Larsen Lecture Notes

177
CHAPTER 1 : NUCLEAR REACTIONS IMPORTANT IN FISSION CHAIN REACTIONS (Chapter 2, Duderstadt and Hamilton) Spontaneous Radioactive Decay - Rate Equations - Half Life Nuclear Collision Reactions - Capture - Scattering (n,γ ) - Fission (n,fission) Microscopic Cross Sections Macroscopic Cross Sections - Number Density - Mean Free Path - Mean Free Time - Mean Collision Frequency Differential Scattering Cross Sections - Scattering Angle - Isotropic Scattering - Linerly Anisotropic Scattering Fission - Fissile Nuclides - Fissionable Nuclides - Fission Cross Sections - Fission Spectrum - Fuels - Fertile Nuclides - Breeding Resonances Doppler Broading 1

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Larsen Lecture Notes

Transcript of Larsen Lecture Notes

Page 1: Larsen Lecture Notes

CHAPTER 1 : NUCLEAR REACTIONS IMPORTANT IN FISSION CHAIN

REACTIONS (Chapter 2, Duderstadt and Hamilton)

Spontaneous Radioactive Decay

- Rate Equations

- Half Life

Nuclear Collision Reactions

- Capture

- Scattering (n,γ)

- Fission (n,fission)

Microscopic Cross Sections

Macroscopic Cross Sections

- Number Density

- Mean Free Path

- Mean Free Time

- Mean Collision Frequency

Differential Scattering Cross Sections

- Scattering Angle

- Isotropic Scattering

- Linerly Anisotropic Scattering

Fission

- Fissile Nuclides

- Fissionable Nuclides

- Fission Cross Sections

- Fission Spectrum

- Fuels

- Fertile Nuclides

- Breeding

Resonances

Doppler Broading

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Spontaneous Radioactive Decay (SRD)

Consider a certain type of atomic nucleus which undergoes SRD. Let:

N(t) = the probable number of these nuclei at time t

Q(t) = the rate (number per second) at which these nuclei are created by sources

Then, the general rate equation:

Rate of change of objects in a given state

= Rate at which objects enter the state

- Rate at which objects leave the state

implies

dN(t)dt

= Q(t) − (rate of SRD)

The term on the left hand side is ’net rate of change’. The first term and the second term on the right

hand side are ’gain term’ and ’loss term’, respectively.

For N>>1, we have the experimental and physically intuitive result

probable rate of SRD = λN(t),

where λ is the ’radioactive decay constant’. Therefore,

dN(t)dt

= Q(t) − λN(t) (1)

dN

dt+ λN(t) = Q

d(eλtN)dt

+ λN(t) = Q

N(t) = N(0)e−λt +∫ t

0

e−λ(t−t)Q(t) dt (2)

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t

N(t)

λ

λ

2

1

1λ > λ( 2)

If Q = 0, then

N(t) = N(0)e−λt

We then have:

(probable) rate of SRD = λN(t) = λN(0)e−λt

(probable) number of SRD′s between t and t+ dt = = [ rate of SRD at time‘t ] dt = λN(0)e−λtdt

probability that a single nucleus will undergo SRD between t and t+dt=

= [λN(0)e−λtdt] / N(0) = [λeλt]dt = p(t)dt (3)

Check:

∫0

p(t) dt =∫

0

λe−λt dt = 1

We also have

probable (or mean) decay time = ζ

=∫ 1

0tp(t) dt

=∫0 λte

−λtdt = 1λ

Inaddition,

t1/2 = half − life = time required for the number of original nuclei to decrease by 1/2

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N(t1/2) = 12N(0)

N(0)e−λt1/2 = 12N(0)

eλt1/2 = 2

t1/2 = ln2λ = 0.693

λ = 0.693t

Now, supposewehavearadioactivedecaychain

[Here X is the chemical symbol for a type of nucleus.] Then, the governing equations are

dN1(t)dt

= −λ1N1(t) + Q1(t)

dN2(t)dt

= −λ2N2(t) + λ1N1(t) + Q2(t)

dN3(t)dt

= −λ3N3(t) + λ2N2(t) + Q3(t)

Nuclear Collision Reactions (neutron-nucleus reactions)

Let

X= chemical symbol (i.e., H or C)

Z= atomic number

N= number of neutrons in a nucleus of X

A=Z+N= mass number of X

Then, the three important neutron-nucleus reactions are:

1) Capture (n,γ)

n01 + XZ

A → ( XZA+1 )∗ → XZ

A+1 + γ

2) Scattering (n,n) or (n,n)

n10 + XA

Z →

⎧⎪⎪⎪⎨⎪⎪⎪⎩

n10 + XA

Z [ elastic scattering (n,n)]

n10 + ( XA

Z )∗ [ inelastic scattering (n,n′)]

n10 + XA

Z + γ [ inelastic scattering (n,n′)]

3) Fission (n,fission)

n01 + X1

Z1A1

→ X2Z2A2

+ X3Z3A3

+ neutrons + energy

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Here ∗ means ’excited state’. In elastic scattering, the kinetic energy of the nucleus is altered, but the nucleus

is not left in an excited state.

We need to describe these interactive processes mathematically, as we did with SRD. To do this, we intro-

duce various types of material cross sections. These cross sections will basically play the same role as the

radioactive decay constants, but with the following difference:

radioactive decay constant = [mean time between events (SRD)]−1

cross section = [mean distance between events (nuclear collisions)]−1

We have already shown the first of these equalities; later we will show the second.

Microscopic Cross Sections Consider a uniform, monoenergetic, pencil beam of neutrons, normally

incident on a very thin target material. Let

I = the number of neutrons / cm2/ sec incident on the target

T = the number of target nuclei / cm2

R = the number of reactions / cm2 / sec

thin target

area = a

I

Then, we define the ’microscopic cross section’ σ by:

R = σ I T

# / cm2 sec = cm2 . # / cm2 sec . # / cm2(4)

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Thus,

σ = microscopic cross section = R / (IT) = (R/T)/I

=number of reactions per target nucleus per sec

number of incident neutrons per cm2 per sec(5)

or

σ

a=

number of reactions per target nucleus per sec

number of incident neutrons per sec

= probable number of reactions per target nucleus per incident neutron

We have the following microscopic scattering cross sections:

σγ : capture

σs : scattering

σf : fission

σe : elastic scattering

σin : inelastic scattering

Then,

σs = σe + σin

Also,

σa = absorption cross section (any event other than scattering)

= σf + σγ

σt = total cross section (any neutron-nucleus event)

= σs + σa

= σe + σin + σfσγ

σne = nonelastic cross section (any event other than elastic scattering)

= σt - σe

These quantities all usually depend on the speed v (or, the kinetic energy E = mv2/2) of the incident

neutrons.

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Macroscopic Cross Sections Now let us consider a normal beam of neutrons incident on a target of

finite thickness, and let us define

N = the number of target nuclei / cm3

= the ’number density’ of target nuclei

I(x) = the probable number of uncollided neutrons / cm2 / sec that passes through x

In a very thin part of this slab, between x and x+dx, we define

dR(x) = the number of reactions / cm2 / sec between x and x+dx

x

I(x)

x x+dx

dT = the number of target nuclei/ cm2 between x and x+dx

= N dx

Then, by Eqn.(4),

dR(x) = σt . I . dT = σt . I . Ndx (6)

However, we also have, by definition,

dR(x) = I(x) − I(x+ dx) = I(x) − [I(x) + I′(x) dx] = − I

′(x) dx

Combining these two equations, we get

dI

dx= − N σt I (7)

We now define

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Σt = N σt = macroscopic total cross section

=number of reactions / cm3

number of incident neutrons / cm2

Then the dimension of Σt is : [Σt] = length−1

Eqn.(7) now implies

I(x) = I(0) e−Σt x

Therefore, Eqn.(6) gives

probable number of interactions between x and x+dx =

= Σt I(x) dx = Σt I(0) e−Σtx dx

and so

p(x) dx = probability that a single neutron undergoes a collision between x and x+dx

= Σt e−Σtx dx

and, as with the case of spontaneous radioactive decay,

∫ ∞

0

p(x) dx = 1

∫ ∞

0

x p(x) dx = mean free path between collisions =1Σt

If v is the speed of a neutron, with

dimension of v = cm / sec dimension of Σt = cm−1

then

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1 / (vΣt) = mean free time between collisions (dimension = sec)

We may define the following three statements

1) vΣt = mean collision frequency (dimension = sec−1)

= collision rate (number of collisions / sec) of a neutron with speed v

2) If there are n neutrons in a system, then n(vΣt) = number of neutrons in the system undergoing

collisions per sec.

3) n(vΣt) dt = number of neutrons in the system that undergoes collision in a time interval dt.

Σf = N σf = macroscopic fission cross section,

Σa = N σa = macroscopic absorption cross section,

etc., so the additive formulas for σ also apply to Σ:

Σf = Σe + Σin ,

Σa = Σγ + Σf , etc.

Also, if we have a homogenous mixture of nuclides X1, X2, ... with number densities N1, N2, ... and

microscopic cross sections σ1, σ2, ..., then for the mixture we have

Σ = N1 σ1 + N2 σ2 + ...

where any subscript (t,a,s, etc.) can be used on Σ and σ. In addition, if Ni depends on x and t, then

Σ(x, t, E) = N1(x, t) σ1(E) + N2(x, t) σ2(E) + ...

so Σ can depend on x and t.

A typical plot of σt versus kinetic energy E of an incident neutron is given below:

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E

(5)

(4)(3)

(2)

(1) σ (Ε)T

(1) wavelength of neutron >> interatomic spacing (σ = E−1/2)

(2) wavelength of neutron = interatomic spacing. (diffraction effects)

(3) potential (billiard ball) scattering; σ = geometrical cross section of the nucleus

(4) resonance effects; incident neutron energy = lowest energy levels in nuclei

(5) wavelength of neutron << 1, so interaction with nuclei is very small

Differential Scattering Cross Sections

Macroscopic cross sections describe the rate at which neutrons with speed v (or kinetic energy E = mv2/2)

undergo various types of interactions. However, they do not describe the results of these interactions. That

is, if a neutron is scattered, the macroscopic cross sections do not give the velocity or direction of the scat-

tered neutron. This information is described (probabilitically) by differential scattering cross sections.

Consider a neutron vith velocity vector

v = vx i + vy j + vz k

The speed of the neutron is

v = |v| = (v · v)1/2 = (v2x + v2

y + v2z)

1/2

and the ‘direction‘ or ‘angle‘ of the neutron is the unit vector

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Ω = v / v = (vx/v) i + (vy/v) j + (vz/v) k

= Ωx i + Ωy i + Ωz k

= sin(θ) cos(φ) i + sin(θ) sin(φ) j + cos(θ) k

=√

1 − μ2 cos(φ) i +√

1 − μ2 sin(φ) j + cos(φ)k

y

x

z

j

i

k

sin cos

cos

sin

φθ

θ

θ φ θ

sin sinθ

φ

Ω

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To compute the increment in area dΩ caused by increments dφ and dθ, we have

y

x

z

area = d Ω

sinsin

sin

θθ

θ

θ

θ+ θ

φ

φ+ φ

φ

d

d

d

φd

Since dμ = sin(θ)dθ,

then dΩ = sin(θ) dφ dθ

For any function of angle

f(Ω) = f(θ, φ) = f(μ, φ)

we have

∫4π

f(Ω) dΩ =∫ 2π

φ=0

∫ π

θ=0

f(θ, φ) sin(θ) dθ dφ =∫ 2π

φ=0

∫ 1

μ=−1

f(μ, φ) dμ dφ

Also,

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E =1/2 m v2 (m = neutron mass) and dE = m v dv

Thus, the variables vx, vy , vz can be replaced by

E, Ω or E, θ, φ or E, μ, φ

Now, consider a neutron with initial energy E and direction ω, which undergoes a scattering event. We

define

ps(E,Ω → E‘,Ω‘) dE‘ dΩ‘ = the probability that the scattered neutron has final energy within dE‘

about E‘, and final direction within dΩ‘ about Ω‘

Then, we must have

∫4π

∫0

ps(E,Ω → E‘,Ω‘) dE‘ dΩ‘ = 1

We now define the macroscopic differential scattering cross section

Σs(E,Ω → E‘,Ω‘) = Σs(E) ps(E,Ω → E‘,Ω‘)

Then the following three statements are equivalent:

1) v Σs(E,Ω → E‘,Ω‘) dE‘ dΩ‘

= v Σs(E) ps(E,Ω → E‘,Ω‘ dE)

= the rate [probable number / sec] at which a neutron initially at state E, Ω, is scattered within dE‘ about

E‘ and within dΩ‘ about Ω‘.

2) If there are n neutrons in a system, then

n v Σs(E,Ω → E‘,Ω‘) dE‘ dΩ‘

= n v Σs(E) ps(E,Ω → E‘,Ω‘ dE)

= the number of neutrons in the system that are scattered from E, ω to within dE‘ and E‘ and within dΩ‘

about Ω‘ per second.

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3) n v Σs(E,Ω → E‘,Ω‘) dE‘ dΩ‘ dt

= n v Σs(E) ps(E,Ω → E‘,Ω‘ dE)

= the number of neutrons in the system that are scattered from E, ω to within dE‘ about E‘ and within

dΩ‘ about Ω‘ in a time interval dt.

Also

∫ ∫Σs(E,Ω → E‘,Ω‘) dE‘ dΩ‘ =

∫ ∫Σsps(E,Ω → E‘,Ω‘) dE‘ dΩ‘ = Σs(E) (8)

This motivates the term ’differential’ scattering cross section; the integral of a ’derivative’ of a fuction is

the function, and the integral of a ’differential’ cross section is a cross section.

We may also define the microscopic differential cross section

σs(E,Ω → E′,Ω

′) = σsps(E,Ω → E

′,Ω

′)

Then

∫ ∫σs(E,Ω → E

′,Ω

′) dE

′dΩ

′= σs(E)

In most cases of interest, neutron scattering does not depend on the four variables that define Ω and Ω‘, but

rather on the single scalar variable Ω Ω‘ = cos(θ0) = μ0 where θ0 is the scattering angle:

all on this cone

are equally probableΩ

Ωθ o

’ Ω ’

_

__

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Hence, we can write

σs(E,Ω → E′,Ω

′) = Σs(E → E

′,Ω ˙Ω

′) = Σs(E → E′ , μ0

If

Σs(E → E′, μ0) = Σso(E → E

′then scattering is isotropic. (9)

If

Σs(E → E′, μ0) = Σso(E → E

′+ 3 μ0 Σs1(E → E

′(10)

then scattering is linearly anisotropic. Otherwise, i.e in general, scattering is anisotropic.

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Nuclear Fission

Fissile nuclides undergo fission when struck by low energy (about 1 ev) neutrons: U-233, U-235, Pu-

239, Pu-241

E (ev)10 1010

10

10

10

10

21−2

0

1

2

3

107

σ f (E) (barns)

Note that σf for small E can be several orders of magnitude larger than σf for large E.

Fissionable nuclides undergo fission when struck by high-energy (1 Mev) neutrons : Th-232, U-238, Pu-240,

Pu-242

E (ev)1. 10 2. 10 6. 10

1

2

7 7 7

thresholdenergy

σ f (E) (barns)

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For fissionable nuclei σf = 1 barn for large E; this is much less than σf for fissile nuclei for low E. Only

fissile nuclei are capable, by themselves, of supporting a chain reaction.

Recall that

σa + σγ + σf

Thus, for a neutron that is absorbed, we can define

(E) = σγ(E) / σf (E) = (rate of capture) / (rate of fission) = capture to fission ratio

E10

5

10 1072−1

α (Ε)

Fission events produce :

(a) fissioned nuclei

(b) neutrons

(c) gammas, betas, neutrinos

(d) energy

In more detail,

(a) Fissioned Nuclei : charged, very energetic, neutron-rich. 80% of yhe energy release consists of kinetic

energy of these nuclei. β-decay accounts for another 4% after a delay time.

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(b) Neutrons : about 99% appear within 10−14 sec (prompt)

about 1% appear within 0.2 to 55.0 sec (prompt)

Let us define

νE = the probable total number of neutrons ( prompt and delay ) released in a fission reaction

initiated by a neutron with energy E.

(11)

Ε (Mev)5 10 15

3

4

5

ν(Ε)

ν(Ε) = ν + ν Ε0 1

Also, let us define

χ(E) dE = the probability that a prompt, fission−produced neutron has energy between E and E+dE.

(12)

(∫ ∞

0

χ(E) dE = 1)

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The prompt fission spectrum χ(E) is independent of the energy of the neutron that caused the fission.

1 2 3 E (Mev)

χ (Ε) (prompt)

Note that prompt fission-produced neutrons are fast, about 1 Mev.

Delayed neutrons are produced by a more complicated process:

X + X + prompt neutrons + neutrinos + A A1

0 1 2 3

An + X

1 2 3

X A

X A

2

2

4

4

− 11n+

prompt

− decay ( 0.2 to 55.0 )

(10 sec)−14

β

β + γ + k

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It is customary to lump delayed neutrons into six precursor groups characterized by half-lives of 55.0,

22.0, 6.0, 2.0, 0.5, and 0.2 sec.

λi = decay constant of the i-th precursor group

βi = fraction of all fission neutrons emitted from i-th precursor group

β = Σ βi = total fraction of delayed neutrons

E (Mev)0.4 1

(composite delayed)χ (Ε)

(c) Gammas, Betas, Neutrinos :

gammas : 4% of energy (prompt)

betas : 4% of energy (delayed)

neutrinos : 5% of energy (lost)

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(d) Energy:

Product % of energy Range Time delay

KE of fission products 80 < 0.01 cm prompt

prompt fast neutrons 3 10 - 100 cm prompt

γ-decay 4 100 cm prompt

fission product beta-decay 4 short delayed

neutrinos 5 infinite

other nonfission reactions 4 100 cm delayed

Note: About 97% of recoverable fission energy is deposited in the fuel material.

Fission Fuels:

Fissile : U-233, U-235, Pu-239, Pu-241

Fissionable : Th-232, U-238, Pu-240, Pu-242

Only U-235 is found in nature. But, fissile fuel can be produced by transmutation of fertile nuclei:

U + n U +

Np

Pu

(23 min)

(2.3 days)

1 239

239

239

238 γ

β

β

Th + n Th +

Pa

U

(22 min)

(27 days)

232 1 233

233

233

(protactinium)

β

β

γ

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Let us define

η = average number of neutrons produced (by fission) per neutron absorbed in fuel.

Then, for a single fuel isotope (nuclide 1)

η1 =ν1σf1

σa1=

ν1σf1

σf1 + σγ1=

ν11 + σγ1

σf1

=ν1

1 + 1

For an infinite, homogenous medium of this material, a steady-state chain reaction is maintained if

η1 =ν1σf1

σa1=

ν1Σf1

Σa1= 1

For a mixture of N types of nuclei, a steady-state chain reaction is maintained if

η =

∑Nj=1 (νΣf )j∑Nj=1 Σaj

= 1

If η1 > 1, then excess neutrons are available for transmutation of fertile nuclei. For example, consider a

mixture of nuclide 1 and a fertile nuclide 2. Then, the mixture is stedy-state if

ν1Σf1Σa1 + Σa2

= 1, then ν1Σf1Σa1 = Σa2

η1 = 1 + Σa2Σa1

Now, if η1 = 2 , then Σa2 = Σa1

so fissile fuel is generated as fast as it is depleted.

If η1 > 2, then Σa2 > Σa1

and the system breeds fissile fuel. This is the principle behind the design of fast breeder reactors.

