( φ

gggsg

trr =∑∑=

rr

5544332211 φφφφφ gsgsgsgsgs ∑+∑+∑+∑+∑

33=g

Upscattering!!??? Skipping!!???


187

Multi-group Model

tφ )()(∑rr

Scattering out

gsggsggsggsggsg

gsg trr

φφφφφ

φ

54321

),()(

∑+∑+∑+∑+∑

=∑

gggggggggg

3=g 3g


188

Multi-group Model5

)()()(1 φνχφ extStrrtr +∑=∂ ∑ rrr

Group 3

31

333

)()()()(

),()(),(\

\\\

φφ

φνχφg

gfgg

trrDtrr

Strrtrtv

∇•∇+∑−

+∑=∂ ∑

=

rrrrrr

[ ][ ]

335334333332331

3333 ),()(),()(

φφφφφφφφφφ

φφ

sssss

a trrDtrr

∑+∑+∑+∑+∑+∑+∑+∑+∑+∑−

∇•∇+∑

[ ]553443333223113 φφφφφ sssss ∑+∑+∑+∑+∑+

Removal cross section

∑+∑+∑+∑+∑=∑−∑+∑≡∑

353432313

33333

ssssa

ssar

∑=

∑+∑=5

133

353432313

\\

ggsa

ssssa


189

≠=

31

\gg

Multi-group Model

actio

n

er s

ourc

es

)()()()()(1 S extGG rrrrr φφφ ∑∑

∂ ∑∑

Total fissionFra Scattering in O

the

)()()()()()(

),()(),()(),(1

11 \\\

\\\\

trrDtrrtrr

Strrtrrtrtv

extg

gggsg

ggfgggg

grrrrrrrr

rrrrr

φφφ

φφνχφ

∇•∇+∑−∑−

+∑+∑=∂∂ ∑∑

==

),()(),()(),()( trrDtrrtrr gggsgggrg φφφ ∇•∇+∑∑

LeakageRemoval In-group Scattering

)()()()()(1 Sttt extGG rrrrr φφφ +∑+∑

∂ ∑∑ ),()(),()(),(\

\\\

\\\\

11

Strrtrrtrtv

extg

ggg

ggsgg

gfggggg

rr

φφνχφ +∑+∑=∂ ∑∑

≠== Net Scattering in


190

),()(),()( trrDtrr gggrgrrrrrr φφ ∇•∇+∑−

Multi-group Model),()(),()(),(1

11 \\\

\\\\ Strrtrrtr

tvextg

G

gggsg

G

ggfgggg

g


==

),()(),()(),()( trrDtrrtrr gggsggagrrrrrrrr φφφ ∇•∇+∑−∑−

Calculate group-averaged:

)(),(),(),(),(,1\\ rDrrrr gsgagf

rrrrr∑∑∑∑

Calculate group averaged:

)(),(),(),(),(, \\v gsgaggsgfg

g

Or for,Or for,),()(),()(),(1

11 \\\

\\\\ Strrtrrtr

tvextg

G

gggsg

G

ggfgggg

g


==

we need group-averaged )(),( rr rr∑∑

),()(),()(),()( trrDtrrtrr gggsgggrgrrrrrrrr φφφ ∇•∇+∑−∑−


191

we need group averaged )(),( rr sggrg ∑∑

Multi-group Model• Group-averaged parameters?• ENDF.

),,(),()()(),,()(

1 \\\\ dEtErErEEt

tErEv f

rsr

φνχφ∑=

∂∂

∫∞

)()(

)(

\\\

0

SdEtErEEr

tEv

extrr φ +→∑+

∂

∫∞ Units!

)()()()(

),,(),(0

tErErtErEr

SdEtErEErs

rrrr φφ

φ

∑−∑−

+→∑+ ∫

),,(),(

),,(),(),,(),(

tErErD

tErErtErEr sarrrr

φ

φφ

∇•∇+

∑∑

• Integrate term by term over groups and equate to equation of multi-group model.


192

Multi-group Model

∫−

≡1

),,(),(gE

Eg dEtErtr rr φφ• Define group flux

gE

∫−∂

=∂

=∂ 1

),,(1),(),(1 gE

gg dEtEr

trtr r

rr φ

φφ ∫∂∂∂

),,()(

),(gEg

gg Evtvttv

φφ

E

∫−1

),,()(

1

1

gE

E

dEtErEv

rφ

∫−

=1

),,(

)(1g

g

EE

g dEtErv rφ


193

∫gE

Multi-group Model

∫−

∇•∇=∇•∇1

)()()()(gE

dEtErErDtrrD rrrrrrrrφφ ∫ ∇•∇∇•∇ ),,(),(),()(

gEgg dEtErErDtrrD φφ

∫−

∇1

)()(gE

dEEED rrr φ∫ ∇

=

),,(),(

)( gE

dEtErErD

rD

rr

rφ

∫−

∇

=1

),,()(

gEg

dEtErrD

rrφ∫

gE


194

Multi-group Model

dEtErErtrrgE

∫−

∑=∑1

)()()()( rrrr φφ dEtErErtrrgE

agag ∫ ∑∑ ),,(),(),()( φφ

gE

∫−1

dEtErEr

r gEa∫ ∑

∑

),,(),(

)(

rr

rφ

dEtErr

g

g

Eag

∫−

=∑1

),,()(

rφgE∫ ),,(φ


195

Multi-group Model

dEtErErtrrgE

∫−

∑=∑1

)()()()( rrrr φφ dEtErErtrrgE

sgsg ∫ ∑∑ ),,(),(),()( φφ

gE

∫−1

dEtErEr

r gEs∫ ∑

∑

),,(),(

)(

rr

rφ

dEtErr

g

g

Esg

∫−

=∑1

),,()(

rφgE∫ ),,(φ


196

Multi-group Model

∫ ∫∑∞−

→∑=∑E

s

G

ggsg

g

dEdEtErEErtrr \\\

1

1

\\\ ),,(),(),()( rrrr φφ

∫ ∑ ∫

=

− −

→∑=E G

E

Eg

g g

g

dEdEtErEEr \\\

01

1 1\

\

)()( rr φ∫ ∑ ∫=

→∑=

E E

E g Es

g g

dEdEtErEEr1

\

\\

),,(),( φ

∑ ∫ ∫=

− −

→∑=G

g

E

E

E

Es

g

g

g

dEdEtErEEr1

\\\

\

1 1\

\

),,(),( rr φg g \

1 1\E E

∫ ∫− −

→∑=∑1 1\

\\

\\\\ ),,()(

),(1)(

g

g

g

g

E

E Es

ggsg dEdEtErEE

trr r

rr φ

φ


197

g g

Multi-group Model

∫ ∫∑∞−

∑=∑E

Ef

G

gfggg

g

dEdEtErEEEtrr \

0

\\\

1

1

\\\\ ),,()()()(),()( rrr φνχφνχ

∫∫∞

=

−

∑= f

E

Eg

g

g

dEtErEEdEE \\\\

01

1

\

),,()()()( rφνχ

∫

∫∫∞

∑=

fE g

dEtErEE \\\\

0

)()()(

)()()()(

rφνχ

φχ

∫−1

)(gE

dEEχχ

∑ ∫

∫=

∑=

GE

fg

g

dEtErEE

\\\\

0

1\

),,()()( φνχ∫= )(gE

g dEEχχ

∑ ∫=

∑=g E

fg

g

dEtErEE1

\\\\

\\

),,()()( rφνχ

∫=

∑=∑1\

\\\\\\ ),,()()(

)(1)(

gE

ffgg dEtErEEtr

r rr

r φνφ

ν


198

∫\

\ ),(g

Ef

gfgg trφ

Multi-group ModelENDF

High G, few meshmesh

points.Flux

Small G

Poison, burnup (or better consumption), Small G,

more mesh points.