Resonances, Doppler Broadening: Various cross sections have resonance regions characterized by large

changes in σ for small changes in E. As the temperature of a material increases, incident neutrons see nuclei

with a larger range of kinetic energies, and the relative velocity between the neutron and nucleus becomes less

specifically tied to the energy of the neutron. Therefore, the fine details of the resonance become smeared out.

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E

σ (Ε,Τ) σ (Ε,Τ)

EEΕ+ΔΕΕ Ε+ΔΕ

Τ

Τ

1

2

Τ < Τ1 2

This process is known as ’Doppler Broadening’. The Breit-Wigner formula (derived using quantum me-

chanics) describes this analytically.

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CHAPTER 2 : FISSION CHAIN REACTIONS (AND NUCLEAR REACTORS)

(Chapter 3, Duderstadt and Hamilton)

Multiplication Factor,k

- Subcritical Reactor

- Critical Reactor

- Supercritical Reactor

Six Factor Formula

- Resonance Escape Probability

- Thermal Utilization Factor

- Fast Fission Factor

- Four Factor Formula

Conversion Ratio

Breeding Ratio

Thermal Reactors

Fast Reactors

Suppose an incident neutron causes a fission. The results are :

a) α, β, γ particles,

b) fissioned nuclei,

c) prompt and delayed neutrons (occuring in a certain ”fission generation”). Some of these (prompt and

delayed) neutrons are eventually absorbed, and some eventually leak out of the system. The rest, after pos-

sibly numerous scattering events, will lead to new fission neutrons occuring in the next ”fission generation”.

We define

k = multilication factor =number of neutrons in a given generation

number of neutrons in the previous generation

Then:

k < 1 : subcritical (number of neutrons → 0 in time)

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k = 1 : critical (steady-state solution exists)

k > 1 : supercritical (number of neutrons → in time)

A more practical definition is

k0 =rate of neutron gain in reactor

rate of neutron loss in reactor=

G(t)L(t)

If neutrons are lost at a rate L(t), and if there are N(t) neutrons, then

l =N(t)L(t)

= mean neutron lifetime

Using the general rate equation

rate of change = rate of gain - rate of loss

we obtain

dNdt = G(t) − L(t) = k0L(t) − L(t) = (k0 − 1)L(t) = k0−1

l N(t)

N(t) = N(0) ek0−1

l t

N(t+ l) = N(0) ek0−1

l (t+l) = N(0) ek0−1

l tek0−1 = N(t) ek0−1

Hence,

k =N(t+ l)N(t)

= ek0−1

We have for k0 = 1,

k = ek0−1 = 1 + (k0 − 1) + 12 (k0 − 1)2 + ..... = k0 + O(k0 − 1)2

Also,

k > 1 if k0 > 1

k = 1 if k0 = 1

k < 1 if k0 < 1

Therefore, for criticality, k0 and k operationally mean the same, are equal for k = 1, and for k0 = 1

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we have

k0 = k, N(t) = ek0−1

l t

The Six Factor Formulas : In thermal teactors, fission neutrons (= 107 ev) are produced that must slow

down to thermal energies (= 0.1 ev) before they are likely to create the next generation of fission neutrons.

The six-factor and four-factor formulas describe qualtitavely how various physical processes affect the mul-

tiplication factor k.

Fast Neutron

Leaks outDoes notleak out

absorbed whileslowing down

slows down tothermal energies

leaks outof system

absorbed insystem

absorbed innon−fuel

absorbed infuel

capturefission(thermal)

capture(fast)fission

neutronsproduced

ν

1− PP

1− pp

P 1− P

P 1− P

P 1− P

FNL

TNLTNL

AF AF

FF

FNL

PFNL = probability that a fast neutron does not leak out of the system,

p = probability that a fast neutron is not absorbed while slowing down = resonance escape probability,

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PTNL = probability that a thermal neutron does not leak out of the system,

PAF = probability that a thermal neutron that is absorbed is absorbed in the fuel

= Σfuelf / ( Σfuelf + Σnonfuelf ) = f = thermal utilization factor

PF = probability that athermal neutron absorbed in the fuel produces a fission

= σfuelf / σfuela ( νPF = ηfuel)

γ = probable number of fast fission produced per fast neutron.

Then, if N1 = the number of neutrons in one generation and N2 = the number in the next generation,

N2 = N1 [ PFNL p PTNL PAF ν PF + γ ]

so

k = N2/N1 = PFNL p PTNL f ηfuel + γ

= PFNL p PTNL fηfuel [ PF NL p PT NL f ηfuel + γ

PF NL p PT NL f ηfuel ]

Thus we have the six-factor formula

k = PFNL p PTNL f ηfuel ε

where

ε = fast fission factor =probable number of neutrons from thermal + fast fissions

probable number of neutrons from thermal fissions

For an infinite medium, PFNL = PTNL = 1

and we get the four factor formula : k∞ = p f ηfuel ε

The six-factor and four-factor formulas show roughly how various different physical phenomena affect the

criticality of the reactor.

Conversion and Breeding; Thermal and Fast Reactors : We know return to the concept of creating fis-

sile fuel by transmutation:

238U + 1n → ... → 239Pu [ works best for fast neutrons]

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232Th + 1n → ... → 233U [ works best for thermal neutrons]

We define:

ConversionRatio = CR =rate of fissile atom production

rate of fissile atom consumption

BreedingRatio = BR = CR if CR > 1

Now, consider an infinite hoogenous mixture of a fissile nuclide (1) and a fertile nuclide (2). Suppose the

mixture is critical. To obtain a formula that determines whether the mixture will breed, we write the criti-

cality condition as

η =ν1 Σf1

Σa1 + Σa2= 1

Then, the system breeds if

CR =Σa2Σa1

> 1 or Σa2 > Σa1

or ν1 Σf1 = Σa1 + Σa2 > 2Σa1

or

η1 =ν1 Σf1

Σa1=

ν1 Σf1

Σf1 + Σγ1=

ν11 + α1

> 2

Thermal Reactors : In thermal reactors, an average energy is comparable to the thermal neutron energy

(= 0.1 ev). Some characteristics of these reactors:

1. σf is largest for thermal neutrons, so it is relatively easy to maintain a chain reaction. Not much fuel

needed.

2. Light, low mass-number materials are used in the construction of these reactors to help moderate

(slow down) the fast neutrons. (In elastic collisions with light nuclei, neutrons lose much more kinetic energy

than they do with heavy nuclei.)

3. Relatively simple to build.

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Page 29: Larsen Lecture Notes

4. η = 2 in fuel, so it is difficult to breed.

Fast Reactors : In fast reactors, an average neutron energy is much higher than in thermal reactors

(= 10−5ev). Some characteristics:

1. σf is small, so it is relatively difficult to maintain a chain reaction. Considerably more fuel (30-40

times as much as in thermal reactors) is needed to maintain chain reaction.

2. Heavy, high mass-number materials are used in the construction of these reactors to inhibit the slowing

down of neutrons.

3. relatively complicated to build.

4. η > 2 in fuel, so it is possible to breed.

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Comparison of Energy-Averaged ν, η, andσf in Thermal and Fast Reactors :

Thermal (LWR) Fast (LMFBR)

U235 Pu239 U235 Pu239

ν 2.4 2.9 2.6 3.1

η 2.0 1.9 2.1 2.6

σf 580 790 1.9 1.8

reflector

core

blanket U−238

(fast) core Pu

(thermal)

thermal reactor fast reactor

7

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CHAPTER 2 PROBLEMS

2.1 D H, p. 100, problem 3-1.

2.2 D H, p. 100, problem 3-3.

2.3 D H, p. 100, problem 3-6.

2.4 D H, p. 100, problem 3-9.

2.5 Measurements on a critical thermal reactor indicate that for every 100 neutrons emitted in fission,

10 leak out while slowing down, 10 are absorbed while slowing down, and 10 leak out afyter being thermal-

ized. Also, 70% of the thermal absoptions occur in the fuel, and 2 fission neutrons are emitted for every

thermal absorption in the fuel. Calculate:

a) the fast non-leakage probability,

b) the resonance escape probability,

c) the thermal utilization factor,

d) the fast fission factor,

e) the infinite medium multiplication factor.

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CHAPTER 3 : NEUTRON TRANSPORT AND DIFFUSION EQUATIONS (Chapter 4, Dud-

erstadt and Hamilton)

Density Fuctions

Angular Neutron Density

- probable number of neutrons in an infinitesimal element of phase space

- probable rate at which neutrons pass through an infinitesimal element of surface area

Delayed Neutrons

- Precursor Groups

Time-Dependent Neutron Transport Equation With Delayed Neutrons

- Initial and Boundary Conditions

Steady-State Transport Equation

Steady-State Transport in a Purely Absorbing Medium - Steady-State Transport in a Vacuum

- Method of Characteristics

- Decay from a Point Source

Angular Flux

Scalar Flux

Current

Discretized Representation of the Transport Equation

- Time Discretization

- Energy Discretization

– Multigroup Approximation

– One-Group Approximation

- Angular Discretization

– SN (Dissrete-Ordinates)

– PN (Spherical Harmonics)

- Representation of Macroscopic Scattering Cross Section

– Scattering Ratio, c

– Mean Scattering Cosine, μ0

- Spatial Discretization

- Iterative Methods

Special Geometries

- One-Dimensional Slab Geometry

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Page 33: Larsen Lecture Notes

– Integral Transport Equation

– SN Equations

– PN Equations

– Closure Relations

- One-Dimensional Spherical Geometry

- Two-Dimensional X,Y-Geometry

N-th Collided Flux Equations

One-Group Time-Dependent Diffusion Equation

- Initial Condition

- Boundary Conditions

– Prescribed Incident Flux

– Reflecting

– Extrapolated Endpoint (Extrapolation Distance)

- Interface Conditions

Energy-Dependent Diffusion Equation

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Density Functions : Consider a bar consisisting of a mixture of materials that vary with position x. We

define the mass density function ρ of the rod by:

ρ(x) dx = the mass between x and x+dx,

then,∫ x2

x1ρ(x) dx = the mass between x1 and x2,

and ρ has units of mass per length. The concept of a density function plays a fundamental role in the

mathematical description of neutron transport and diffusion processes.

Angular Neutron Density : We define the angular density function n(r, E,Ω, t) by:

n(r, E,Ω, t) d3rdEdΩ (1)

= the probabale number of neutrons in d3r about , having energy within dE about E,

travelling with direction in the solid angle dΩ about Ω, at time t.

x

y

z

x

y

z

E E+dE

energy

dy dx

dz

d 3r = dx dy dz

Ωd

Then, for example, if D is some domain in physical space, we have

[∫0 n(r, E,Ω, t) d3r ] dE dΩ = the probable number of neutrons in D, having energies within dE about

E and directions dΩ about Ω, at time t.

and

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∫4π

∫ E2

E1

∫0n(r, E,Ω, t) d3r dE dΩ = the total number of neutrons in D, with energies between E1 and

E2, at time t.

Also, let us consider a surface with an area increment dA and a unit normal vector n at the point r:

r

n

area=dA

Ω

y

x

z

We wish to compute the rate at which neutrons at ( r, E,Ω, t) pass through dA. To do this, let us consider

neutrons at r, travelling in the direction Ω, with energy E [or speed v = (2E/m)1/2]. In time dt, the neutrons

travel a distance dτ = v dt:

n

r

volume=dV=dAds

ds

area=dA

Ω

τ

θ

θd

dsdτ = |cosθ| = |Ω · n|

dV = dA dS = dA |Ω · n| dτ = dA |Ω · n| v dt

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Since neutrons travel a distance dτ in time dt, then every neutron in the volume dV at time t, having

direction Ω and speed v, will exit through dA between t and t+dt. Therefore,

number of neutrons that pass through dA in time dt = number of neutrons initially

in the volume ( and in dE about E, dΩ about Ω ) = ( n dE dΩ ) dV = ( n dE dΩ ) dA| Ω · n| v dt

and

v |Ω · n| n(r, E,Ω, t) dE dΩ dA

= the rate at which neutrons at r, within dE about E, within dΩ about Ω, pass through dA(2)

Now let us consider an arbitrary volume D with surface ∂D. At each point r ∈ ∂D letn= n(r) be the unit

outer normal vector. Then for fixed Ω, E, and t,

∂D∓ : that part of ∂D for which Ω · n >< 0

n

n

[∫∂D

vΩ · n n(r, E,Ω, t) dA ] dE dΩ

= [∫∂D+ vΩ · n n(r, E,Ω, t) dA ] dE dΩ − [

∫∂D− vΩ · n n(r, E,Ω, t) dA ] dE dΩ

= [ the rate at which neutrons flow out of D ] − [ the rate at which neutrons flow into D ]

or

[∫∂D

v Ω · n n(r, E,Ω, t) dA ] dE dΩ

= the net rate at which neutrons within dE about E and dΩ about Ω leak out of D = net leakage rate

(3)

Thus, integral can be positive or negative. If it is positive, then the rate of flow out of D is greater than the

rate of flow in. If it is negative, then the rate of flow in of D is greater than the rate of flow out.

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Similarly, if S is a surface with a continuously varying normal vector n(r):

n

n

then

[∫S

v Ω · n n(r, E,Ω, t) dA ] dE dΩ (4)

= the net rate at which neutrons within dE about E and dΩ about Ω flow through S.

This integral can be positive or negative. If it is positive, then the net neutron flow is in the direction

of the normal vectors; otherwise, it is in the positive direction.

If the above expression is integrated over Ω, we also have:

[∫

∫S

v Ω · n n(r, E,Ω, t) dA dΩ ] dE (5)

= the net rate at which neutrons within dE about E flow through S.

Again, this integral can be positive or negative, with the same meaning as before.

Now we shall derive an equation for n(x,E,Ω, t). We have, for any volume D, and for fixed E and Ω,

the general rate equation:

Rate of change of the neutrons for [r ∈ D, energies within dE about E, and directions dΩ about Ω]

= Rate of gain of neutrons [ r ∈ D, energies within dE about E, and directions dΩ about Ω ]

− Rate of loss of neutrons [ r ∈ D, energies within dE about E, and directions dΩ about Ω ](6)

However,

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Rate of change of neutrons [r ∈ D, energies within dE about E, and directions dΩ about Ω]

=d

dt[

∫D

n(r, E,Ω, t) dr3 ] dE dΩ] = [∫D

n

t(r, E,Ω, t) dr3 ] dE dΩ ] (7)

Also

Rate of loss of neutrons in dE about E and direction dΩ about Ω =

Rate at which neutrons in dE about E and direction dΩ about Ω undergo collisions

+ leakage rate of neutrons in dE about E and direction dΩ about Ω out of D

= [∫D

v ΣT (E) n(r, E,Ω, t) d3r] dE dΩ + [∫∂D

v Ω · n n(r, E,Ω, t) d2r] dE dΩ (8)

Here we have set dr2 = dA , dr3 = dV

However, the Divergence Theorem (or Green’s Theorem) gives for a general vector function f(r):

∫∂D n · f(r)d2r =

∫D∇ · f(r)d3r ,

where

∇ = i ∂∂x + j ∂

∂y + k ∂∂z = gradient operator

Therefore,

∫∂D

n · [ v Ω n ] dr2 =∫D∇ · v Ω n dr3 =

∫Dv Ω · ∇ n dr3 ,

so Eqn.(3.8) can be written

Rate of loss of neutrons in dE about E and direction dΩ about Ω

= [∫D

[ v Ω · ∇ n(r, E,Ω, t) + v ΣT (E) n(r, E,Ω, t)] d3r ] dE dΩ (9)

Next,

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Rate of gain into dE about E and direction dΩ about Ω

= Rate of gain into dE about E and direction dΩ about Ω due to scattered neutrons

+ Rate of gain into dE about E and direction dΩ about Ω due to prompt fission neutrons

+ Rate of gain into dE about E and direction dΩ about Ω due to delayed fission neutrons

+ Rate of gain into dE about E and direction dΩ about Ω due to interior sources

(10)

However,

Rate of gain due to scattered neutrons

= [∫D

[∫

∫ ∞

0

v′Σs(E

′ → E,Ω′ · Ω) n(r, E

′,Ω

′, t) dE

′dΩ

′] d3r] dE dΩ (11)

Note : see Eqn.(1.9). Also,

Rate of gain due to prompt fission neutrons

= [∫D

χp(E)4π

[∫

∫ ∞

0

[ 1 − β(E′) ] ν(E

′) v

′Σf (E

′) n(r, E

′,Ω

′, t) dE

′dΩ

′] d3r] dE dΩ (12)

Notes:

v′Σf (E

′) n(r, E

′,Ω

′, t) dE

′dΩ

′d3r = the fisison rate in dE

′about E

′, about dΩ

′about Ω

′, and d3r

about r,

ν(E′) = the total number of neutrons (prompt and delayed) produced in a fission event that is caused

by a neutron with energy E′,

β(E′) = the fraction of neutrons in a fission event, caused by a neutron with energy E

′, that are delayed,

[1 − β(E′)] ν(E

′) v

′Σf (E

′) n(r, E

′,Ω

′, t) dE

′dΩ

′] d3r = the rate at which prompt fission neutrons

are created by neutrons in dE′about E

′, about dΩ

′about Ω

′, and d3r about r,

χp(E) = prompt fission spectrum [∫ ∞0χp(E)dE = 1]

χp(E)4π dE dΩ = probability that a prompt neutron is emitted in dE about E, about dΩ about Ω.

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Rate of gain due to delayed fission neutrons

= [∫D

Qd(r, E, t) d3r] dE dΩ (Qd to be determined) (13)

Rate of gain due to interior sources

= [14π

∫D

Q(r, E, t) d3r] dE dΩ (14)

Note: Interior sources are usually isotropic. The factor 4π is included as a normalization factor, so that

[∫

∫ ∞

0

14π

∫D

Q(r, E, t) d3r] dE dΩ =∫ ∞

0

∫D

Q(r, E, t) d3r dE

= the total rate at which source neutrons are introduced in D.

Combining Eqns.(3.10)-(3.14), we get

Rate of gain into dE′about E

′, about dΩ

′about Ω

′, and d3r about r,

= [∫D

[∫ ∫

v′Σs n dE

′dΩ

′+χp(E)

∫ ∫(1 − β) ν v

′Σf n dE

′dΩ

′+ Qd +

14πQ] d3r] dE dΩ (15)

Finally, combining Eqns.(3.6), (3.7), (3.9), and (3.15), we obtain

∫D

[dn

dt+ vΩ·∇n + vΣTn −

∫ ∫v

′ΣsndE

′dΩ

′ − p

∫ ∫(1 − β)νv

′ΣfndE

′dΩ

′ − Qd − 14πQ]d3r = 0

Because D is arbitrary, the integrand [...] = 0. Thus, defining

ψ(r, E,Ω, t) = v n(r, E,Ω, t) = angular neutron flux, (16)

we obtain

1v

∂ψ

∂t+ Ω ·∇ψ + Σtψ =

∫ ∫Σs ψ dE

′dΩ

′+ p

∫ ∫(1 − β) ν Σf ψ dE

′dΩ

′+ Qd +

14πQ (17)

Now we shall derive an expression for Qd:

Qd(r, E, t) d3r dE dΩ = the rate at which delayed neutrons are emitted in dE′

about E′, about dΩ

about Ω′, and d3r about r.

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Recalling that delayed neutrons arise from fissioned nuclei that undergo spontaneous radioactive decay,

we define:

Cj(r, t)d3r = the probable number of fissioned nuclei in precursor group j, in d3r about r at time t. (The

decay constant for these nuclei is λj .)