)temperature, control rod position, etc…

Flux


199

Flux

Multi-group ModelWhat could we make life a little easier?!• No upscattering .for0)( \

\ ggr >=∑rNo upscattering

set group G to include neutrons up to ~1 eV..for 0)(\ ggr

gsg>∑

( ))()()()()()(1gG rrrrrr φφφ ∑∑∑ ∑∑

−

( )),()(),()(),()(11 \

\\\

\\ trrtrrtrr gsggg

ggsgg

ggsg

rrrrrr φφφ ∑+∑⇒∑ ∑∑==

Your choice of how to

• No group skipping when scattering down (directly

Your choice of how to tackle in-scattering.

No group skipping when scattering down (directly coupled).

( )G

( )),()(),()(),()( 1)1(1\

\\ trrtrrtrr gsgggggsg

ggsg

rrrrrr φφφ ∑+∑⇒∑ −−=∑HWHW 2727 H l d thi ? Wh t b t H?


200

HW HW 2727 How can we pledge this? What about H?

Multi-group ModelCriticalityCriticality

1Not all sources,

only fission.φφ F

kM 1

=Not only sinks

y

yIterations.Iterations.

),()(),()(),(1\\\\\ Strrtrrtr ext

g

G

ggsg

G

gfggggrrrrr φφνχφ +∑+∑=

∂∂ ∑∑

No upscatter

),()(),()(

),()(),()(),(\

\\ 11

trrDtrr

tv

gggrg

g

ggg

ggsgg

gfggggg

rrrrrr φφ

φφχφ

∇•∇+∑−

∂ ∑∑≠==

∑−

=

∑−∑+∇•∇−g

gggsggrggg trrtrrtrrD

1

1\\\ ),()(),()(),()( rrrrrrrr

φφφupscatter

∑

≠

∑=G

gfggg

ggg

trrK

\\\

\

),()(1 rr φνχRedundant when no upscatter.


201

∑=g

gfgggK 1\

)()( φχ

Multi-group Model

⎤⎡rr

No upscatter

⎥⎥⎥⎤

⎢⎢⎢⎡

∑+∇•∇−∑−∑+∇•∇−

= rrL

rrL

rr

2212

11

000

rs

r

DD

M

⎥⎥⎥

⎦⎢⎢⎢

⎣

∑+∇•∇−∑−∑−OMMM

L332313 rss D

⎥⎤

⎢⎡ 1φ⎥

⎤⎢⎡ ∑∑∑ L331221111 fff νχνχνχ

⎥⎥⎥

⎢⎢⎢

= 2

1

φφφ

φ⎥⎥⎥

⎢⎢⎢

∑∑∑∑∑∑

=L332222112 fffF

νχνχνχνχνχνχ

⎥⎥

⎦⎢⎢

⎣ M3φ

⎥⎥

⎦⎢⎢

⎣

∑∑∑OMMM

L333223113 fff νχνχνχ


202

⎦⎣Iterations.Iterations.

Multi-group Model

⎤⎡rr

No upscatter

⎥⎥⎥⎤

⎢⎢⎢⎡

∑+∇•∇−∑−∑+∇•∇−

= rrL

rrL

rr

2212

11

000

rs

r

DD

M

⎥⎥⎥

⎦⎢⎢⎢

⎣

∑+∇•∇−∑−OMMM

L33230 rs DDirectly coupled

⎥⎤

⎢⎡ 1φ⎥

⎤⎢⎡ ∑∑∑ L331221111 fff νχνχνχ

p

⎥⎥⎥

⎢⎢⎢

= 2

1

φφφ

φ⎥⎥⎥

⎢⎢⎢

∑∑∑∑∑∑

=L332222112 fffF

νχνχνχνχνχνχ

⎥⎥

⎦⎢⎢

⎣ M3φ

⎥⎥

⎦⎢⎢

⎣

∑∑∑OMMM

L333223113 fff νχνχνχ


203

⎦⎣Iterations.Iterations.

Multi-group ModelMultiMulti--group group ⇒⇒ oneone--groupgroup

E

∫∫∞

≡⇒≡−

0

),,(),(),,(),(1

dEtErtrdEtErtrgE

Eg

rrrr φφφφ0gE

E

∫∫∞−

0

),,()(

11

),,()(

1

1

1

dEtErEv

dEtErEv

g

g

E

E

rrφφ

∫∫∞=⇒=

−

0

),,(

)(

),,(1

dEtErv

dEtErv g

g

Eg rr φφ∫

0gE


204

Multi-group Model

∫∫∞

∇∇−

),,(),(),,(),(1

dEtErErDdEtErErDgE

rrrrrrφφ

∫

∫

∫

∫∞

∇=⇒

∇

=−

0

),,()(

),,()(

1

dEtErrD

dEtErrD

g

g

EE

grr

r

rr

r

φφ ∫∫0

),,(gE

φ

dEtErErdEtErErE g

∫∫∞

∑∑−

)()(),,(),(1

rrrrφφ

dEtEr

dEtErErr

dEtEr

dEtErEr

ra

aEE

a

ag g

g

∫

∫

∫

∫∞

∑=∑⇒

∑

=∑−

0

)(

),,(),()(

)(

),,(),(

)(1 r

r

r

r

φ

φ

φ

φ

dEtErdEtErE g

∫∫0

),,(),,( φφ


205

Multi-group Model

1 when 0),()(),()(\

\\ =⇒∑−∑∑ Gtrrtrr gsg

G

ggsg

rrrr φφ1\ =g

ggg

1

∫∫∞−gE

1)()(0

==⇒= ∫∫ dEEdEEgE

g χχχχ

1 when ),()(),()(1\

\\\ =∑⇒∑∑=

Gtrrtrr f

G

ggfgg

rrrr φνφνg


206

Multi-group ModelSubstituting all of the above into

)()()()()(1 Sttt extGG rrrrr φφφ +∑+∑

∂ ∑∑

),()(),()(),()(

),()(),()(),(11 \

\\\

\\\

trrDtrrtrr

Strrtrrtrtv

extg

gggsg

ggfgggg

grrrrrrrr φφφ

φφνχφ

∇•∇+∑−∑−

+∑+∑=∂ ∑∑

==

yields

),()(),()(),()( trrDtrrtrr gggsggag φφφ ∇∇+∑∑

yields

),()(),(1 Strrtr extf

rrr φνφ +∑=∂∂

),()(),()(

)()()(

trrDtrrtv

a

f

rrrrrr φφ

φφ

∇•∇+∑−∂

which is the one-group diffusion equation.

)()()()(a φφ

c s t e o e g oup d us o equat o


207

Multi-group Model

Project Project 33

Work out the multimulti--group to twogroup to two--groupgroup collapsingWork out the multimulti--group to twogroup to two--group group collapsing and investigate criticality.

Write down the appropriate matricesWrite down the appropriate matrices.


208

Poisoning135Xe106 b

S t tSaturates


209

Poisoning149Sm105 b

Continuously accumulates


210

accumulates

Poisoning• Not anticipated! Reactor shut down! Time scale:Time scale:

Hours and days.Hours and days.135Xe 149SmXe106 b

Sm105 b

φσmXe

a

XeγIγ≈ I

I φXe


211

φσ Ia φσ Xe

a

PoisoningHW HW 2828 k

kk reactor). (Infinite use uslet ,1Reactivity −

=≡ ∞ρ

eratorcladfuel

fuelaf

k

mod1 (critical) ∑+∑+∑

∑=

itl df l

fuela

aafa

f d2∑

=

∑+∑+∑

poisona

poisona

eratora

clada

fuela

f mod2

thatShow ∑==∆

∑+∑+∑+∑

ρρρ eratora

clada

fuela

mod12 that Show∑+∑+∑

−=−=∆ ρρρ

Negative reactivity due to poison buildup. It is proportional to the amount of poison.


212

p opo t o a to t e a ou t o po so

Poisoning),(),(),(),(),( trtrItrItr

ttrI I

aIfIrrrr

r

φσλφγ −−∑=∂

∂ small

),(),(),(),(),(),( trtrXetrXetrItrt

trXet

XeaXeIfXe

rrrrrr

φσλλφγ −−+∑=∂

∂∂

t∂Initial conditions?Initial conditions?• Clean Core StartupClean Core Startup Assume no spacialClean Core Startup.Clean Core Startup.• Shutdown (later).

Assume no spacial dependence.