χj(E) = the spectrum of neutrons emitted from precursor group j (∫ ∞0χjdE = 1)

βj(E) = the fraction of all fission neutrons, caused by a neutron with energy E, that are emitted from

the j-th precursor group.

β =∑6

j=1 βj(E) = the total fraction of delayed neutrons in a fission event caused by a neutron with

energy E.

Then, nuclei are introduced into j-th precursor group in d3r about r at the rate

[∫ ∞

0

∫4π

βj(E′) ν(E

′) v

′Σf (E

′) n(r, E

′,Ω

′, t) dE

′dΩ

′] d3r]

= [∫ ∞

0

∫4π

βj(E′) ν(E

′) Σf (E

′) ψ(r, E

′,Ω

′, t) dE

′dΩ

′] d3r]

Hence,

∂tCj(r, t) + λj Cj(r, t) =

∫ ∞

0

∫4π

βj(E′) ν(E

′) v

′Σf (E

′) n(r, E

′,Ω

′, t) dΩ

′dE

′(18)

The rate at which precursor group j nuclei spontaneously decay is λj Cj , and hence this is the rate at which

group j neutrons are produced. Since the spectrum for these neutrons is χj(E) and

χj(E)4π dEdΩ = probability that a group j delayed fission neutron is emitted in dE about E and dΩ about Ω.

[∫ ∞0

∫4π

χj(E)4π dEdΩ =

∫ ∞0χj(E)dE = 1]

then

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χj(E)4π λjCj(r, t)d3rdEdΩ = the rate at which delayed neutrons are emitted from the j-th precursor group

into d3r, dE about E and dΩ about Ω at time t.

Hence,

Qd(r, E, t) =∑6

j=1χj(E)

4π λjCj(r, t) ,

and Eqn.(3.17) becomes:

1v

∂tψ(r, E,Ω, t) + Ω·∇ψ(r, E,Ω, t) + Σtψ(r, E,Ω, t) =

∫ ∞

0

∫4π

Σs(E′ → E,Ω

′ · Ω)ψ(r, E,Ω, t)dE′dΩ

+ p(E)4π

∫ ∞

0

∫4π

(1 − β(E′))νΣf (E

′)ψ(r, E,Ω, t)dE

′dΩ

′+ Qd(r, E, t) =

6∑j=1

χj(E)4π

λjCj(r, t) +14πQ(r, E, t)

(19)

This is the full time-dependent neutron transport equation with delayed neutrons; it is coupled with

Eqn.(3.18), which govern the precursor densities. (Note: in some formulations of these equations, the

4π factors are absent, but then χp(E), χj(E), and Q(r,E,t) have different normalizations.)

Some physics which is omitted from these equations :

1) Certain quantum mechanical effects,

2) Motion of the host material,

3) Statistical fluctuations in the neutron density n,

4) Neutron-neutron and ather rare interactions,

5) Forces (for example, gravity) on neutrons,

6) Temperature feedback (σt depends on temperature which depends on ψ).

Initial and Boundary Conditions : Eqns.(3.18) and (3.19) do not, by themselves, describe ψ; we also need

initial and boundary conditions. Physically, we expect that given all (interior and boundary) sources of

neutrons, the initial values of the angular flux and precursor densities, we should be able to determine ψ

and Cj uniquely. This expectation is correct, and it gives us the appropriate initial and boundary conditions.

Interior source of neutrons: 14π Q(r, E, t) r ∈ D, 0 < E <∞, t > 0

This known term is already in Eqn.(3.19).

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Initial source of neutrons : we prescribe

ψ(r, E,Ω, 0) = ψi(r, E,Ω) r ∈ D, 0 < E <∞, |Ω| = 1, (20)

where ψi is known.

Boundary source of neutrons: we prescribe

ψ(r, E,Ω, 0) = ψb(r, E,Ω, t) r ∈ ∂D, 0 < E <∞, t > 0, Ω · n < 0, (21)

where ψb is known. Note that since n is the unit outer normal, Ω · n < 0 corresponds to all directions

Ω pointing into the spatial domain D. Therefore, we are free to prescribe the incoming or incident flux,

corresponding to Ω ·n < 0, but we cannot prescribe the outgoing or exiting flux, corresponding to Ω ·n < 0.

Initial values of the precursor densities:

Cj(r, 0) = Cij(r) r ∈ D, 1 < j < 6 (22)

where Cij are all known.

Eqns.(3.18)-(3.22) uniquely determine ψ and Cj in a given physical system D.

Note: an assumption implicit in this formulation is that nrutrons entering D through its outer boundary can

be arbitirarily chosen and are independent of the exiting fluxes. That is true only if the boundary of D is

convex:

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Page 44: Larsen Lecture Notes

D

exterior of D = ’vacuum’

1) convex

2) non-reentrant boundary

3) incident flux does not depend on the exiting flux

D

exterior of D= ’vacuum’

1) non-convex

2) reentrant boundary

3) incident flux does depend on the exiting flux

This problem can be cured by enlarging our definition of D to include some exterior points, so that the

new boundary is convex:

newboundary

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Page 45: Larsen Lecture Notes

Neutron transport problems are always solved in physical systems that are convex.

Transport Equation Without Delayed Neutrons: Set βj = β = Cj = 0. Then, Eqn.() becomes

1v

∂ψ

∂t+ Ω · ∇ψ + Σtψ =

∫ ∫Σsψ dE

′dΩ

′+

x

∫ ∫ν Σf ψ dE

′dΩ

′+

14πQ (23)

+ initial condition (3.20) + boundary condition (3.21)

Steady-State Transport Equation Without Delayed Neutrons:

Ω · ∇ψ + Σtψ =∫ ∫

Σs ψ dE′dΩ

′+

x

∫ ∫ν Σf ψ dE

′dΩ

′+

14πQ (24)

+ boundary condition (3.21)

Steady-State Transport Equation in a Purely Absorbing Medium: Set s = Σf = 0 ; then

Ω · ∇ψ + Σt ψ =14πQ (25)

+ boundary condition (3.21)

Steady-State Transport Equation in a Vacuum:

Ω · ∇ ψ = 0 (26)

+ boundary condition (3.21)

Equations (3.25) and (3.26) can be explicitly solved because these equations do not couple angle or en-

ergy. Eqn.() couples angle and energy, can be explicitly solved only in very special idealized cases.

Solution of the Steady-State Transport Equation in a Purely Absorbing Homogenous Medium: We consider

a convex homogenous domain D and

Ω · ∇ψ(r, E,Ω, t) + Σtψ(r, E,Ω, t) =14πQ(r, E, t) r ∈ D, |Ω| = 1

ψ(r, E,Ω) = ψb(r, E,Ω) r ∈ ∂D, Ω · n < 0 (27)

Eqn.(3.27) can be written

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Page 46: Larsen Lecture Notes

Ωx∂ψ

∂x+ Ωy

∂ψ

∂y+ Ωz

∂ψ

∂z+ ΣT ψ =

14πQ (28)

This is a first-order differential equation that can be solved by the method of characteristics. To do this, we

define a curve

r(s) = x(s) i + y(s) j + z(s) k

and

ψ(s, E,Ω) = ψ [ r(s), E,Ω ] = ψ [ x(s), y(s), z(s), E,Ω ]

Q(s, E) = Q[ r(s), E ] = Q[ x(s), y(s), z(s), E ]

Then,

∂ψ

∂s=

∂sψ [ x(s), y(s), z(s), E,Ω ] =

dx

ds

∂ψ

∂x+

dy

ds

∂ψ

∂y+

dz

ds

∂ψ

∂z

Let us require x(s), y(s), and z(s) to satisfy:

dx

ds= Ωx,

dy

ds= Ωy,

dz

ds= Ωz , (29)

Then, ∂ψ∂s = Ωx∂ψ∂x + Ωy

∂ψ∂y + Ωz

∂ψ∂z

Hence, using Eqn.(3.28), we obtain the system of equations (3.29) and

∂ψ

∂s(s, E,Ω) + Σt(E) ψ(s, E,Ω) =

14πQ(s, E,Ω) (30)

We must impose ’initial conditions’ for this system. Let r0 = x0i+ y0j + z0k be any point on the boundary

of D for which Ω · n < 0. Then, we set

x(0) = x0, y(0) = y0, z(0) = z0 (31)

and

ψ(0, E,Ω) = ψb(ro, E,Ω) (32)

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Page 47: Larsen Lecture Notes

Solving Eqns.(3.29) and (3.31), we get

x(s) = x0 + Ωx , y(s) = y0 + Ωy , z(s) = z0 + Ωz

so r(s) = r0 + Ωs

Thus, for s > 0, r(s) tracks into D. In fact, the characteristic line r(s) for s > 0 traces the physical path of a

neutron as it enters the system at r0 and propagates inward in the direction Ω.

n

s

D

.

Ω

Ω

dD

r +so

Eqns.(3.30) and (3.32) now give

∂ψ

∂s+ Σtψ =

14πQ

∂ψ

∂seΣts =

14πQ eΣts

ψ(s) eΣts − ψ(0) =∫ s

0

Q(s′)

4πeΣts

′ds

ψ(s) = ψ(0) eΣts +∫ s

0

Q(s′)

4πe−Σt(s−s′ ) ds

= ψ(0) eΣts +∫ s

0

Q(s− τ)4π

e−Σtτ dτ

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Page 48: Larsen Lecture Notes

or

ψ(r0 + s, E,Ω) = ψb(r0, E,Ω) eΣt(E)s +∫ s

0

Q(r0 + Ω s− Ω τ, E,Ω)4π

e−Σt(E)τdτ (33)

Let us now change notation a bit. For s fixed point r in domain D and direction Ω, let

r0 = the intersection of ∂D with the line r + Ω τ, τ < 0

s = | r − r0 | (34)

s

ro

dD

r

Ω

Then, making contact with our previous notation, we have

r = r0 + sΩ , and Eqn.(3.33) becomes

ψ(r, E,Ω) = ψb (r0, E,Ω) e−Σt(E)s +∫ s

0

Q(r − Ωτ, E,Ω)4π

e−Σt(E) τ dτ (35)

Note that the above equation gives the exiting flux from a system.

Thus, uncollided neutrons decay exponentially with rate Σt(E). (We predicted this earlier, in Chapter

1 of these notes.) Now for some special cases:

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Page 49: Larsen Lecture Notes

’Vacuum’ Boundary Condition: ψb = 0. Eqn.(3.35) becomes

ψ(r, E,Ω) =∫ s

0

Q(r − Ω τ, E,Ω)4π

e−Σt(E) τ dτ

Therefore

φ(r, E) = scalar flux =∫

∫ s

0

Q(r − Ω τ, E,Ω)4π

e−Σt(E) τ dτ

The variables s, Ω constitute a set of polar ccodinates with origin at the point r. Let us convert this set to

Cartesian ccordinates. We take

r′= r − sΩ (variable of integration)

τ = |r − r′ | (radius)

d3r′

= τ2 dτ dΩ = |r − r′ |2 dτ dΩ

and then we obtain

φ(r, E) =∫r′∈D

Q(r′, E)

4πe−Σt(E)| r−r′ |

| r − r′ |2 d3r′

(36)

Isotropic Point Source: If Q(r,E) is a delta-fuction source at r = r0, E = E0, i.e.,

Q(r, E) = δ(x− x0) δ(y − y0) δ(z − z0) δ(E − E0),

then

φ(r, E) =e−Σt(E0) |r−r0|

4π |r − r0|2δ(E − E0)

Therefore, the decay of the scalar flux away from a point source is as e−Σtr/r2, where r is the distance to the

point source.

Pure Streaming in a Vacuum: (Σt = Q = 0). Eqn.(3.35) reduces to

ψ(r, E,Ω) = ψb (r0, E,Ω)

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Page 50: Larsen Lecture Notes

Ω

r

ro

dD

D

Thus, the angular flux inside D consists of sloy of the free-streaming neutrons that enter D through its outer

boundary, and there is no exponential attenuation away from the boundary.

One-Dimensional Half-Space: Let D consist of the half-space z > 0, and let all quantities be indepen-

dent of x and y. Then for μ > 0, Eqn.(3.35) becomes

z

s=z/

r=xi + yj + zkz

z/s =cos

Ω

θθ

μ θ =μ

ψ(z, E, μ) = ψ(0, E, μ) e−Σt(E)s +∫ s

0

Q(z − μτ)4π

e−Σt(E) τ dτ

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Page 51: Larsen Lecture Notes

= ψb (E, μ) e−Σt(E)z/μ +∫ z /μ

0

Q(z − μτ)4π

e−Σt(E) τ dτ

= ψb (E, μ) e−Σt(E)z/μ +∫ z

0

Q(z − τ′)

4πe−Σt(E) τ

′/μ dτ

′(37)

For μ < 0, z = ∞, so

ψ(z, E, μ) =∫ ∞

0

Q(z + τ′, E)

4πe−Σt(E) τ

′/|μ|

|μ| dτ′

Some definitions:

n(r, E,Ω, t) = angular neutron density

ψ(r, E,Ω, t) = v n(r, E,Ω, t) = angular flux

φ(r, E, t) =∫4πψ(r, E,Ω, t) dΩ = scalar flux

J(r, E, t) =∫4π Ω ψ(r, E,Ω, t) dΩ = current

If e is a unit vector, then

J±(r, E, t) =∫Ω·e0 |Ω · e| ψ(r, E,Ω, t) dΩ = partial current

Note that if e is perpendicular to an area element dA, then

J+(r, E, t)dA = rate at which neutrons at (r,E,t) flow through dA in the direction of e.

J−(r, E, t)dA = rate at which neutrons at (r,E,t) flow through dA in the direction of -e.

ee

dAdA

J J+ −

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Page 52: Larsen Lecture Notes

J± > 0 [and J± are not vectors]

J± can not be determined without first specifying e.

J · e dA = (J+ − J−) dA = net rate at which neutrons flow through dA (+ if net flow is in direction of

e, - if net flow is in direction of -e)

Discretized Representation of the Transport Equation:

1) Time: The following scheme is stable for all Δt. Also, if Q is independent of t, it is accurate for

very large Δt.

ψn − ψn−1

v Δt+ Ω · ∇ ψn + Σt ψn =

∫ ∫Σs ψn dE

′dΩ

′+

14πQ

or

Ω · ∇ ψn + (Σt +1

v Δt) ψn −

∫ ∫Σs ψn dE

′dΩ

′=

14πQ +

ψn−1

v Δt

Thus, the time-dependent problem reduces to solving a steady-state problem within each time step. For

very large Δt,

Ω · ∇ ψn + Σt ψn −∫ ∫

Σs ψn dE′dΩ

′=

14πQ

and we obtain the correct equation for the steady-state solution.

2) Energy: (Multigroup approximation) We assume Emin ≤ E ≤ Emax and divide this large interval

into G smaller energy groups:

Emin = Ea < Ea−1 < ........ < E0 = Emax

Then, taking the transport equation

Ω · ∇ ψ + Σt(E) ψ =∫ Emax

Emin

∫4π

Σs(E′ → E,Ω

′ · Ω) ψ dΩ′dE

′+

14πQ

=G∑

g′=1

∫4π

∫ Eg′−1

Eg′

Σs(E′ → E,Ω

′ · Ω) ψ dE′dΩ

′+

14πQ,

21

Page 53: Larsen Lecture Notes

we operate by∫ Eg−1

Eg(.) dE and we get

Ω · ∇(∫ Eg

Eg+1

ψ dE ) + (

∫ Eg

Eg+1Σt(E) ψ dE∫ Eg

Eg+1ψ dE

)∫ Eg

Eg+1

ψ dE

=G∑

g′=1

∫4π

(

∫ Eg

Eg+1

∫ Eg

Eg′+1

Σs(E′ → E,Ω

′ · Ω) ψ dE′dE

∫ Eg

Eg+1ψ dE′ )

∫ Eg

Eg+1

ψ dE′dΩ

′+

∫ Eg

Eg+1

Q dE

Thus, if we define

ψg (r,Ω) =∫ Eg

Eg+1

ψ (r, E,Ω) dE, (38)

Qg(r) =∫ Eg

Eg+1

Q(r, E) dE, (39)

Σtg(r) =

∫ Eg

Eg+1Σt(E)Ψ dE∫ Eg

Eg+1Ψ dE

(40)

Σs,g′→g(r,Ω′ · Ω) =

∫ Eg

Eg+1

∫ Eg′

Eg′+1

Σs(E′ → E,Ω

′ · Ω) ΨdE′

∫ Eg

Eg+1Ψ dE′

(41)

where Ψ(E) is a suitably approximate shape function, then we get the multigroup transport equations

Ω · ∇ ψg + Σt(E) ψg =G∑

g′=1

∫4π

Σs,g′→g(Ω′ · Ω) ψg′ (r,Ω) dΩ

′+

14πQg, g = 1, 2, ..., G(42)

Note: These equations are exact if for each g and g′

Σt(E) = constant [= Σtg] for Eg+1 < E < Eg,

Σs(E′ → E,Ω

′ · Ω) = function of Ω′ · Ω)

[= Σs,g′→g(Ω′ · Ω)ΔEg ] for Eg+1 < E < Eg, Eg′+1 < E < Eg′ ,

because the shape fuction then cancels out of the expression for Σtg and Σs,g′→g. They are also exact if

ψ(r, E,Ω) = ψ(r,Ω) Ψ(E),

where Psi(E) is the shape function, because then psi(E) cancels out of the expression for Σtg and Σs,g′→g.

This shows that to get good accuracy, the shape fuction Psi(E) must be representative of the energy-

dependence of ψ(r, E,Ω).

22

Page 54: Larsen Lecture Notes

One-Group Approximation:

Ω · ∇ ψ(r,Ω) + Σt(r) ψ(r,Ω) =∫

Σs(Ω′ · Ω) ψ(r,Ω) dΩ

′+

14πQ(r) (43)

3) Angle (SNandPN ) : There are two widely-used ways to discretize in angle. The first is the discrete-

ordinates or SN approximation, in which beutrons are assumed to travel only in discrete directions Ωm, m

= 1,2,....,N. With each of these discrete directions (or ordinates) we associate a section of the unit sphere

wm, and we make the approximation

∫f(Ω) dΩ �

N∑m=1

f(Ωm) wm

where, by the definition of the wm, 4π =∑Nm=1 wm

The set Ωmwm is a ’quadrature set’, and Eqn.(3.42) becomes

Ω·∇ ψg + Σt(E) ψg =G∑

g′=1

N∑m′=1

Σs,g′→g(Ωm′ ·Ωm) ψg′m′ wm′ +14πQg′ , 1 ≤ g ≤ G , 1 ≤ m ≤ N (44)

There are three discrete-ordinates or SN equations. Eqn.(3.44) holds in Cartesian geometries only. Boundary

conditions are obvious.

An alternative to SN is the spherical harmonics or PN method. Consider the various spherical harmonic

functions

Υlm(Ω) , −l ≤ m ≤ l , l = 0, 1, 2, ..... ,N

It is known that any reasonable function f(Ω) can be expanded as a linear combination of these func-

tions:

f(Ω) =∞∑l=0

l∑m=−l

flmΥlm(Ω)

Let us multiply Eqn.() by Υlm(Ω) and integrate over Ω. This produces a system of

23

Page 55: Larsen Lecture Notes

[1 + 3 + 5 + ...... + (2N+1)] G = (2N + 1)2 G equations.

Next we insert the approximate expansion

ψg(r,Ω) �∞∑l=0

l∑m=−l

flmg(r)Υlm(Ω)

into this system. [Note that there are (2N + 1)2 G unknowns, flmg(r).] One then has a system with the

same number of equations as unknowns. [We will explicitly carry out this derivation later, in slab geometry.]

Note: Unlike the SN method, the derivation of boundary conditions for the PN equations is somewhat

problematical.