)()(ldFuel.Fresh 0)0()0( ==φφ

XeIconstant.)0()( assumeuslet and ==φφ t


213

Poisoning

)1()( 0 tfI IetI λ

λφγ −−

∑=HW HW 2929 Show that:

Iλ

)(∞I )(∞I)(∞Xe

)1()(

)( )(0 0 tXe

fXeI XeaXeetXe φσλ

φλφγγ +−−

∑+=and )()(

)(0

0

ttfI

XeaXe

Xe λφλφγφσλ

+∑+

)( )(

0

0 0 ttXeaIXe

fI IXeaXe ee λφσλ

φσλλφγ −+− −

+−

∑+


214

Poisoning

)(∞I )(I)(∞Xe

• Now, we know Xe(t)

eratorcladfuel

Xea

eratorcladfuel

poisona tXet

modmod

)()(∑+∑+∑

−=∑+∑+∑

∑−=∆

σρ


215

aaaaaa ∑+∑+∑∑+∑+∑

Poisoning•• Shutdown. Shutdown. After the reactor has been operating for a “long” time.

)()()()(),( ttItIttrI I rrrrr

φλφ∑∂

)()()()()(),(

),(),(),(),(),(

ttXtXtIttrXe

trtrItrItrt

Xe

IaIfI

rrrrrr

φλλφ

φσλφγ

+∑∂

−−∑=∂

),(),(),(),(),(),( trtrXetrXetrItrt

XeaXeIfXe φσλλφγ −−+∑=

∂

)()0( ∞= II ),(),( trItrII

rr

λ−=∂

0)0()()()0(

==∞=

φφ tXeXe

)()(),(

),(

tXtItrXet I

rrr

λλ∂∂

.0)0()( ==φφ t ),(),(),( trXetrIt XeI λλ −=∂


216

PoisoningHW HW 3030 Show that

)()()()(

)()(

ttIt

tI

IXX

eItI

λλλ

λ

λ

−

∞∞=

)()()()( tt

XeI

It IXeXe eeIeXetXe λλλ

λλλ −−− −−

+∞=

> 0 ?

Height of the peak depends on I(∞)depends on I(∞)and Xe(∞), i.e. depends on φ


217

depends on φ.

PoisoningShutdown Xe negative ∆ρ try to add positive reactivity move control rods out need to have

If, the available excess

yenough reserve costly to do that.

reactivity can compensate for less than 30 minutes ofthan 30 minutes of poison buildup, can’t startup again after ~30startup again after 30 minutes of shutdown, because you can’t achieve criticality. Need to wait long hours for Xeto decay down


218

to decay down.

Poisoning


219

PoisoningStrategiesStrategies• If you plan to shut down for “short maintenance”, think y pabout stepback.• Examine different scenarios using a code from g

http://www.nuceng.ca/• Prepare your own report, code, calculations, graphs, p y p g pcomments, conclusions etc…..• Be creative.


220

PoisoningXeXe OscillationsOscillations• φ(r,t) (spacial dependence) flux locally Xeburnup ρ (reactivity) flux further control rods globally in flux elsewhere Xe burnup ρ ….. limited by opposite effect due to increase (decrease) of I in the high (low) flux region.• In large reactors (compared to neutron diffusion length) local flux, power and temperature could reach unacceptable values for certain materials safety issues. • Think of one sensor and one control rod feel average flux apparently OK more sensors and


221

control rods to locate and deal with local changes.

Poisoning

Permanent PoisonsPermanent PoisonsPermanent PoisonsPermanent Poisons• 149Sm has sizeable but lower cross section than 135Xe.• It does not decayIt does not decay.

)()(),( trtrtrSm rrr

φγ ∑≈∂ ????

• Accumulates with time

....................).........,(),( trtrt fSm φγ ∑≈∂

????

Accumulates with time.• Consequences?????????


222

Fuel DepletionTime scale:Time scale:

Days and months.Days and months.

1214322 10~,10~ −−− scmcmN φ• More depletion increase steady state flux by means of reducing absorbers.

)(N r∂• For a given fuel isotope ),(),(

),(trtrN

ttrN f

aff rrr

φσ−=∂

∂

• For constant flux constant flux φφ00 the solution is),()( )0()0()( 0 trtr f

af

a erNerNtrNrr rrr Φ−− == σφσ

Exponential burnup

• For time varying fluxtime varying flux

)0,()0,(),( 0fff

aa erNerNtrN ==

Neutron fluencet

∫ ),(),(

)0,()0,(),( 0

\\

trf

dttr

ff

fa

fa

erNerNtrNr

r

rrr Φ−−

=∫

= σφσ


223Solve numerically.

Fuel Depletion•• Constant power.Constant power.

)()0,(),(),(),( 0 rPrPtrtrwNtrP fff

rrrrr=== φσ

Energy released per

Fission rate )0,()0,(),(),(

)0,()0,(),(),(

rrtrtr

rrNtrtrN

ff

ffrrrr

rrrr

φφ

φφ

∑=∑

=

• Power ~ flux only over short time periods during which Nf is constant.

released per fission

N )( rr∂

ff

wrPtrtrN

ttrN f

aff )(),(),(

),( 0r

rrr

−≈−=∂

∂φσ Linear

depletion!• The solution is obviously

trPrNtrN ff)()0,(),( 0r

rr−≈

fa

ff σσ ≈


224

tw

NtN ff )0,(),(

Fuel Depletion

Do the

HW HW 3131

Do the calculations for differentfor different

flux and power levels.


225

Poisoning and Fuel DepletionInfinite, critical homogeneous reactor.Infinite, critical homogeneous reactor.

)(tf∑)()()()(

)(mod tttt

tfk controla

poisona

eratora

clada

fa

fa

∑+∑+∑+∑+∑∑

==∞ εηρρεη

thus

trPrNtrN ff)()0,(),( 0r

rr−≈Constant powerConstant power

ttrtrNrN

tw

rNtrN

faff

ff

),(),()0,(

)0,(),(rrr φσ−=

Constant powerConstant power

rrrN

tr f )0,()0()0,(

)(r

rr

r φφφ ==

[ ]trrNrN f

aff

aff

)0,()0,()0,(

),(),(),(rrr φσ

φ

−=tr

rtrN

tr faf )0,(1

)0,(),(

),( rr φσφφ

−

[ ]trrN faf )0,(1)0,( rr φσ−=

[ ]tt fff )0(1)0()( rrr φ∑∑Nuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).226

[ ]trrtr fa

fa

fa )0,(1)0,(),( φσ−∑=∑

Poisoning and Fuel Depletion

)( )(0fXI Xeφλφγγ ∑+

∞XeConstant

)1()(

)( )(

0

0 0 tXeaXe

fXeI XeaXeetXe φσλ

φγφσλφγγ +−

∑

−+

∑+=

)( )(

0

0 0 ttXeaIXe

fI IXeaXe ee λφσλ

φσλλφγ −+− −

+−

∑+

Constant

)0,()0,()(),(),(

rrtrXetr fXeIXe

aXea

rrrr

λφγγ

σ∑+

==∑ ∞

),()()(

trXea

Xeaa rφ

σλ

+∞

trrtr fSmSma

Sma )0,()0,(),( rrr φγσ ∑≈∑ tr

rtr fa )0,(1

)0,(),( r

rr

φσφφ

−=

Oth fi i d t ( i ) ith l t tiNuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).227

• Other fission products (poisons) with less capture cross sections.

Poisoning and Fuel Depletion• Now we know all macroscopic cross sections.

)()()()()(

mod tttttfk control

apoisona

eratora

clada

fa

fa

∑+∑+∑+∑+∑∑

==∞ εηρρεη)()()()( aaaaa

• When there are no absorbers left to Until = 0.

Solve for t to get remove, we need to refuel.• Absorbers are not only control rods.

upper limit for “core loading

lif ti ”

y• All fuel nuclei should be considered.• For each species, all sources and

lifetime”.Damaged

fuel !

sinks should be taken into account.• Online loading environmental.


228

fuel…!• 3H.