Explicit Representation for Σs(Ω′ · Ω) : In a given group, for the important linearly anisotropic scat-

tering, we set

Σs(Ω′ · Ω) =

14π

(Σs0 + 3 Ω · Ω′Σs1) (45)

To define Σs0 and Σs1, we shall need the identities

∫4π

dΩ = 4π,∫

Ω′ · ΩdΩ =

∫4π

Ω′ · ΩdΩ′

= 0,∫

(Ω′ · Ω)2dΩ =

∫4π

(Ω′ · Ω)2dΩ

′=

4π3,

We then have:

∫4π

Σs(Ω′ · Ω) dΩ =

14π

∫4π

Σs0 dΩ = Σs0

and

∫4π

(Ω′ · Ω) Σs(Ω

′ · Ω) dΩ =14π

∫4π

[ (Ω′ · Ω) Σs0 + 3 (Ω

′ · Ω)2] dΩ′

= Σs1 (46)

Now let ψ(Ω) be the angular flux at a fixed spatial point in a given group. Then,

[∫

Σs(Ω′ · Ω) ψ(Ω

′) dΩ

′] dΩ = rate at which neutrons are scattered into dΩ about Ω,

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Page 56: Larsen Lecture Notes

so

∫4π

[∫

Σs(Ω′ · Ω) ψ(Ω

′) dΩ

′] dΩ = rate at which neutrons are scattered

=∫

[∫

Σs(Ω′ · Ω) dΩ] ψ(Ω

′) dΩ

′=

∫4π

Σs0 ψ(Ω′) dΩ

′= Σs0

∫4π

ψ(Ω′) dΩ

Dividing this equation by Σt∫4πψ(Ω

′)dΩ

′, we get

Σs0Σt

=∫ ∫

Σs(Ω′ · Ω) ψ(Ω

′) dΩ

′dΩ

Σt∫4π ψ(Ω

′) dΩ′

=rate at which neutrons are scattered

rate at which neutrons undergo collisions

= mean number of scattered neutrons per collision = c. (47)

Next, we define

μ = mean scattering cosine

=∫ ∫

(Ω′ · Ω) Σs(Ω

′ · Ω) ψ(Ω′) dΩ

′dΩ∫ ∫

Σs(Ω′ · Ω) ψ(Ω

′) dΩ′ dΩ

=∫

[∫[(Ω

′ · Ω) Σs0 + 3 (Ω′ · Ω)2 Σs1] dΩ] ψ(Ω

′) dΩ

Σs0∫ψ(Ω

′) dΩ′ , for Ω

′ · Ω = 0,

=Σs1Σs0

Therefore, if Σs(Ω′ · Ω) can be written in the form of Eqn.(3.45), then

Σs0 = c Σt

Σs1 = μ0 Σs0 = c μ0 Σt

Σs(Ω′ · Ω) =

c Σt4π

(1 + 3 μ0 Ω′ · Ω) (48)

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Page 57: Larsen Lecture Notes

Usually, μ > 0, which is a consequence of the fact that forward-scattering is more probable than back-

scattering.

4) Spatial Discretizations: A lengthy topic. See NE 542.

5) Efficient Iterative Methods for Solving the Fully Discretized SN Equations: Another lenghty topic.

See also NE 542.

Special Geometries: The general steady-state one-group transport equation depends on five independent

variables: x, y, z, and Ω. This can be require an enormous amount of computer storage, even for prob-

lems with moderate numbers of discrete values of x, y, z, and Ω. However, in specialgeometries for which

symmetries occur, the number of independent variables can be reduced to a manageable number. We will

now describe one-dimensional slab geometry in detail, and then we will very briefly describe one-dimensional

spherical geometry and two-dimensional x, y-geometry.

26

Page 58: Larsen Lecture Notes

One-Dimensional Slab Geometry: ψ = ψ(z, μ). Let us consider a situation in which all geometrical quan-

tities and all boundary conditions are independent of x and y, and all boundary conditions depend only on

the polar angle μ and not on the azimuthal angle φ:

kz

x,y−plane

Ω

θ

Ω =√

1 − μ2cosφi +√

1 − μ2sinφj + μk

μ = cosθ , ψ is independent of x, y, and φ

We have Eqn.(3.43):

Ω · ∇ ψ + Σt(E) ψ =14π

∫4π

Σs0 + 3 Ω · Ω′Σs1 ψ dΩ

′+

Q

4π(49)

However,

Ω · ∇ ψ = (Ωx∂

∂x+ Ωy

∂y+ Ωz

∂z) ψ(z, μ) = μ

∂ψ

∂z(z, μ)

and

14π

∫4π

(Σs0 + 3 Ω · Ω′Σs1) ψ dΩ

=14π

∫ 1

μ′=−1

∫ 2π

φ′=0

(Σ0 + 3 [ sqrt1 − μ2 sqrt1 − μ′2 (cosφ

′cosφ + sinφ

′sinφ) + μ

′μ] Σs1) ψ(z, μ

′) dφ

′dμ

27

Page 59: Larsen Lecture Notes

=12

∫ 1

μ′=−1

(Σs0 + 3 μ · μ′Σs1) ψ(z, μ

′) dμ

Therefore, Eqn.(3.49) becomes

μ∂ψ

∂z(z, μ) + Σt ψ(z, μ) =

12

∫−1

1( Σs0 + 3 μ · μ′Σs1 ) ψ(z, μ

′) dμ

′+

Q(z)4π

(50)

It is customary to define

Ψ(z, μ) =∫ 2π

0

ψ(z, μ) dφ = 2 π ψ(z, μ),

and then

Ψ(z, μ) dμ dz = the total number of neutrons in dμ about μ, in dz about z, per unit area in the x, y−plane,

and Eqn.(3.50) becomes

μ∂Ψ∂z

(z, μ) + Σt Ψ(z, μ) =12[

∫−1

1( Σs0 + 3 μ · μ′Σs1 ) Ψ(z, μ

′) dμ

′+ Q(z) ] (51)

For the case of general anisotropic scattering, this equation generalizes to

μ∂Ψ∂z

(z, μ) + Σt Ψ(z, μ) =12(∫−1

1[∞∑n=0

(2n+ 1) Pn(μ) Pn(μ′) Σsn ] Ψ(z, μ

′dμ

′+ Q(z)) (52)

where Pn(μ) are the Legendre Polynomials:

P0(μ) = 1 , P1(μ) = μ , P2(μ) = 12 (3μ2 − 1) , . . . . . .

which satisfy the recursion relations

μPn(μ) = n2n+1Pn−1(μ) + n+1

2n+1Pn+1(μ)

and the orthogonality relations

∫ 1

−1

Pn(μ) Pm(μ) dμ =2

2n+ 1δnm (53)

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Page 60: Larsen Lecture Notes

Eqns.(3.51) and (3.52) typically hold on a slab −Z < z < Z. Using

μ > 0μ < 0

z

(neutrons flow to the right)(neutrons flow

to the left)

we see that incident boundary conditions for Ψ must be assigned as follows:

−Z Z

μ < 0

Ψ(−Ζ,μ) Ψ(Ζ ,μ)

μ > 0

so

Ψ(−Z, μ) = Ψb (μ) μ > 0

Ψ(Z, μ) = Ψb (μ) μ < 0(54)

If Ψb = 0 on either edge, then that edge is termed a vacuum boundary.

29

Page 61: Larsen Lecture Notes

μ

z

−1

+1

Z−Z

boundary conditions imposedon edges denoted by

Other boundary conditions, based on physical symmetries, are also possible. For example, suppose that

Ψb = 0, Q(z) = Q(−z), Σt(z) = Σt(−z), Σs0(z) = Σs0(−z), and Σs1(z) = Σs1(−z) :

z = −Z z = 0 z = Z

.. .−z z

Q(z) = Q(−z)Σ (z) = Σ (−z)

Then, Ψ must have a similar type of symmetry about z = 0:

Ψ(z, μ) = Ψ(−z,−μ) [⇒ Ψ(0, μ) = Ψ(0,−μ) ] (55)

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Page 62: Larsen Lecture Notes

z = −Z z = 0 z = Z

.. .−z z

−μ μ

Thus, we can reformulate the problem for the smaller slab 0 < z < Z and assign reflecting boundary

conditions on the left edge:

0 Ζ

’vacuum’ boundary :’reflecting’ boundary :

Ψ(Ζ,μ) = 0 ,−1 < μ < 0

Ψ(0,μ) = Ψ(0,−μ)

Ψ(0, μ) = Ψ(0,−μ) [reflecting boundary]

Ψ(z, μ) = 0 , − 1 ≤ μ < 0 [vacuum boundary](56)

This reduces by a factor of two the amount of storage and arithmetic required to solve a problem.

Integral Transport Equation: Let us consider the following slab-geometry problem with Σt = 1 and Σs1 = 0 :

μ∂ψ

∂z(z, μ) + ψ(z, μ) =

c

2

∫−1

1ψ(z, μ′) dμ

′+

q(z)2

0 < z < L (57)

ψ(0, μ) = 0 , 0 < μ ≤ 1

ψ(L, μ) = 0 , − 1 ≤ μ < 0

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Page 63: Larsen Lecture Notes

Defining the scalar flux

φ(z) =∫−1

1 ψ(z, μ′) dμ

′(58)

we can write Eqn.(3.57) as

∂ψ∂z + 1

μ ψ = 12 μ(c φ + q)

∂∂zψ ez/μ = ez/µ

2μ (c φ + q)

For μ > 0,∫ z0 (·)dz′

gives

ψ(z, μ) ez/μ − ψ(0, μ) =∫ z

0

ez′/μ

2 μ[ c φ(z

′) + q(z

′) ] dz

for ψ(0, μ) = 0 ,

ψ(z, μ) =∫ z

0

e− (z−z′)/μ

2 μ[ c φ(z

′) + q(z

′) ] dz

′(59)

For μ < 0,∫ Lz

(·)dz′gives

ψ(L, μ) eL/μ − ψ(z, μ) =∫ L

z

ez′/μ

2 μ[ c φ(z

′) + q(z

′) ] dz

′for

ψ(L, μ) = 0 ,

ψ(z, μ) =∫ L

z

e(z′−z)/μ

−2 μ[ c φ(z

′) + q(z

′) ] dz

′(60)

Combining Eqns.(3.58)-(3.60), we obtain

φ(z) =∫ 1

0

ψ(z, μ) dμ +∫ 0

−1

ψ(z, μ) dμ

=∫ z

0

[∫ 1

0

e−(z−z′)/μ

2 μdμ] [ c φ(z

′) + q(z

′) ] dz

+∫ L

z

[∫ 0

−1

e−(z′−z)/−μ

−2 μdμ] [ c φ(z

′) + q(z

′) ] dz

However, for z′< z,

∫ 1

0

e−(z−z′)/μ

2 μdμ =

∫ 1

∞e−(z−z′)t(

−dtt

) =∫ ∞

1

e−(z−z′)tt dt

32

Page 64: Larsen Lecture Notes

= E1( z − z′) = E1( | z − z

′ | ) (61)

and for z′> z,

∫ 0

−1

e−(z′−z)/−μ

−2μdμ =

∫ 0

1

e−(z′−z)/s

s(−ds) =

∫ 1

0

e−(z′−z)/s

sds

= E1( z′ − z ) = E1( | z − z

′ | )

Combining the last three equations, we find

φ(z) =12

∫ L

0

E1( | z − z′ | ) [ c φ(z

′) + Q(z

′)] dz

′(62)

This is the integral transport or Peierls equation for the scalar flux. It can be derived in any geometry and

with inhomogenous media, and non-vacuum boundary conditions. The only requirements is that scattering

be isotropic.

The Peierls equation has the advantage that is Ω is removed. However, it has the disadvantage that every

spatial point is explicitly coupled to every other point; the equation is spatially global, unlike the integro-

differential form of the transport equation, which is spatially local. Therefore, the Peierls equation is normally

used in computer codes only for small physical systems for which ona can safely make the approximation

that scattering is isotropic.

Discrete-Ordinates or SN Equations : Consider an angular quadrature set consisting of angles μm and

angular weights wm, 1 ≤ m ≤ N , satisfying

∑Nm=1 wm = 2 , wm > 0

μm = − μN+1−m , wm = wN+1−m (symmetry)

wN/2 + ....... + wm−1 < μm < wN/2 + ....... + wm N/2 + 1 ≤ m ≤ N

. . . .μ μ μ μ

ω ω ω ω

1 2

1 2

N−1 N

NN−1

The Gauss-Legendre Quadrature Sets satisfy these conditions; some of these sets are given at the end of this

33

Page 65: Larsen Lecture Notes

chapter. For μ � μm, we can make the approximations

ψ(z, μ) � ψ(z, μm) ≡ ψm(z)

and

∫ 1

−1

ψ(z, μ′) dμ

′ � ψ(z, μ1) w1 + . . . . . + ψ(z, μN ) wN =N∑n=1

ψ(z, μn) wn

Therefore, the problem (3.57) can be approximated by

μmdψmdz

(z) + ψm(z) =12[c

N∑n=1

ψn(z) wn + q(z)] 0 < z < L (63)

ψm(0) = 0 1 ≤ m ≤ N/2 (μm > 0)

ψm(L) = 0 N/2 + 1 ≤ m ≤ N (μm < 0)

which is termed the discrete-ordinates or SN approximation. This approximation can be developed for any

transport geometry. Physically, it amounts to constraining neutrons to travel in only a finite set of directions

μ1, ...., μN , rather than an infinite set −1 ≤ μ ≤ 1.

Spherical Harmonics or PN Equations : We now make explicit use of the Legendre polynomials defined in

Eqn.(3.53):

P0(μ) = 1 , P1(μ) = μ , P2(μ) = 12 (3μ2 − 1) , .........

μPn(μ) = n2n+1Pn−1(μ) + n+1

2n+1Pn+1(μ) n ≥ 1 [P−1 = 0]

∫ 1

−1 Pn(μ)Pm(μ)dμ = 22n+1δnm n,m ≥ 0

It is known that for a reasonable function f(μ), one has the following Legendre polynomial expansion

f(μ) =∞∑n=0

2n+ 12

fn Pn(μ)

where

fn =∫ 1

−1

Pn(μ) f(μ) dμ = ′expansion coefficients′

34

Page 66: Larsen Lecture Notes

Let us now define the angular flux moments

φn(z) =∫ 1

−1

Pn(μ) ψ(z, μ) dμ, n ≥ 0

Note that φ0 = scalar flux and φ1 = current.

We multiply Eqn.(3.57) by Pn(μ) and obtain

μ Pn(μ)∂ψ

∂z+ Pn(μ) ψ = Pn(μ)

12

[ c φ(z) + q(z)]

or

∂z[

n

2n+ 1Pn−1(μ)ψ +

n+ 12n+ 1

Pn+1(μ) ψ] + Pn(μ) ψ = Pn(μ)12

[ c φ(z) + q(z)]

Now we operate by∫ 1

−1(·)dμ. For n = 0, we obtain

d

dzφ1(z) + (1 − c) φ0(z) = q(z)

For 1 ≤ n ≤ N , we obtain

n

2n+ 1dφn−1

dz(z) +

n+ 12n+ 1

dφn+1

dz(z) + φn(z) = 0

These are N+1 equations in N+2 unknowns φ0, φ1, ...., φN+1. The standart closure relation is simply to take

φN+1 = 0.

Then, we obtain the standart slab geometry PN equations, which have the same number of equations as

unknowns:

dφ1

dz+ (1 − c) φ0 = q

n

2n+ 1dφn−1

dz+

n+ 12n+ 1

dφn+1

dz+ φn = 0 1 ≤ n ≤ N − 1

N

2N + 1dφN−1

dz+ φN = 0 (64)

Normally, N is odd, so there are an even number of equations and unknowns φ0, φ1, ...., φN .

In principle, one solves these equations, and then ψ is obtained from

ψ(z, μ) �N∑n=0

2n+ 12

Pn(μ) φn(z)

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Page 67: Larsen Lecture Notes

However, in practice, one is usually only interested in φ0(x) and φ1(x), and the higher order moments are

usually discarded at the end of the calculation.

P3 Equations:

dφ1dz + (1 − c) φ0 = q

13dφ0dz + 2

3dφ2dz + φ1 = 0

25dφ1dz + 3

5dφ3dz + φ2 = 0

37dφ2dz + φ3 = 0

(65)

Eliminating φ1 and φ3, we obtain two coupled diffusion equations:

−13d2φ0dx2 + (1 − c) φ0 = q + 2

3d2φ2dx2 ,

−1121

d2φ2dx2 + φ2 = 2

15d2φ0dx2

(66)

P1 Equations:

dφ1dz + (1 − c) φ0 = q ,

13dφ0dz + φ1 = 0

(67)

Eliminating φ1, we obtain the single diffusion equation:

−13d2φ0

dx2+ (1 − c) φ0 = q

P1 or diffusion theory is a very common and useful approximation to transport theory, and later we will

discuss it in detail.

Note: Boundary conditions for the diffusion (P1) equations are known and well-accepted, and we will derive

them in the next chapter. However, boundary conditions for the general PN equations with N odd and geq3

are controversial.

One-Dimensional Spherical Geometry : We take Eqn.(3.49)

Ω · ∇ ψ + Σt(E) ψ =14π

∫4π

(Σs0 + 3 Ω · Ω′Σs1) ψ dΩ

′+

Q

4π(68)

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Page 68: Larsen Lecture Notes

with ψ = ψ(r, μ)

where

r = x i + y j + z k = spatial position

r = | r | = ( x2 + y2 + z2 )1/2 = distance to origin (69)

Ω = direction of neutron travel

μ = cosθ = Ω · ( r/r ) = ( x Ωx + y Ωy + z Ωz )/r (70)

z

x

y

r

r / r

θΩ

We note that if Ω is fixed and r is varied in any non-radial direction, then μ varies. Therefore, an uncollided

particle moves along a path in which r and μ both change. We have

Ω · ∇ ψ(r, μ) =∂ψ

∂r(Ω · ∇ r) +

∂ψ

∂μ(Ω · ∇ μ) (71)

However,

r2 = x2 + y2 + z2

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Page 69: Larsen Lecture Notes

2r ∂r∂x = 2x 2r ∂r∂y = 2y 2r ∂r∂z = 2z∂r∂x = x

r∂r∂y = y

r∂r∂z = z

r

so

Ω · ∇r = Ωx ∂r∂x + Ωy ∂r∂y + Ωz ∂r∂z= Ωx xr + Ωy yr + Ωz zr = μ

Also, by Eqn.(3.70),

∂μ∂x = ∂μ

xΩx

+ yΩy

+ zΩz

r == Ωx

r − xΩx+yΩ

y+ zΩ

z

r2xr = Ω

x

r − μxr2

∂μ∂y = . . . . . . . . . =

Ωy

r − μyr2

∂μ∂z = . . . . . . . . . = Ω

z

r − μzr2

so

Ω · ∇ μ = Ωx∂μ∂x + Ωy

∂μ∂y + Ωz

∂μ∂z

= Ωx(Ω

x

r − μxr2 ) + Ωy(

Ωy

r − μyr2 ) + Ωz(

Ωz

r − μzr2 )

= 1r − μ

r

xΩx

+ yΩy

+ zΩz

r = 1−μ2

r

Hence, Eqn.(3.71) becomes

Ω · ∇ ψ(r, μ) = μ∂ψ

∂r+

1 − μ2

r

∂ψ

∂μ(72)

Next, we consider the scattering term

14π

∫( s0 + 3 Ω · Ω′

Σs1) ψ(r,Ω′ · r/r)dΩ′

The value of this integral cannot depend on our choice of angular coordinate system. Therefore, we can

choose this coordinate system to simplify the evaluation of the integral. In particular, for r fixed, we set

k = r/r

and we choose i and j in any suitable way. Then,

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Page 70: Larsen Lecture Notes

Ω =√

1 − μ2cosφi +√

1 − μ2sinφj + μk

Ω′=

√1 − (μ′)2cosφ

′i +

√1 − (μ′)2sinφ

′j + μ

′k

and dΩ′

= dφ′dμ

so

14π

∫(s0 + 3 Ω · Ω′

Σs1) ψ(r,Ω′ · r/r) dΩ′

= 14π

∫ 1

μ′=−1

∫ 2π

φ′=0(Σ0 + 3 [ sqrt1 − μ2 sqrt1 − μ

′2(cosφ′cosφ + sinφ

′sinφ) + μ

′μ ]‘Σs1) ψ(z, μ

′) dφ

′dμ

= 12

∫μ′=−1 1(Σs0 + 3 μ · μ′

Σs1) ψ(z, μ′) dμ

(73)

Using Eqns.(3.72) and (3.73), Eqn.(3.68) becomes:

μ∂ψ

∂r+

1 − μ2

r

∂ψ

∂μ+ Σt ψ =

12

∫ 1

−1

(Σs0 + 3 μ · μ′Σs1) ψ dμ

′+

Q(r)4π

(74)

Again, it is customary to define

Ψ(r, μ) =∫ 2π

0 ψ(r, μ)dφ = 2π ψ(r, μ)

so that

Ψ(r, μ)dμdV = Ψ(r, μ)dμ(4πr2)dr = the total number of neutrons in dμ about μ, in dr about r,

and Eqn.(3.74) becomes

μ∂Ψ∂r

(r, μ) +1 − μ2

r

∂Ψ∂μ

+ Σt Ψ(r, μ) =12[∫−1

1(Σs0 + 3 μ · μ′Σs1) Ψ(r, μ

′) dμ

′+ Q(r)] (75)

This equation holds for 0 < r < R and −1 ≤ μ ≤ 0 corresponds to directions pointing toward the center of

the sphere, and 0 < μ ≤ 1 corresponds to directions pointing away from the center of the sphere. Boundary

conditions to go with Eqn.(3.75) are therefore:

Ψ(R,μ) = Ψb(μ) , − 1 ≤ μ ≤ 0 (76)

[This prescribes the incident angular flux at the outer boundary.]

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Page 71: Larsen Lecture Notes

μ

rR

1

0

−1

boundary conditionsare applied here

Note: SN (discrete-ordinates) is trickier to apply here because of the angular derivative term. However, PN

(spherical harmonics) is just as easy as to apply. As before, P1 theory leads to a standart diffusion equation.

Two-Dimensional X,Y-Geometry: We again take Eqn.(3.49):

Ω · ∇ ψ + Σt(E) ψ =14π

∫4π

Σs0 + 3 Ω · Ω′Σs1 ψ dΩ

′+

Q

ψ = ψ(x, y,Ω) [independent of z]

and ψ(x, y,Ω) = ψ(x, y,Ωr)

y

x

z

Ω

Ω θ

θr

cos

cos k

k−

40

Page 72: Larsen Lecture Notes

Ω = sin(θ) cos(φ) i + sin(θ) sin(φ) j + cos(θ) k

Ωr = sin(θ) cos(φ) i + sin(θ) sin(φ) j − cos(θ) k

= reflection of Ω across the x,y-plane

Then,

Ωx∂ψ

∂x+ Ωy

∂ψ

∂y+ Σtψ =

14π

∫4π

Σs0 + 3 Ω · Ω′Σs1 ψ dΩ

′+

Q

4π(77)

where

Ωx = sin(θ) cos(φ) = μ

Ωy = sin(θ) sin(φ) = η

( Ωz = cos(θ) = ζ )

(78)

The surface of the unit sphere is defined by

1 = f( μ, η, ζ ) = μ2 + η2 + ζ2

and the unit outer normal for a point on the sphere is

n =∇f

| ∇f | =2 μ i + 2 η j + 2 ζ k2 (μ2 + η2 + ζ2)1/2

= μ i + η j + ζ k

z

x

y

n

μ η

Ω

= μ η ζ

ζ

μη

d d

d

i j k+ +

41

Page 73: Larsen Lecture Notes

In the plane generated by n and k, we have

kn

d

d d

α

α

Ω

μ η

Therefore,

dμ dηdΩ = cosα = k · n = ζ =

√1 − μ2 − η2

Thus, dΩ = dμ dη√1 − μ2 − η2

and Eqn.(3.77) becomes:

μ∂ψ

∂x+ η

∂ψ

∂y+ Σt ψ

=14π

∫ ∫[ Sigmas0 + 3 (μ μ

′+ η η

′+ ζ ζ

′) Σs1 ]

ψ dμ′dη

√1 − (μ′)2 − (η′)2

+14π

∫ ∫[ Sigmas0 + 3 (μ μ

′+ η η

′+ ζ ζ

′) Σs1 ]

ψ dμ′dη

√1 − (μ′)2 − (η′)2

+Q

=12π

∫ ∫[ Sigmas0 + 3 (μ μ

′+ η η

′) Σs1 ]

ψ dμ′dη

√1 − (μ′)2 − (η′)2

+Q

4π(79)

It is customary to define

Ψ(x, y, μ, η) = 2 ψ(x, y, μ, η)

so that

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Page 74: Larsen Lecture Notes

Ψ(x, y, μ, η) dx dy = dμ dη√1 − μ2 − η2

= the number of neutrons in dμ about μ, dη about η, dx about

x, dy about y, per unit length in z,

and Eqn.(3.79) becomes:

μ∂Ψ∂x

(x, y, μ, η) + η∂Ψ∂y

(x, y, μ, η) + Σt Ψ(x, y, μ, η)

= [12π

∫ ∫[ Sigmas0 + 3 (μμ

′+ ηη

′) Σs1 ] Ψ(x, y, μ

′, η

′)

dμ′dη

√1 − (μ′)2 − (η′)2

+ Q(x, y) ] (80)

This equation holds for −X < x < X and −Y < y < Y . Using

μ > 0η > 0

μ > 0η < 0

μ < 0η < 0

μ < 0η > 0

x

y

43

Page 75: Larsen Lecture Notes

we find that incident boundary conditions for Ψ must be assigned as follows:

(0,0) μ < 0μ > 0

η < 0

η > 0

Ψ ( μ,η)μ,η)

Ψ ( μ,η)

μ,η)

YX X,y,

x,Y,

b

b

X, y,

Ψ ( b

x,−Y,

Ψ (−b

The function Ψb is defined at each point on the boundary, and for each incoming direction (μ, η). Then we

set Ψ = Ψb for all such points and directions.

If Ψb = 0 along a section of the boundary, then this section is termed a ’vacuum boundary’.

Other boundary conditions are possible, based on symmetry considerations. For example, suppose the

outer boundaries of the system are vacuum and

X

Y

.

. .

.(−x,y) (x,y)

(−x,−y) (x,−y)

44

Page 76: Larsen Lecture Notes

Q(x,y) = Q(x,-y) = Q(-x,y) = Q(-x,-y), and similarly for Σt,Σs0,Σs1.

Then, Ψ must have a similar type of symmetry:

Ψ(x, y, μ, η) = Ψ(x,−y, μ,−η) = Ψ(−x, y,−μ, η) = Ψ(−x,−y,−μ,−η)

. .

..

Y

X(x,y)

(x,−y)(−x,−y)

(−x,y)

(−μ,−η) (μ,−η)

(μ,η)(−μ,η)

Thus, we can reformulate the problem in the firts quadrant and assign reflecting boundary conditions on the

left and bottom edges:

y

x

(’vacuum’)

(’vacuum’)

(’reflecting’)

(’reflecting’)

Ψ( μ,η)

Ψ(

Ψ(Ψ(

= Ψ(μ,η)

μ,η) = μ,η)Ψ( μ,η)

= 0

= 0μ < 0

η < 0Y

X

x, Y,

X, y,

x, 0, x, 0,μ, −η)

0, y,−

0, y,

45

Page 77: Larsen Lecture Notes

This reduces by a factor of four the amount of storage and arithmetic required to solve this problem.

Finally, Sn and PN versions of Eqn.(3.80) are relatively easy to formulate.

N-th Collided Flux Equations: Suppose we wish to solve the transport problem

μ∂ψ

∂z+ Σt ψ =

Σs02

∫ 1

−1

ψ(z, μ′) dμ

′+

Q(z)2

0 < z < L (81)

ψ(0, μ) = ψl(μ) 0 < μ ≤ 1

ψ(L, μ) = ψr(μ) − 1 ≤ μ < 0

Let us consider the following recursive sequence of problems for functions ψ0, ψ1, ψ2, .....

We define ψ0 by the problem

μ∂ψ0

∂z+ Σt ψ0 =

Q(z)2

0 < z < L (82)

ψ0(0, μ) = ψl(μ) 0 < μ ≤ 1

ψ0(L, μ) = ψr(μ) − 1 ≤ μ < 0,

and for n ≥ 1, we define ψn in terms of ψn−1 by

μ∂ψn∂z

+ Σt ψn =Σs02

∫ 1

−1

ψn−1 dμ′

0 < z < L (83)

ψn(0, μ) = 0 0 < μ ≤ 1

ψn(L, μ) = 0 − 1 ≤ μ < 0

Then, adding up all of Eqns.(3.82) and (3.83), we obtain

μ ∂∂z [ ψ0 + ψ1 + ψ2 + . . . . . . ]

+ Σt [ ψ0 + ψ1 + ψ2 + . . . . . . . ]

= Q2 + Σs0

2

∫ 1

−1[ ψ0 + ψ1 + . . . . . . . ] dμ

c

Likewise,adding up all of the first boundary conditions of Eqns.(3.82) and (3.83), we obtain

[ψ0(0, μ) + ψ1(0, μ) + . . . . ] = ψl(μ) 0 < μ ≤ 1,

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Page 78: Larsen Lecture Notes

and adding up all of the second boundary conditions of Eqns.(3.82) and (3.83), we obtain

[ ψ0(L, μ) + ψ1(L, μ) + . . . . ] = ψr(μ) − 1 ≤ μ < 0

Therefore, if we define

ψ(z, μ) =∞∑n=0

ψn(z, μ) , (84)

then ψ satisfies the original problem (3.81). We have from problems (3.82) and (3.83):

ψn = the angular flux due to neutrons that have undergone exactly n collisions.

Thus, ψ0 = uncollided flux, ψ1 = once-collided flux, ....., ψn = n-th collided flux. These interpretations

are consistent with Eqn.(3.85), because then ψ is the sum of the angular fluxes of neutrons that have under-

gone all possible numbers of collisions.

The One-Group Diffusion Equation: Let us consider the following general-geometry one-group transport

problem:

1v

∂ψ

∂t+ Ω · ∇ ψ + Σt ψ =

14π

∫4π

(Σs0 + 3 Ω · Ω′Σs1) ψ dΩ

′+

Q

=Σs04π

∫ψ dΩ

′+

3 Σs14π

Ω ·∫

Ω′ψ dΩ

′+

Q

=Σs04π

φ +3 Σs14π

Ω · J +Q

=14π

(Σs0 φ + 3 Σs1Ω · J + Q) (85)

where

φ =∫ψ dΩ

′= scalar flux

J =∫

Ω ψ dΩ′

= current(86)

Eqn.(3.86) holds for points r in a spatial domain D. The initial and boundary conditions to go with Eqn.(3.86)

are

ψ(r,Ω, 0) = ψi(r,Ω) (87)

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Page 79: Larsen Lecture Notes

ψ(r,Ω, t) = ψb(r,Ω, t) , r ∈ ∂D , Ω · n < 0 (88)

We wish to derive the full P1 approximation to this general problem. To do this, we need the identities

∫dΩ = 4π

∫Ω dΩ = 0∫

Ω Ω dΩ = 4π3 I

∫Ω Ω Ω dΩ = 0

(89)

Now, we first operate on Eqn.(3.86) by∫

(·)dΩ and get

1v

∂φ

∂t+ ∇ · J + (Σt − Σs0) φ = Q (90)

This is the ’balance’ or ’conversation’ equation. If we multiply Eqn.(3.91) by a small volume element d3r,

then each term has the following physical interpretation:

1v

∂φ

∂td3r + ∇ · J d3r + ( Σt − Σs0) φ d3r = Q d3r

- First term on the left: rate of change of φ in d3r

- Second term on the left: rate of leakage out of d3r

- Last term on the left: rate of absorption in d3r

- Term on the right: production rate due to source in d3r

Next, we operate on Eqn.(3.86) by∫

Ω(·)dΩ and get

1v

∂tJ + ∇ ·

∫Ω Ω ψ dΩ + ( Σt − Σs1 ) J = 0 (91)

Eqns.(3.91) and (3.92) are exact. To close them (so that there are the same number of equations as un-

knowns) we must approximate the integral in Eqn.(3.92) using φ and J . We do this using

ψ(r, J, t) � 14π

[ φ(r, t) + 3 Ω · J(r, t) ] (92)

This approximates ψ in terms of spherical harmonic functions of order 0 and 1 in such a way that Eqn.()

are preserved. [Thus, this approximation sets to zero all spherical harmonic moments of ψ of order ≥ 2.]

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Page 80: Larsen Lecture Notes

Introducing Eqn.(3.93) into the integral in Eqn.(3.92), we obtain

∇ · ∫ Ω Ω ψ dΩ � ∇ · ∫ Ω Ω ( φ + 3 Ω · J ) dΩ , for Ω · J = 0 ,

= 14π∇ · ( 1

∫Ω Ω dΩ) φ = 1

4π ∇ · (4π3 I) φ

= 13∇ · I φ = 1

3 ∇ φ

Hence, Eqn.(3.92) becomes

1v

∂tJ +

13∇ φ + ( Σt − Σs1 ) J = 0

Defining

Σa = Σt − Σs0 ( absorption cross section )

ΣtR = Σt − Σs1 ( transport cross section )(93)

we obtain the general− geometryP1equations:

1v

∂tφ + ∇ · J + Σa φ = Q , (94)

1v

∂tJ +

13∇ φ + Σtr J = 0 (95)

The approximation that lead to these equations is valid provided ψ is nearly isotropic (of the form of Eqn.()

with |J | � φ), absorption is low, and space and time derivatives are weak:

φ = 0(1)

| J | = 0(∈)

| ∇ | = 0(∈)

∂/∂t = 0(∈2)

Σa = 0(∈2)

Q = 0(∈2) all for ∈� 1

(96)

This scaling is such that each term in Eqn.(3.95) is 0(∈2). However, applying this scaling to Eqn.(3.96), we

find

Σtr J = 0(∈)13 ∇ φ = 0(∈)

1v

∂∂t J = 0(∈3)

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Page 81: Larsen Lecture Notes

Thus, we can delete ∂J/∂t in Eqn.() [note that this is not an approximation for steady-state problems] and

obtain

J = − 13 Σtr

∇ φ = −D ∇ φ (97)

where

D(r) =1

3 Σtr(r)= diffusioncoefficient (98)

Now, using Eqn.(3.98) to eliminate J from Eqns.(3.95) and (3.96), we obtain the time-dependent diffusion

equation

1v

∂tφ − ∇ ·D ∇ φ + Σa φ = Q (99)

[note that the second term on the left side of the above equation is approximate.]

and

ψ(r,Ω, t) � 14π

[ φ(r, t) − 1Σtr

Ω · ∇ φ(r, t) ] (100)

Note that the scaling (3.97) implies that each term in Eqn.(3.100) is O(∈2), and that the second term on

the right side of (3.101) is smaller than the first.

Eqn.(3.100) does not fully determine φ; initial and boundary conditions must also be derived. To obtain

the initial condition, we operate on Eqn.(3.88) by∫

(·)dΩ and obtain

φ(r, 0) =∫ψi (r,Ω, 0) dΩ (101)

Next, let us attempt to solve Eqn.(3.89) with the approximatio (3.101), i.e.,

ψb (r,Ω, t) =14π

[ φ(r, t) − 1Σtr

Ω · ∇ φ(r, t) ] r ∈ ∂D , Ω · n < 0 (102)

This boundary condition clearly cannot be satisfied, in general, for every direction Ω. Therefore, we shall

approximately satisfy it: we operate by

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Page 82: Larsen Lecture Notes

∫Ω·n<0

| Ω · n |(·) dΩ (103)

and obtain

∫Ω·n<0

| Ω · n | ψb(r,Ω, t) dΩ = J−(r, t)

= −∫

Ω·n<0

Ω · n 14π

( φ(r, t) − 1Σtr

Ω · ∇ φ(r, t) ) dΩ

= − 14π

∫Ω·n<0

Ω · n [ φ − 1Σtr

[ Ω − ( Ω · n ) n + ( Ω · n ) n ] · ∇ φ ] dΩ

= − 14π

∫Ω·n<0

Ω · n [ φ − 1Σtr

(Ω · n)(n · ∇) φ ] dΩ +1

4π Σtr[

∫Ω·n<0

Ω · n [ Ω − (Ω · n) n ] dΩ ]·∇ φ

for k = n and Ω · n by problem (3.18)

= − 14π

∫ 0

μ=−1

∫ 2π

φ=0

μ [ φ − 1Σtr

μ n · ∇ φ ] dφ dμ

= − 12

∫ 0

−1

[ μ φ − 1Σtr

μ2 n · ∇ φ ] dμ

= − 12

[ − 12φ − ‘

13 Σtr

n · ∇ φ ] =14

[ φ +2

3 Σtrn · ∇ φ ]

This yields the ’incident current’ or ’mixed boundary condition’

4 j−(r, t) = φ(r, t) +2

3 Σtr(r)n · ∇ φ(r) , r ∈ ∂D , n = unit outer normal (104)

The weight function |Ω · n| in Eqn.(3.104) is chosen for physical reasons (the incident partial current is

preserved). However, other weight functions can lead to slightly better results. For example, for ψb = 0, it

can be shown that ’the transport-corrected’ boundary condition

0 = φ(r, t) +0.7104Σtr

n · ∇ φ(r, t) r ∈ ∂D (105)

is more accurate than (3.105).

At a reflecting transport boundary, we can derive an exact diffusion boundary condition. Let us consider a

point r on the boundary of D at which

ψ(r,Ω) = ψ(r,Ωr) Ω · > 0 (106)

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Page 83: Larsen Lecture Notes

where Ωr is the reflection of Ω of the boundary:

n

Ω

Ω

Ω−(Ω.

Ω−(Ω

Ω.(

n)n

.n)n

n)n

r

Ωr = [ Ω − (Ω · n) n ] − (Ω · n) n = Ω − 2 (Ω · n) n

Ωr · n = − Ω · n (107)

Then, the transport current at this point satisfies

n · Jtr =∫n · Ω ψ dΩ

=∫n·Ω>0 [ n · Ω ψ(Ω) + n · Ωr ψ(Ωr) ] dΩ

=∫n·Ω>0

[ n · Ω ψ(Ω) − n · Ω ψ(Ω) ] dΩ = 0

Requiring the diffusion current to also satisfy this, we get, from (3.98), the following reflecting boundary

condition:

n · ∇ φ(r, t) = 0 r ∈ ∂D (108)

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Page 84: Larsen Lecture Notes

Finally, if the domain D has material discontinuities across an interface Γ:

then ψ is continuous across Γ, so [by Eqn.(3.87)] φtransport and J transport will be continuous:

material 1 material 2

rr

Γ

− +. .