Poisoning and Fuel Depletion

dN )(tFNNNNdt

dNC

CBBAAAA

A +++−−= γφσλφσλ


229

dtFuel loading

Poisoning and Fuel Depletion• Some poisons are intentionally introduced into Some poisons are intentionally introduced into the reactor.the reactor.• Fixed burnable poisons.B, Gd. More uniform distribution than rods, more intentionally localized than shim.• Soluble poisons (chemical shim) with caution.Boric acid (soluble boron, solbor) in coolant.Boration and dilution.Scram emergency shutdown (sodium polyborate or gadolinium nitrate).• Non-burnable poisons.C f fNuclear Reactor Theory, JU, First Semester, 2010-2011

(Saed Dababneh).230

Chain of absorbers or self shielding.

Delayed Precursors

1 GG∂

)()()()()()(

),()(),()(),(1

11 \\\

\\\\

tDtt

Strrtrrtrtv

extg

gggsg

ggfgggg

grrrrrrrr

rrrrr

φφφ

φφνχφ

∇∇∑∑

+∑+∑=∂∂ ∑∑

==

),()(),()(),()( trrDtrrtrr gggsggag φφφ ∇•∇+∑−∑−

1 ∂• For one-group

),()(),(1 Strrtrtv

extf

rrr φνφ +∑=∂∂

),()(),()( trrDtrrarrrrrr φφ ∇•∇+∑−

Wh t b t d l d t ?• What about delayed neutrons?


231

Delayed PrecursorsDelayed

neutron emitter

+One of 66 delayed

t

dp ννν +=

Delayed neutron fractionνβ d=neutron

precursors known so far.

Delayed neutron fractionν

β =

Data for all precursors are not accurately

kknown.


232

Delayed PrecursorsFissile nucleus Delayed neutron / 100 fissions

233U 0 667233U235U

238U*

0.6671.6214 39

Increases with N.238U* 4.39

239Pu240P *

0.6280 95

t

240Pu*241Pu

242P *

0.951.522 21242Pu* 2.21

Data for thermal neutron induced fission, except for , p* fast neutron induced fission.


233

Delayed Precursors

(s)

β < 0.7% ≅ 0.016 / ν235U

pν

),()()1(),(1 6

1SCtrrtr

tvext

iiif

rrr λφνβφ ++∑−=∂∂ ∑

),()(),()(1

trrDtrrtv

a

irrrrrr φφ ∇•∇+∑−

∂ =

),()(),(),( trrtrCt

trCfiii

i rrrr

φνβλ ∑+−=∂

∂


234

t∂

Delayed Precursors• The multi-group equation now becomes

Different energy spectra

),(),()()1(),(1 6

11\\\\ trCtrrtr

tv iii

Cg

G

ggfgg

pgg

g

rrrr λχφνβχφ +∑−=∂∂ ∑∑

==

),()(1\

\\ Strr extg

G

gggsg

rr φ +∑+ ∑=

),()(),()(),()( trrDtrrtrr gggsggag

grrrrrrrr φφφ ∇•∇+∑−∑−

∂ G)( r ∑=

∑+−=∂

∂ G

ggfggiii

i trrtrCt

trC

1\\\\ ),()(),(),( rrr

r

φνβλ


235

=g 1

Full Blown Diffusion Equation• In steady statesteady state

∑G

∑=

∑=g

gfggiii trrtrC1\

\\\ ),()(),( rrr φνβλ

)()()()()1(011 \

\\\\

\\\ rrrr

G

G

ggfgg

Cg

G

ggfgg

pg

rrrr φνβχφνβχ ∑+∑−= ∑∑==

)()()()()()()()(1\

\\ rrDrrrrSrr gggsggagextg

G

gggsg

rrrrrrrrrr φφφφ ∇•∇+∑−∑−+∑+ ∑=

GC [ ] )()()(01\

\\\ rr

G

G

ggfgg

pg

Cg

pg

rr φνβχχχ ∑−+= ∑=

Significance of ggg depends on whether

Cgχ

)()()()(1\

\\ rrSrr gagextg

G

gggsg

rr

rrrr φφ ∑−+∑+ ∑=

pwe have fine or

course energy groups.


236

)()()()( rrDrr gggsgrrrrrr φφ ∇•∇+∑−

gy g p

Reactor Physics 1

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