φtransport (r+, t) = φtransport (r−, t)

J transport (r−, t) = J transport (r−, t)

We also require the diffusion scalar flux and current to be continuous across an interface.

Thus, using Eqn.(3.98), we obtain the interface conditions:

φ(r+, t) = φ(r−, t)

D(r+)∇ φ(r+, t) = D(r−) ∇ φ(r−, t)

Therefore, across an interface, φ will be continuous, but ∇ φ will be discontinuous:

Q = 0 Q = 0

Q > 0

Σ << 1

∼ ∼Σ ∼ 1Σ ∼ 1a

a

a

Eqns.(3.100), (3.102),(3.105),(3.109), and (3.110) now fully specify φ.

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We have already stated that ψb = 0, then the boundary condition (3.105) can be replaced by the more

accurate condition (3.106). These can be written as

0 = φ(r, t) + λ n · ∇ φ(r, t) (110)

where

λ =

⎧⎨⎩

2/(3 Σtr) standart diffusion theory

0.7104/Σtr transport-corrected diffusion theory

Using

φ(r + λ∇, t) � φ(r) + λ n · ∇ φ(r, t) + λ2

2 (n · ∇)2 φ(r) + . . . .

= λ2

2 (n · ∇)2 φ(r) = 0(∈2) [ see Eqn. (3.97) ] ,

we see that Eqn.(3.110) can be approximated by the ’extrapolated endpoint condition’:

0 = φ(r + λ∇, t) r ∈ ∂D , n = unit outer normal (111)

true boundaries

extrapolated boundaries

λ λ

φ>0 , φ = φ

φ>0 ,

φ = − φ <0 < 0ddx

ddx

n.Vn.V

The constant λ in Eqns.(3.110)-(3.112) is termed the extrapolation distance.

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Remarks: Because Eqn.(3.109) cannot be satisfied, the diffusion solution will be inaccurate at material

boundaries. However, a few mean free paths away from boundaries, it is often much more accurate.

transport solution

diffusion solution

vacuum boundary

dDD

z

Therefore, if a system is many mean free paths thick, the diffusion solution can be quite accurate in ’most’

of the system (i.e., away from boundaries and interfaces, where transport effects can dominate).

Derivation of One-Group Diffusion Theory by the Method of Successive Approximations:

To conclude this section on the one-group diffusion equation, we shall discuss the validity of the P1 method

for deriving this equation. The idea is to derive the diffusion equation by a different method that more

clearly demonstrates the validity of the assumptions used in the P1 method. We begin this derivation with

the slab geometry transport equation

μ∂ψ

∂z(z, μ) + Σt ψ(z, μ) =

12

[ Σs0∫ 1

−1

ψ(z, μ′) dμ

′+ Q(z) ] (112)

and we assume that the leakage term is small compared to the collision and scattering terms, i.e.,

μ∂

∂z� Sigmat or

μ

Σt∂

∂z� 1 (113)

Then, just using the inequality (3.114), we can derive the standart diffusion approximation

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Page 87: Larsen Lecture Notes

− d

dz

13Σt

d

dzφ(z) + Σa φ(z) = Q(z) (114)

to Eqn.(3.113) using the method of successive approximations. This derivation is of onterest because it

provides insight into the P1 method.

The successive approximations idea is simple: It says that if the leakage term is small, then it can be

approximated more crudely than the other terms, and in thr first approximation, we can ignore it altogether.

Doing this, Eqn.(3.113) reduces to

Σt ψ(z, μ) =12

[ Σs0∫ 1

−1

ψ(z, μ′) dμ

′+ Q(z) ] (115)

Thus, defining the scalar flux

φ(z) =∫ 1

−1

ψ(z, μ) dμ

we get from Eqn.(3.116)

Σt φ = Σs φ + Q ⇒ Σa φ = Q , (116)

and using this to eliminate Q from Eqn.(3.116), we obtain

Σt φ =12

[ Σs φ + Σa φ ] =12

[ Σt φ ]or

ψ(z, μ) = 12 φ(z)

(117)Eqns.(3.117) and (3.118) are the zero-th order (infinite medium) result.

To obtain a better approximation, we introduce the result (3.118) into the leakage term in Eqn.(3.113)

and obtain

μd

dz

φ

2+ Σt ψ =

12

[ Σs φ + Q ] (118)

Integrating over μ , (∫ 1

−1(·)dμ), we obtain

Σtφ = Σs φ + Q ⇒ Σa φ = Q (119)

and using this to eliminate Q in Eqn.(3.119), we get

μd

dz

φ

2+ Σt ψ =

12

[ Σs φ + Σa φ ] =12

Σt φ

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Page 88: Larsen Lecture Notes

Thus,

ψ(z, μ) =12

[ φ(z) − μ

Σtdφ

dz(z) ] (120)

This is the first order-result. We note that Eqns.(3.117) and (3.120) agree, but Eqn.(3.121) contains a (small)

correction term to (3.118).

To do netter, we use the result (3.121) in the leakage term in Eqn.(3.113) and get:

μ

2d

dz( φ − μ

Σtdφ

dz) + Σt ψ =

12

[ Σs φ + Q ] (121)

Integrating over μ, we obtain

− d

dz

13 Σt

d

dzφ(z) + Σa φ = Q , (122)

and using this result to eliminate Q from Eqn.(3.122), we get

μ

2d

dz(φ − μ

Σtdφ

dz) + Σt ψ =

12

[ Σs φ − d

dz

13 Σt

dz+ Σa φ =

12

[ Σt φ − d

dz

13Σt

dz]

or

ψ =12

[ φ − μ

Σtdφ

dx+

3μ2 − 13

(1Σt

d

dx) (

1Σt

d

dx) φ ] (123)

Eqn.(3.123) is the standart diffusion equation, and using

P0(μ) = 1 , P1(μ) = μ , P2(μ) = 3μ2−12

We see that Eqn.(3.124) is a Legendre polynomial expansion of the angular flux, with the coefficient of

P2 small compared to that of P1, and the coefficient of P1 small compared to that of P0. This justifies the

closure relation used in the P1 method; when space derivatives are weak, the P2 moment of the angular flux

is small, and it is consistent to set it to zero.

Finally, operating on Eqn.(3.124) by∫ 1

−1 μ (·) dμ ,

we obtain

J(z) =∫ 1

−1

μ′ψ(z, μ

′) dμ

′= − 1

3 Σtdφ

dz(z) ,

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Page 89: Larsen Lecture Notes

which is just Fick’s Law.

In summary, we have shown that in regions where space derivatives are weak; i.e., where scattering dom-

inates leakage, the assumptions underlying the P1 approximation are valid, and diffusion theory should be a

good approximation to transport theory.

The Energy-Dependent Diffusion Equation: Now let us consider the energy-dependent transport equa-

tion

1v

∂ψ

∂t+ Ω · ∇ ψ + Σt ψ

=14π

∫ ∫[ Σs0(E

′ → E) + 3 Ω′ · Ω Σs1(E

′ → E) ] ψ(r, E′,Ω

′, t) dΩ

′dE

′+

14π

Q(r, E, t) (124)

We wish to derive the P1 approximation for this more general equation. As before, we operate by∫

(·) dΩand

∫Ω (·) dΩ to obtain exactly

1v

∂φ

∂t+ ∇ · J + Σt φ =

∫[ Σs0(E

′ → E) φ(r, E′, t) dE

′+ Q(r, E

′, t) (125)

1v

∂J

∂t+ ∇ ·

∫Ω Ω ψ dΩ + Σt(E) J =

∫[ Σs1(E

′ → E) J(r, E′, t) dE

′(126)

To close these equations, we take

ψ(r, E′,Ω

′, t) � 1

4π[ φ(r, E, t) + 3 Ω · J(φ(r, E, t) ]

so that, as before,

∇ ·∫

Ω Ω ψ dΩ � 13∇ φ

and then we obtain the ’energy-dependent P1 equations’:

1v

∂φ

∂t(r, E, t) + ∇·J(r, E, t) + Σt φ(r, E, t) =

∫Σs0(E

′ → E) φ(r, E′, t) dE

′+ Q(r, E, t) , (127)

1v

∂J

∂t(r, E, t) +

13∇ φ(r, E, t) + Σt(E) J(r, E, t) =

∫Σs1(E

′ → E) J(r, E′, t) dE

′(128)

To proceed, approximations to terms (A, the first term in the left hand side of Eqn.(3.129)) and (B, the

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Page 90: Larsen Lecture Notes

right hand side of Eqn.(3.129)) are introduced:

(A): As in the one-group case, it is customary to set

1v

∂J

∂t= 0

( B): Here, one takes

Σs1(E′ → E) � Σs1(E

′) δ(E

′ → E) (129)

where

Σs1(E′) =

∫Σs1(E

′ → E) dE (130)

Then,

∫Σs1(E

′ → E) J(r, E′, t) dE

′ �∫

Σs1(E′) δ(E

′ → E) J(r, E′, t) dE

′= Σs1(E) J(r, E, t)

Although this is approximate,∫

(·) dE of this equation is exact because of Eqn.(3.131). Eqn.(3.129) now

becomes

13∇ φ + Σt J = Σs1 J

so, once again, we obtain Fick’s law

J(r, E, t) = −D(r, E) ∇ φ(r, E, t) (131)

D(r, E) =1

3 [ Σt ( D(r, E) ) − Σs1 ( D(r, E) ) ], (132)

and Eqn.(3.128) becomes the standart ’time and energy-dependent diffusion equation’:

1v

∂φ

∂t(r, E, t) − ∇ ·D(r, E) ∇ φ(r, E, t) + Σt(r, E) φ(r, E, t)

=∫

Σs0(r, E′ → E) φ(r, E

′, t) dE

′+ Q(r, E, t) (133)

Note that to derive this equation, we need to make an approximation [Eqn.(3.130)] that is not required for

the derivation of the one-group diffusion equation.

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Page 91: Larsen Lecture Notes

CHAPTER 4 : ONE-GROUP DIFFUSION THEORY (Chapter 5, Duderstadt and Hamilton)

Analytic One-Group Diffusion Theory

- Diffusion Coefficient

- Diffusion Length

- Jump Condition (across a delta source)

- Green’s Functions

Plane Source, Infinite Medium

Point Source, Infinite Medium

- Geometric Buckling

- Material Buckling

- k-Eigenvalue

- Reactivity

- Perturbation Theory

Inner Product

Adjoint Operator

Adjoint Boundary Conditions

Self-Adjoint Operator

Perturbation of k-Eigenvalue Problems

Numerical One-Group Diffusion Theory

- Fixed Source Problems

Spatial Mesh

Tridiagonal Matrix

Diagonally Dominant Matrix

Gaussian Elimination

- k-Eigenvalue (Criticality) Problems

Inverse Power Iteration Method

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Page 92: Larsen Lecture Notes

In this chapter we shall consider analytical and numerical solution methods for fixed source and eigen-

value problems in the one-group diffusion approximation derived in the previous chapter. First we shall work

out some fixed source problems that can be solved analtically.

Example 1: Plane source in an infinite homogeneous medium

Let us consider a delta-function source at x = x0, of strength S0:

− Dd2φ

dx2(x) + Σ φ(x) = S0 δ(x− x0)

or

φ′′(x) − 1

L2= − S0

Dδ(x− x0), (1)

where

D =1

3 Σtr= diffusion coefficient (2)

L =√D

Σa=

1√3 Σtr Σa

= diffusion length (3)

At x = x0, we want φ(x) to be continuous, but we shall allow φ′(x) to be discontinuous. Then, operating

on Eqn.(1) by∫ x0+∈x0−∈ (·) dx

we obtain

φ′(x0+ ∈) − φ

′(x0− ∈) − 1

L2

∫ x0+∈

x0−∈φ(x) dx = − S0

D

Now, letting ∈→ 0, we obtain the ’jump condition’

φ′(x0) − φ

′(x0) = − S0

D= 0 (4)

The diffusion problem can now be written

φ′′(x) − 1

L2= 0 x = x0 (5)

φ(x0) − φ(x0) = 0 (6)

φ′(x0) − φ

′(x0) = − S0

D(7)

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φ(±∞) = 0 (infinite medium) (8)

Solutions of Eqn.(5) are φ = e±x/L. Therefore, Eqns.(5),(6),and (8) imply

φ(x) =

⎧⎨⎩

A e−(x−x0)/L x > x0

A e−(x0−x)/L x < x0

Then

φ′(x) =

⎧⎨⎩

− AL e−(x−x0)/L x > x0

AL e−(x−x0)/L x < x0

so Eqn.(7) implies

(− AL ) − (AL ) = − S0

D or A = S0L2D

and

φ(x) =S0L

2De| x−x0 |/L (9)

xxo

φ

If we have a source which is a sum of plane delta functions:

S(x) =∑Ii=1 Si δ(x− xi) then φ(x) = L

2D

∑Ii=1 Si e

−| x−xi |/L

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Page 94: Larsen Lecture Notes

Likewise, if the source is distributed,

S(x) =∫ ∞−∞ S(x

′) δ(x− x

′) dx

then φ(x) = L2D

∫ ∞−∞ S(x′) e−| x−x′ |/L dx

′=

∫ ∞−∞ G(x, x

′) S(x

′) dx

where

G(x, x′) =

L

2De−| x−x′ |/L (10)

is the ’infinite medium Green’s function for a plane source’.

Physical Interpretation of the Jump Condition: Using

J(x) = − Ddφ

dx(x) i = current,

the jump condition (4) can be written

[ −D dφdx (x0 + 0) ] + [ −D dφ

dx (x0 − 0) ] = S0

or [ i · J(x0 + 0) ] + [ − i · J(x0 − 0) ] = S0

or J+(x0 + 0) + J−(x0 + 0) S0 (J± = partial currents)

−i i

x xxo

o o + εε−

Thus, the rate at which neutrons are produced by the source (S0) equals the rate at which neutrons leak

away from the point x0.

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Page 95: Larsen Lecture Notes

Example 2: Point source in a finite homogeneous slab: (See Chapter 4 problems.)

Example 3: Point source in an infinite medium:

− 1r2

d

drD r2

dr+ Σa φ = S0

δ(r)4πr2

(11)

φ(∞) = 0 (12)

In spherical geometry, dV = 4πr2dr. Therefore, for all R > 0,

∫ R

0

S0δ(r)4πr2

dV = S0

∫ R

0

δ(r) dr = S0

so we have a point source of strength S0. Operating on Eqn.(11) by∫ ε0

(·) dV , we get

− 4π D r2 dφdr |ε0 + 4π

∫ ε0

Σa φ r2 dr = S0

Letting ε → 0, we obtain

S0 = limr→0

(4πr2)[ − Ddφ

dr(r) ] = lim

r→0(ar)J+(r) (13)

This says that the rate (S0) at which neutrons are produced by the delta function source at r=0 equals the

rate at which neutrons flow out of the sphere centered at r=0 with infinitely small radius. The problem thus

can be rewritten as

1r2

d

drr2

d

drφ − 1

L2φ = 0 0 < r < ∞ (14)

limr→∞φ(r) = 0 (infinite medium) (15)

limr→0

r2dφ

dr= − S0

4πD(limr→0

4πr2J+ = S0) (16)

Solutions of Eqn.(15) are:

φ(r) =e±r/L

r⇒

⎧⎨⎩

B er/L

r , r > 0

A e−r/L

r , r < 0

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Page 96: Larsen Lecture Notes

To satisfy Eqn.(16), we must ignore the solution that grows as r → ∞ and take

φ(r) =e−r/L

r

Then,

r2 dφdr = r2 A [ − 1

r2 e−r/L − 1rL e−r/L ]

= − A (1 + rL) e−r/L → −A forr = 0

so −A = − S04πD

or

φ(r) =S0

4πDe−r/L

r(17)

In cartesian geometry, with a delta function source of unit strength at r = r0, we obtain from Eqn.(18)

φ(r) = e−|r−r0|/L4πD |r − r0| = G(r, r0) = infinite medium Green′s function for a point source

Remark 1: Let us consider one neutron emitted per second, S0 = 1. Then, the probability that the neutron

is absorbed in dr about r is

Σa φ(r) dV = Σa φ(r) 42 dr = Σa ( 14πD

e−r/L

r ) 4πr2 dr

= Σa

D r e−r/L dr = rL2 e

−r/L dr ( L =√

DΣa

)

Thus, the root mean square distance between emission and absorption, < r2 >1/2, satisfies

< r2 > =∫ ∞0r2 [ r

L2 e−r/L ] dr = L2

∫ ∞0

( rL)3 e−r/L d( rL )

for t = rL ,

= L2∫ ∞0t3 e−t dt = L2 Γ(4) = 6 L2

so

< r2 >1/2 =√

6 L � 2.45 L (18)

Remark 1: For the case of a unit source (S0 = 1) and a pure absorber we have

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Page 97: Larsen Lecture Notes

D = 13 Σtr

= 13 (Σt−Σs1)

= 13 Σt

Σa = Σt − Σs0 = Σ , Σtr = Σt − Σs1 = Σt − μ0 Σs

L =√

DΣa

=√

13 Σt

1Σt

= 1√3 Σt

and Eqn.(18) gives

φdiff (r) =1

4π (1/3Σt)e−

√3Σt r

r=

3 Σt4π

e−√

3Σt r

r(19)

The exact transport scalar flux, from Eqn.(3.36), is

φtransp(r) =e− Σt r

4πr2(20)

Therefore, in this problem, diffusion theory is quite inaccurate.

Example 4: Shell source in a finite sphere: (See Chapter 2 problems)

Example 5: Finite slab geometry with a plane source:

The problem schematically has the form

vacuumboundary

vacuumboundary

material 2 material 2material 1

0 a / 2 a / 2 + b− a / 2 − b − a / 2

plane sources ofstrength So

xx o o−

and the specific equations are:

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Page 98: Larsen Lecture Notes

d2φdx2 − 1

L2 φ = 0 0 < x < x0 , x0 < x < a/2d2φdx2 − 1

L2 φ = 0 a/2 < x < a/2 + b [ + 23 Σtr

]dφdx (0) = 0 reflecting boundary condition

φ(x0 + 0) − φ(x0 − 0) , φ′(x0 + 0) − φ

′(x0 − 0) = − S0

D1jump conditions

φ(a2 + 0) = φ(a2 − 0) , D1dφdx (a2 + 0) = D2

dφdx (a2 − 0) interface conditions

φ( a2 + b + 2

3 Σtr) = 0 extrapolated endpoint boundary condition

φ( a2 + b ) + 2

3 Σtr

dφdx ( a

2 + b ) zero incident current boundary condition

a/2 a/2+b0 xo

The solution of this problem has the form

φ(x) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

A1 e−x/L1 + B1 e

x/L1 0 < x < x0

A2 e−x/L1 + B2 e

−x/L1 x0 < x < a/2

A3 e−x/L2 + B3 e

−x/L2 a/2 < x < a/2 + b

Now, we have two boundary conditions, two jump conditions, and two interface conditions to detrmine the

six constants Ai, Bi. Thus, the diffusion problem has been reduced to a purely algebric problem of solving

a linear system of six equations and six unknowns. (We will not carry out the details here.)

Example 6: Distributed source in a finite homogenous medium:

Now, we wish to solve

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Page 99: Larsen Lecture Notes

∇2 φ(r) − 1L2

φ(r) = − Q(r)D

r ∈ D+ [extrapolated domain] (21)

φ(r) = 0 r ∈ ∂D+ [extrapolated endpoint boundary condition] (22)

To do this, we construct eigenfunctions ψn(r) and eigenvalues B2n satisfying, for n ≥ 1,

∇2 ψn(r) + B2n ψ(r) = 0 r ∈ ∂D+ (23)

ψn(r) = 0 r ∈ ∂D+ (24)∫D+

ψ2n d

3r = 1 (25)

It is now possible to show (but we will not do this here) that

Having ψnandBn, we expand

Q(r) =∞∑n=1

Qn ψn(r) , (26)

where, by Eqn.(26),

Qn =∫D+

ψn(r′) Q(r

′) d3r

′(27)

We also expand

φ(r) =∞∑n=1

φn ψn(r) (28)

where φn are to be determined. Note that by Eqn.(25), the ’guess’ (29) satisfies the boundary condition

(23). Thus, it remains to shoose the constants φn so that the differential equation (22) is satisfied.

Introducing Eqns.(27) and (28) into (22), we obtain

∑∞n=1 φn (∇2 − 1

L2 ) ψn(r) =∑∞n=1 (− Qn

D ) ψn(r)

or∑∞

n=1 φn (B2n + 1

L2 ) ψn(r) =∑∞n=1 (Qn

D ) ψn(r)

Therefore,

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Page 100: Larsen Lecture Notes

φn = Qn

D1

B2n + 1/L2 = Qn

Σa + D B2n

for DL2 = Σa

so φ(r) =∑∞

n=1Qn

Σa + D B2nψn(r) =

∑∞n=1 ( 1

Σa + D B2n) [

∫D+ ψn(r

′) Q(r

′) d3r

′] ψn(r)∫

D+ [∑∞

n=1ψn(r) ψn(r

′)

Σa + D B2n

] Q(r′) d3r

∫D+ G(r, r

′) Q(r

′) d3r

(29)

where

G(r, r′) =

∞∑n=1

ψn(r) ψn(r′)

Σa + D B2n

= finite medium Green′s function

The eigenvalues B2n and eigenfunctions ψn can be explicitly constructed for certain geometries. For example,

for a slab with a reflecting boundary at x=0 and a vacuum boundary at x=a/2, we have

0 = ψ”n(x) + B2

n ψn(x) 0 < x < a2 + λ

= ψ′n(0) = 0

= ψn(a2 + λ) = 0∫ a

2 + λ

0 ψ2n(x) dx = 1

Then,

ψn(x) = 2√a + 2 λ

cos Bn(x)

Bn = (2 n − 1) πa + 2 λ n = 1, 2, 3, ........

(30)

We note that B1 < B2 < B3 < ..........

One-Speed Diffusion Theory for a Nuclear Reactor:

Let us now consider the following transport problem for a spatial domain D:

1vdψdt + Ω · ∇ ψ + Σt(E) ψ = 1

∫Σs0 + 3 Ω · Ω′

Σs1 ψ(r,Ω′, t) dΩ

+ ν Σf

∫ψ(r,Ω

′, t) dΩ

′+ Q(r,t)

4π , r ∈ D ,

ψ(r,Ω, 0) = ψi(r,Ω) r ∈ D

ψ(r,Ω, t) = 0 r ∈ ∂D , Ω · n < 0

The diffusion approximation to this problem is

1vdφdt − D ∇2 φ + Σa φ = ν Σf φ + Q ( r ∈ D+ )

φ(r, 0) = φi(r) =∫ψi(r,Ω) dΩ , ( r ∈ D+ )

φ(r, t) = 0 ( r ∈ ∂D+ ) ,

(31)

where

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Page 101: Larsen Lecture Notes

D = 13 Σtr

= 13 ( Σt − Σs1 )

Σa = Σt − Σs0 , Σtr = Σt − Σs1

D+ = extrapolated distance

As before, we consider eigenfunctions ψn(x) and eigenvalues B2n satisfying

∇2 ψn(r) + B2n ψn(r) = 0 r ∈ D+

ψn(r) = 0 r ∈ ∂D+

∫D+ ψn(r

′) ψm(r

′) d3r

′= δnm

B1 < B2 < B3 < . . . . . .

(32)

To solve problem (32), we write

Q(r, t) =∑∞

n=1Qn(t) ψn(r)

( Qn(t) =∫ψn(r) Q(r, t) d3r = known )

(33)

φ(r, t) =∑∞

n=1 φn(t) ψn(r)

( φn(t) =∫ψn(r) φ(r, t) d3r = unknown )

(34)

We note that because each eigenfunction ψn satisfies the boundary condition of problem (32), then φ sat-

isfies this boundary condition also. It remains to make φ satisfy the diffusion equation (32) and the initial

condition of this equation. Thus, introducing (34) and (35) into (32), we find

∑∞n=1[

1v φ

′n(t) ψn(r) − D φn(t) ∇2 ψn(r) + Σa φn(t) ψn(r)

− ν Σf φn(t) ψn(r) − Qn(t) ψn(r) ] = 0

or∑∞

n=1[1v φ

′n(t) + ( D B2

n + Σa − ν Σf ) φn(t) − Qn(t) ] ψn(r) = 0

Hence,

dφndt

(t) + λn φn(t) = v Qn(t) , (35)

λn = v ( D B2n + Σa − ν Σf ) (36)

Solving Eqn.(36), we get

φn(t) = φn(0) e−λnt + v

∫ t

0

Qn(t) e−λn(t−t′) dt′

(37)

Now, Eqn.(32) is satisfied, and initial condition of Eqn.(32) is satisfied if

φi(r) =∑∞

n=1 φn(0) ψn(r)

so φn(0) =∫D+ ψn(r) φi(r) d3r =

∫D+

∫4π ψn(r) ψi(r,Ω) dΩ d3r

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Page 102: Larsen Lecture Notes

Hence, the solution of the diffusion problem (32) is:

φ(r, t) =∑∞

n=1[ φn(0) e−λnt + v∫ t0Qn(t) e−λn(t−t′ ) dt

′] ψn(r) ,

where λn = v ( D B2n + Σa − ν Σf ) ( λ1 < λ2 < . . . . )

φn(0) =∫D+

∫4πψn(r) ψi(r,Ω) dΩ d3r

Qn(t) =∫ψn(r) Q(r, t) d3r

In a source-free medium, we set Q = Qn = 0 and get

φ(r, t) = φ1(0) e−λ1t ψ1(r) +∑∞n=2[ φn(0) e−λnt ψn(r)

= e−λ1t [ φ1(0) ψ1(r) +∑∞n=2[ φn(0) e−(λn−λ1)t ψn(r)

� e−λ1t φ1(0) ψ1(r) for t 1

Thus, λ1 determines the long-time behaviour of the system.

Definitions:

1.Geometric Buckling = B2g = B2

1 = the smallest eigenvalue of the diffusion problem (33 ) (38)

Examples:

B2g =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

(πa )2 slab of width a, a = a+ 2λ (Eqn.(31))

(πa )2 + (πb)2 rectangle of width a, length b

(πa )2 + (πb)2 + (πc )

2 rectangular parallel piped of width a, length b, height c

2.Material Buckling = B2m = νΣf−Σa

D

3. A reactor is critical when a steady-state neutron flux can be sustained in the absence of sources, is

subcritical when the neutron flux tends to zero as t → ∞, and is supercritical when the neutron flux → ∞as t→ ∞. Because

λ1 = v ( D B21 + Σa − ν Σf ) = v D ( B2

1 − νΣf − ΣaD

) = v D ( B2g − B2

m )

then

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Page 103: Larsen Lecture Notes

supercritical iff λ1 < 0 iff B2g < B2

m

critical iff λ1 = 0 iff B2g = B2

m

subcritical iff λ1 > 0 iff B2g > B2

m

(39)

Criticality (k-Eigenvalues): Let us consider the transport k-eigenvalue problem

Ω · ∇ ψ + Σt(E) ψ − 14π

∫Σs0 + 3 Ω · Ω′

Σs1 ψ dΩ′

= 1kν Σf

∫ψ dΩ

′r ∈ D

ψ(r,Ω, 0) = 0 r ∈ ∂D , Ω · n < 0

We seek the largest eigenvalue k for which a nonzero solution exists. Clearly,

supercritical iff k > 1

critical iff k = 1

subcritical iff k < 1

(40)

The diffusion approximation to this (homogenous medium) problem is:

− D ∇2 φ(r) + Σa φ(r) =1kν Σf φ(r) r ∈ D+ (41)

φ(r) = 0 ( r ∈ ∂D+ ) (42)

To solve this problem, let us again consider ψn(r), Bn satisfying (33). Then, ’guessing’ φ = ψn, Eqn.(43) is

satisfied and Eqn.(42) becomes:

− D ∇2 ψn + Σa ψn = 1kn

ν Σf ψn

or D B2n ψn + Σa ψn = 1

knν Σf ψn

therefore, kn = ν ΣfΣa + D B2

n

The largest kn occurs for the smallest Bn = B1 = Bg. Therefore, we obtain the explicit formula

kn =ν Σf

Σa + D B2g

=Σa + D B2

m

Σa + D B2g

(43)

We note that Eqns.(40) and (41) are consistent, because of Eqn.(44).

Remark: We can now make a connction between the largest k-eigenvalue of the diffusion problem (42),(43)

[or of the proceeding transport problem] and the multiplication factor k defined in Chapter 2. Using the

notation in Eqns.(42) and (43), let us suppose that the n-th fission generation is φ(n)(r) and let us define

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Page 104: Larsen Lecture Notes

the (n+1)-st fission generation to be the solution of the problem

− D ∇2 φ(n+1)(r) + Σa φ(n+1)(r) = ν Σf φ(n)(r) r ∈ D+

φ(n+1)(r) = 0 ( r ∈ ∂D+ )(44)

Then, φ(n+1) represents all the neutrons in the system that result from the fission source νΣfφ(n). According

to the definition on page 2-1, we should take

MF = multiplication factor =

∫D+ φ(n+1)(r) d3r∫D+ φ(n)(r) d3r

This definition, however, depends on the spatial variation of φ(n)(r); as we continuously change φ(n)(r), and

we would expect to continuously change MF. Therefore, let us modify the above definition as follows:

MF = largest possible ratio

∫D+ φ(n+1)(r) d3r∫D+ φ(n)(r) d3r

(45)

Now, how do we determine this? Using some higher-level mathematics that we cannot go into here, one

can show that it suffices to look only at fission-generations that reproduce their spatial shape, i.e.,that satisfy

φ(n+1)(r) = k φ(n)(r)

Then, Eqns.(45) and (46) become:

k [ − D ∇2 φ(n)(r) + Σa φ(n)(r) ] = ν Σf φ(n)(r) r ∈ D+

φ(n+1)(r) = 0 ( r ∈ ∂D+ )(46)

multiplication factor = largest k (47)

Now, because the problems (47) and (45) are identical, we have shown that the largest eigenvalue of either

of these problems is, in fact, the multiplication factor.

Definition:

reactivity = ρ =k − 1k

= 1 − 1k

(48)

Then,

supercritical iff ρ > 0

critical iff ρ = 0

subcritical iff ρ < 0

Perturbation Theory: One can obtain exact ’pencil-and-paper’ solutions of one-group diffusion problems

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Page 105: Larsen Lecture Notes

only for fairly special problems, typically having good spatial symmetries and (at worst) only a few spatial

regions. However, using perturbation theory, we can obtain approximate ’pencil-and-paper’ solutions of

problems that are nearly, but not exactly, of this type. The idea is to first obtain the exact solution of the

’nice’ problem that is close to the desired problem, and then to generate approximate correction terms. To

do this for eigenvalue problems, we must first intriduce the concepts of inner product, adjoint operator, and

adjoint boundary conditions.

Let L be a linear operator and let B(L) be the set of boundary conditions for L. So, let (,) denote an inner

product. Then tha adjoint operator L∗, together with the adjoint boundary conditions B(L∗), is defined to

satisfy

(g, Lf) = (L∗g, f)

Example: Let L be the ordinary differential operator

Lf(x) = − ddxD(x) d

dxf(x) + u(x) ddxf(x) + Σ(x)f(x) 0 < x < a (slab geometry)

Boundary Conditions

B(L): f(0) = 0, f(a) = 0

the inner product

(f, g) =∫ a0 f(x)g(x)dx

We will now construct L∗ and B(L∗). To do this we let f(x) be any function satisfying

B(L) : f(0) = f(a) = 0

(g, Lf) =∫ a0g(x) [ − d

dxD(x) dfdx + u(x) df

dx + Σ(x) f(x) ] dx

=∫ a0g ddxD

dfdx dx +

∫ a0g u df

dx dx +∫ a0g Σ f dx

= − [ g (D dfdx) |a0 − ∫ a

0dgdx D

dfdx dx ] + [ g u f |a0 − ∫ a

0d gudx f dx ] + [

∫ a0 (Σ g) f dx ]

= − g D dfdx |a0 + (D dg

dx) f |a0 − ∫ a0

( ddx D

dgdx ) f dx − ∫ a

0( ddxg u ) f dx +

∫ a0

(Σ g) f dx

= − [ g(a) D(a) f′(a) − g(0) D(0) f

′(0) ] +

∫ a0 [ − d

dx Ddgdx − d

dxu g + Σ g ] f dx

= (L∗g, f) ⇒ ( g , Lf ) = ( L∗g , f )

provided we define L∗ and B(L∗) as

L∗g(x) = − ddx D(x) d

dx g(x) − ddx u(x) g(x) + Σ(x) g(x)

B(L∗) : g(0) = 0 , g(a) = 0 ( Boundary condition )

We note that if u(x)=0, then L = L∗andB(L) = B(L∗). In this case, the operator L is self-adjoint.

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Page 106: Larsen Lecture Notes

L = L∗ ⇒ L is self − adjoint

L = [− ddx D

ddx + (ν Σf − Σa) ] = one speed diffusion equation

Theorem: Let L be self-adjoint, and let there exists a nonzero function f such that

if Lf = 0 (49)

then, the problem

Lg = Q (50)

has a solution if and only if (f,Q) = 0

Proof (one way only): Suppose Eqn.(51) has a solution. Then,

(f,Q) = (f, Lg) = (L∗f, g) = (Lf, g) = (0, g) = 0 for Lf = 0

Application: Let us consider the following ’unperturbed’ eigenvalue problem:

L φ0 = − ddx D

ddxφ0 + ( a − ν Σf

k0) φ0 = 0 0 < x < a for slab plane

φ0 = φ0(a) = 0 ,

which is solved by φ0, k0. We wish to compute the change in reactivity if νΣf is perturbed by a small

amount, i.e.,if

ν Σf (x) → ν Σf (x) + Δ ν Σf (x) ,

whereΔ ν Σf

ν Σf� 1

To solve this problem, we set

φ = φ0 + Δφ1k0

= 1 − ρ0

1k = 1 − ρ = 1 − ( ρ0 + Δρ0 ) = 1

k0− Δρ

Then, we want

− ddx D

ddxφ + ( Σa − ν Σf

k ) φ = 0

or 0 = − ddx D

ddx(φ0 + Δφ) + [Σa − ( 1

k0− Δρ) (ν Σf + Δ ν Σf )] (φ0 + Δφ)

= − ddx D

ddxφ + [ Σa − ν Σf

k0+ (Δρ)(ν Σf ) − 1

k0(Δ ν Σf ) + (Δρ)(Δ ν Σf ) ] (φ0 + Δφ)

=

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Page 107: Larsen Lecture Notes

Thus, we have

Lφ0 = 0 (L is self − adjoint)

and we want

L Δφ = ( Δ ν Σf

k0− (Δρ) ν Σf ) φ0

( Δ ν Σf

k0− (Δρ) ν Σf ) = Q

By the theorem, the existence of Δφ follows only if

0 = ( φ0, [ ΔνΣf

k0− (Δρ)νΣf ] φ0 )

=∫ a0φ0(x) [ ΔνΣf

k0− (Δρ)νΣf ] φ0(x) dx

Hence, we get the desired result:

Δρ =1k0

∫ a0 φ

20 ΔνΣf (x) dx∫ a

0 φ20 νΣf (x) dx

Remark: The general one-group diffusion operator

L φ(r) = − ∇ ·D(r) ∇ φ(r) + Σ(r) φ(r) r ∈ D

with boundary conditions

B(L) : φ(r) + b(r) n · ∇ φ(r) = 0 r ∈ ∂D

where b(r) is arbitrary, L is self-adjoint. Therefore, this type of theory can be applied in very general ways

in one-group diffusion calculations.

Numerical One-Group Diffusion Theory: We have seen that the use of ’analytical’ or ’pencil-and-paper’

methods for solving one-group diffusion problem is applicable only to a limited class of geometrically simple

(or, nearly simple) problems. For ’realistic’ problems that have complicated geometrical features, one must

obtain numerical solutions. Wewill now discuss how to do this in slab geometry.

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Page 108: Larsen Lecture Notes

To begin, let us consider the ’fixed-source’ diffusion problem:

L = − d

dxdd

dxφ(x) + Σa(x) φ(x) = Q(x) = Q0(x) + Q1(x) (51)

dx(0) = 0 (reflecting) (52)

φ(a) + 2 D(a)dφ

dx(a) = 0 (zero incident current) (53)

We impose a spatial mesh on the physical system:

x x x x x xΔ Δ Δ

Σ

1i I

I−1/2 I+1/2i+1/2i−1/23/21/2

ai

Q

D

=a0=

i

oi

where any material discontinuities coincide with the cell edges, xi+1/2. Thus, the cross-sections and the

source Q0i are assumed constant within each cell, as indicated above. We define fluxes φi+1/2 at cell edges,

and we assume that φ and Q1 are constant on each ith ’fivtitious cell’ (’centered’ at xi+1/2).

Δi = xi+1/2 − xi−1/2

Δi+1 = xi+3/2 − xi+1/2

xi = 12 (xi+1/2 + xi−1/2) = xi+1/2 − 1

2Δi

xi = 12 (xi+3/2 + xi+1/2) = xi+1/2 + 1

2Δi+1

Δi−1/2 = xi+1 − xi = 12 (Δi+1 − Δi) 1 ≤ i ≤ I − 1

Δ1/2 = 12 Δ1

ΔI+1/2 = 12 ΔI

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Page 109: Larsen Lecture Notes

Q

D

Q

D

Q

D

Q

DΣ Σ Σ Σ

Δ Δ ΔΔ

Δ Δ Δ

1

1 a1

o1

i ai

oi

i+1a

i+1

o,i+1

I aI

oI

i i+1 I

1/2 i+1/2

x x x x x x x

x x x x x x x1/2 3/2 i−1/2 i+1/2 i+3/2 I I+1/2

1/2 1 2 i i+1 I I+1/2

−1/2

I+1/2

Q Q Q

Φ Φ Φ

=a

=a

0=

0=

1,1/2

1/2 i+1/2

1,i+1/2

I+1/2

1,I+1/2

Now we intwgrate Eqn.(52) over xi < x < xi+1: (the i-th fictitious cell)

− [ D(xi+1)dφ

dx(xi+1) − D(xi)

dx(xi) ] +

∫ xi+1

xi

Σa(x) φ(x) dx =∫ xi+1

xi

[ Q1(x) + Q0(x) ] dx (54)

However, consistent with the above assumptions, we have at an ’interior’ (1 ≤ i ≤ I − 1) cell,

D(xi+1) dφdx (xi+1) = Di+1

φi+3/2 − φi+1/2

Δi+1,

D(xi) dφdx (xi) = Di

φi+1/2 − φi−1/2

Δi,∫ xi+1

xiΣa(x) φ(x) dx =

∫ xi+1

xiΣa(x) dx φi+1/2

= ( 12 Δi Σai + 1

2 Δi+1 Σa,i+1 ) φi+1/2

= (Δi Σai + Δi+1 Σa,i+1Δi + Δi+1

) (Δi + Δi+12 ) φi+1/2

= Σa,i+1/2 Δi+1/2 φi+1/2 ,∫ xi+1

xiQ0(x) dx = 1

2 Δi Q0i + 12 Δi+1 Q0,i+1

(Δi Q0i + Δi+1 Q0,i+1Δi + Δi+1

) (Δi + Δi+12 )

Q0,i+1/2 Δi+1/2 ,∫ xi+1

xiQ1(x) dx = Q1,i+1/2 Δi+1/2

Introducing these approximations into Eqn.(55), we obtain

− Di+1

Δi+1(φi+3/2 − φi+1/2) +

Di

Δi(φi+1/2 − φi−1/2) + Σa,i+1/2 Δi+1/2 φi+1/2 = (Q0,i+1/2 +Q1,i+1/2) Δi+1/2

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Page 110: Larsen Lecture Notes

or

− [Di

Δi] φi−1/2 + [

Di

Δi+Di+1

Δi+1+ Σa,i+1/2 Δi+1/2] φi+1/2 − [

Di+1

Δi+1] φi+3/2 = (Q0,i+1/2 +Q1,i+1/2) Δi+1/2 , 1 ≤ i ≤ I−1

(55)

Now we must consider the boundary cells and use the assigned boundary conditions. First, we integrate

Eqn.(52) over the leftmost cell 0 < x < x1 = Δ1/2 and use the boundary condition (53):

− D(x1)dφ

dx(x1) +

∫ x1

0

Σa(x) φ(x) dx =∫ x1

0

(Q0(x) + Q1(x)) dx

Consistent with the earlier discretization, we take:

- D1φ3/2−φ1/2

Δ1+ Σa1 φ1/2 Δ1/2 = (Q0,1 + Q1,1/2) Δ1/2

or

[D1

Δ1+ Σa1 Δ1/2] φ1/2 − [

D1

Δ1] φ3/2 = (Q0,1 + Q1,1/2) Δ1/2 (56)

Finally, we integrate Eqn.(52) over the rightmost cell xI < x < xI+1/2 and obtain

− D(a)dφ

dx(a) + D(xI)

dx(xI) +

∫ xI+1/2

xI

Σa(x) φ(x) dx =∫ xI+1/2

I

[Q0(x) + Q1(x)] dx (57)

However, the boundary condition (54) implies:

− D(a)dφ

dx(a) = − 1

2φ(a) (58)

Therefore, Eqn.(58) becomes

12φ(xI+1/2) + D(xI)

dx(xI) +

∫ xI+1/2

xI

Σa(x) φ(x) dx =∫ xI+1/2

I

[Q0(x) + Q1(x)] dx

Consistent with the earlier discretization, we take:

12φI+1/2 + DI

φI+1/2 − φI−1/2

ΔI+ ΣaI φI+1/2 ΔI+1/2 = (Q0I + Q1,I+1/2) ΔI+1/2

or

[12

+DI

ΔI+ ΣaI ΔI+1/2] φI+1/2 − [

DI

ΔI] φI−1/2 = (Q0I + Q1,I+1/2) ΔI+1/2 (59)

Eqns.(56),(57), and (60) can now be written in the following tridiagonal form: We define

Φi = φi−1/2 1 ≤ i ≤ I + 1

Bi = Ai + Ci + Σa,i−1/2 Δi−1/2 1 ≤ i ≤ I + 1(60)

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Page 111: Larsen Lecture Notes

Then,

(57) : B1 Φ1 − C1 Φ2 = S1

(56) : −Ai Φi− 1 + Bi Φi − Ci Φi+1 = Si 2 ≤ i ≤ I

(60) : −AI+1 ΦI + BI+1 ΦI+1 = SI+1

If we write this system in matrix notation, we find

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

B1 −C1 0 0 . . . . . .

−A2 B2 −C2 0 . . . . . .

0 −A3 B3 C3 0 . . ....

. . . . . . . . . . . .

. . . . . . . . . −AI BI CI

. . . . . . . . . . . . −AI+1 BI+1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

Φ1

Φ2

. . .

. . .

. . .

ΦI+1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

S1

S2

. . .

. . .

. . .

SI+1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

This system is called tridiagonal, because inly the main diagonal and the two diagonals immediately above

and below the main diagonal have non-zero elements. Also, this matrix is diagonally dominant because each

diagonal element is greater then the sums of the absolute values of the off-diagonal elements on the same row:

Bi ≥ |Ai| + |Ci|

For these reasons, we can use Gaussian elimination to explicitly solve this system. To describe the logic

behind this method, let us consider a small 3x3 system:

B1 Φ1 − C1 Φ2 = S1 (61)

−A2 Φ1 + B2 Φ2 − C2 Φ3 = S2 (62)

−A3 Φ2 + B3 Φ3 = S3 (63)

The Gaussian elimination method has two parts. In part 1, this system is reduced to an ’upper triangle’

system. Then, in part 2, the upper triangle system is explicitly solved.

Part 1:

1. Define

u1 = B1 , v1 = S1 (64)

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Page 112: Larsen Lecture Notes

and write Eqn.(62) as

u1 Φ1 − C1 Φ2 = v1 (65)

2. Multiply by A2/u1:

A2 Φ1 − A2 C1

u1Φ2 =

A2

u1v1

Add this to Eqn.(63):

(B2 − A2 C1

u1) Φ2 − C2 Φ3 = S2 +

A2

u1v1

Define

u2 = B2 − A2 C1

u1, v2 = S2 +

A2

u1v1 (66)

and then

u2 Φ2 − C2 Φ3 = v2 (67)

3. Multiply (68) by A3/u2:

A3 Φ2 − A3 C2

u2Φ3 =

A3

u2v2

Add this to (64):

(B3 − A3 C2

u2) Φ3 = S3 +

A3

u2v2

Define

u3 = B3 − A3 C2

u2, v3 = S3 +

A3

u2v2 (68)

and then

u3 Φ3 = v3 (69)

This system (62)-(64) has now been converted to the upper triangle system:

u1 Φ1 − C1 Φ2 = v1 (70)

u2 Φ2 − C2 Φ3 = v2 (71)

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u3 Φ3 = v3 (72)

Part 2:

4. Eqn.(66) → Φ3 = v3 / u3 ,

5. Eqn.(68) → Φ2 = (v2 + C2 Φ3) / u2 ,

6. Eqn.(70) → Φ1 = (v1 + C1 Φ2) / u1

This describes in detail the Gaussian elimination procedure for the simple 3x3 system. The algorithm

is simple and mechanical and can be applied to tridiagonal diagonally dominant system of any size. The

general Gaussian elimination procedure for such a system of I+1 equations

B1 Φ1 − C1 Φ2 = S1

−A2 Φ1 + B2 Φ2 − C2 Φ3 = S2

−Ai Φi−1 + Bi Φi − Ci Φi+1 = Si

−AI ΦI−1 + BI ΦI − CI ΦI+1 = SI

−AI+1 ΦI + BI+1 ΦI+1 = SI+1

(73)

is as follows:

1. Define

u1 = B1 , v1 = S1

and then, for i=2,3,....,I+1, define recursively

ui = Bi − Ai Ci−1

ui−1, vi = Si +

Ai Vi−1

ui−1

2. Now, take

ΦI+1 =1

uI+1vI+1

and then, for i=I, I-1, ....., 2,1, compute recursively

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Page 114: Larsen Lecture Notes

ΦI =1ui

(vi + Ci Φi+1)

This procedure explicitly solves Eqn.(71). Thus, we can explicitly obtain a numerical solution

Φi+1/2 = Φi+1 0 ≤ i ≤ I

of the fixed-source diffusion problem.

Remark: If the incident partial current on the right boundary is

I−(a) =14

[φ(a) + 2 D(a)dφ

dx(a)] = 0 [byEqn.(54)]

then the exiting partial current is

J+(a) =14

[φ(a) − 2 D(a)dφ

dx(a)] =

14

[φ(a) + φ(a)] =12φ(a) =

12φI+1/2

and the total current is:

J(a) = i · j(a) = J+(a) − J−(a) =12φI+1/2

Eigenvalue (Criticality) Calculations: In the above, we have considered the numerical solution of fixed source

problems. Let us now consider obtaining numerical solutions of the k-eigenvalue problem

− d

dxD(x)

d

dxφ(x) + Σa(x) φ(x) =

1kν Σf (x) φ(x) 0 < x < a (74)

dx(0) = 0 (75)

φ(a) + 2 D(a)dφ

dx(a) = 0 (76)

Note: The right hand side of equation (72) is discretized exactly like the absorption term Σaφ(x), or, like the

source Q1(x) in our treatment of fixed-source problems. We note that if φ(x) is a solution (an eigenfunction)

of this problem, then Aφ(x) is also a solution, for any constant A. Therefore, with no loss of generality, we

require φ(x) to satisfy the normalization condition

1a

∫ a

0

ν Σf φ(x) dx = 1 (77)

24

Page 115: Larsen Lecture Notes

xa

reflectorreflector

core

zero incident current bcreflecting bc

The problem (72)-(74) is equivalent to the fixed source problem (52)-(54) if we set

Q(x) =1kν Σf φ(x)

However, we do not know k or φ, so we do not know Q. Nevertheless, we can determine Q by oterating.

Inverse Power Iteration Method: We construct a sequence of functions

φ0(x), φ1(x), φ2(x), . . . .

and constants k0, k1, k2, . . . .

that have the following properties:

1a

∫ a

0

ν Σf (x′) φn(x

′) dx

′= 1 n ≥ 0 (78)

φn(x) → φ(x) for n → ∞kn → k for n → ∞

We (typically) begin with the guesses

φ0(x) = [ 1a∫ a0 ν Σf (x

′) dx

′]−1 = constant

k0 = 1

Then, for any n ≥ 0, havinh an estimate φn(x) for the eigenfunction that satisfies Eqn.(76), and an

estimate kn for the eigenvalue, we generate improved estimates φn+1(x) and kn+1 as follows:

25

Page 116: Larsen Lecture Notes

First, we solve the fixed source problem

− d

dxD(x)

d

dxφn+1/2(x) + Σa(x) φn+1/2(x) =

1kn

ν Σf (x) φn(x) 0 < x < a (79)

dφn+1/2

dx(0) = 0

φn+1/2(a) + 2 D(a)dφn+1/2

dx(a) = 0

Integrating Eqn.(77) over 0 < x < a and then rearranging, we obtain

kn =

∫ a0 ν Σf (x

′) φn(x

′) dx

∫ a0 ( − d

dxD(x′ ) ddxφ

n+1/2(x′) + Σa(x′ ) φn+1/2(x′ )) dx′

Now, using the most recent estimate of the scalar flux in this formula, we define

kn+1 =

∫ a0ν Σf (x

′) φn+1/2(x

′) dx

∫ a0 ( − d

dxD(x′) ddxφ

n+1/2(x′) + Σa(x′ ) φn+1/2(x′)) dx′

Dividing these last two equations, we obtain the simplier result

kn+1 = kn (1a

∫ a

0

ν Σf (x′) φn+1/2(x

′) dx

′) (80)

Finally, we renormalize φn+1/2(x′) by taking

φn+1(x′) =

φn+1/2(x′)

1a

∫ a0 ν Σf (x

′) φn+1/2(x′) dx′ (81)

so that Eqn.(76) is satisfied. The inverse power iteration method is now fully specified by Eqns.(77)-(79).

We stop iterating when k(n) and φ(n) are sufficiently converged, i.e., when

|k(n+1) − k(n)| < ε

and

max |φ(n+1)(x) − φ(n)(x)| < δ , 0 ≤ x ≤ a

Here ε and δ are preassigned numbers, typically ε = 10−5 and δ = 10−4.

Plane source in infinite medium: One of the most basic problem in neutron diffusion theory is the cal-

culation of the neutron flux from a plane source in an infinite medium.

The equation can be written as

Dd2φ

dx2− Σa φ = 0

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Page 117: Larsen Lecture Notes

with the boundary condition φ → 0 as x → ∞ and the source condition

12S0 = − D

dx, x > 0

where S0 is the source strength.

Now we set κ2 = Σa/D, we get the diffusion equation as

φ′′

= κ2 φ = 0

The general solution is

φ(x) = A e−κx + B eκx

Now B=0 to prevent φ → ∞ at x → ∞. To get A, we use the source condition. A and then φ are

determined as

A =S0

2 κ D, φ =

S0

2 κ De−κx

If we set L=1/κ , this is called the diffusion length. L is a measure of the distance they travel before

they are absorbed.

medium L (cm)

H2O 2.88

D2O 100

Be 23.6

C 50

The mean distance of travel from the source is

x =∫ ∞

−∞x φ(x) dx /

∫ ∞

−∞φ(x) dx = 0

The mean square distance of travel is

x2 =∫ ∞

−∞x2 φ(x) dx /

∫ ∞

−∞φ(x) dx = 2 L2

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CHAPTER 5 : MULTIGROUP DIFFUSION THEORY (Chapter 7, Duderstadt and Hamil-

ton)

- Group Structure

Logarithmic

Directly Coupled

- g-th Energy Group

- Scalar Flux for the g-th Energy Group

- Spectral Function

- Multigroup Diffusion Approximation

- Group Collapse

- Removal Cross Section

- Fixed Source Problem

Power Iteration

Fission Source

Inner Iteration

Outer Iteration

- k-Eigenvalue Problems

Inverse Power Iterations

- One-Group Diffusion Approximation

- Two-Group Diffusion Approximation

- One and One-Half Group Approximation

1

Page 119: Larsen Lecture Notes

We begin this chapter with the energy-dependent diffusion equation derived earlier in chapter-3:

1v

dt−∇·D∇φ+ Σtφ(r, E, t) =

∫Σs(E

′ → E)φ(r, E′, t)dE

′+ χ(E)

∫νΣf (E

′)phi(r, E

′, t)dE

′+Q(r, E, t)

(1)

which we assume to hold for points r in a spatial domain D, t > 0, and Emin < E < Emax.

The initial and boundary conditions to go with this equation are

φ(r, E, t) = φi(r, E) (2)

φ(r, E, t) + 2D(r, E) n · ∇ φ(r, E, t) = 0 r ∈ ∂D (3)

We shall now discretize the energy variable using the multigroup approximation. [This derivation will be

similar to the derivation of the multigroup transport equations outlined in chapter-3.]

To begin, we introduce a group structure

Emin = EG < ........ < Eg < Eg−1 < ....... < E0 = Emax

with the g-th energy group defined by

Eg < E < Eg−1

Often the group structure is chosen to be logarithmic, i.e., Eg−1/Eg = r > 1 (all g)

Then,

(E0E1

)(E1E2

)...........(EG−2EG−1

)(EG−1EG

) = rG = ( E0EG

)

so r = ( E0EG

)1/G

and hence we obtain the explicit formula

Eg = 1r Eg−1 = 1

r2 Eg−2 = ......... = 1rg E0 = (EG

E0)g/G E0

or

Eg = Eg/GG E

1−g/G0 , 0 ≤ g ≤ G (4)

Eqn.(1) can now be written

1v

dt−∇·D∇φ + Σtφ =

G∑g′=1

∫ Eg′−1

Eg′

Σs(E′ → E) φ dE

′+ χ(E)

G∑g′=1

∫ Eg′−1

Eg′

νΣf (E′) phi dE

′+Q (5)

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Page 120: Larsen Lecture Notes

We define the scalar flux for the g-th energy group

φg(r, t) =∫ Eg−1

Eg

phi(r, E′, t)dE

and integrate Eqn.(5) over Eg < E < Eg−1, obtaining

∂∂t

∫ Eg−1

Eg

1v φ dE − ∇ · ∫ Eg−1

EgD ∇ φ dE +

∫ Eg−1

EgΣt φ dE

=∑G

g′=1

∫ Eg′−1

Eg′ [

∫ Eg−1

EgΣs(E

′ → E) dE] φ(r, E′, t) dE

+ [∫ Eg−1

Egχ dE]

∑Gg′=1

∫ Eg′−1

Eg′ νΣf (E

′) φ dE

′+

∫ Eg−1

EgQ(r, E, t) dE

(6)

Now, we use the approximation

φ(r, E, t) � ψ(r, t) ϕ(E)

where ϕ(E) is a precomputed spectral function, to get, for each term in Eqn.(6),

∫ Eg−1

Eg

1v φ dE = [

∫ Eg−1

Eg

1v φ dE /

∫ Eg−1

Egφ dE] φg(r, t)

= [∫ Eg−1

Eg

1v ϕ dE /

∫ Eg−1

Egϕ dE] φg(r, t) = 1

vgφg(r, t)

(7)

∫ Eg−1

EgD ∇ φ dE � ∫ Eg−1

EgD ∇ ϕ ψ dE

= [∫ Eg−1

EgD ϕ dE] ∇ ψ(

∫ Eg−1

EgD ϕ dE /

∫ Eg−1

Egϕ dE) ∇ ∫ Eg−1

Egϕ ψ dE

� (∫ Eg−1

EgD ϕ dE /

∫ Eg−1

Egϕ dE) ∇ ∫ Eg−1

Egφ dE

= Dg(r) ∇ φg(r, t) ,

(8)

∫ Eg−1

EgΣt φ dE = (

∫ Eg−1

EgΣt φ dE /

∫ Eg−1

Egφ dE) φg

= (∫ Eg−1

EgΣt ϕ dE /

∫ Eg−1

Egϕ dE) φg

= Σtg(r) φg(r, t) ,

(9)

∫ Eg′−1

Eg′

∫ Eg−1

EgΣs(E

′ → E) dE φ(r, E′t) dE

′(∫ ∫

Σs(E′ → E) φ dE dE

′/

∫ g′−1

g′ φ dE′) φg′

� (∫ g′−1

g′∫ g−1

gΣs(E

′ → E) ϕ dE dE′/

∫ g′−1

g′ ϕ dE′) φg′

= Σs;g′ ,g(r) φg′ (r, t) ,

(10)

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Page 121: Larsen Lecture Notes

∫ Eg−1

Eg

χ dE = χg , (11)

( Note that∫ E0

EG

χ(E) dE = 1 which implies

G∑g=1

= 1 ) (12)

∫ Eg′−1

Eg′ νΣf (E

′) φ dE

′= (

∫ g′−1

g′ νΣf φ dE′/

∫ g′−1

g′ φ dE′) φg′

(∫ g′−1

g′ νΣf ϕ dE′/

∫ g′−1

g′ ϕ dE′) φg′

= (νΣf )g′ φg′ (r, t) ,

(13)

∫ Eg−1

Eg

Q(r, E, t) dE = Qg(r, t) (14)

Combining Eqns.(6)-(14), we obtain the following system of multigroup diffusion equations:

1vg

∂tφg − ∇·Dg ∇φg + Σtg φg(r, t) =

G∑g′=1

Σsg′g φg′ + χg

G∑g′=1

(νΣf )g′ φg′ + Qg , 1 ≤ g ≤ G (15)

Summary:

1. Determine the spectral function ϕ(E) [D and H, Chapters 8, 9, and 10]

2. Determine group structure [EG < EG−1 < ..... < E0] [D and H, Chapter 8]

3. Determine multigroup constants [D and H, Chapter 7] [Eqns.(7)-(14)]

4. Solve Eqn.(13) [DH, Chapter 7. See below.]

Remark 1: The above procedure can also be used to performe a group collapse. The starting point is

not the energy-dependent diffusion equation (1), but rather a system of multigroup diffusion equations [of

the form (13)] with a ’large’ number G1 of groups, and the result is a similar system of multigroup diffusion

equations with fewer (G2) groups. To perform this group collapse , we need a spectral function χg defined

on the fine group structure. In doing this collapse, there is a trade-off: there is a loss in accuracy, due to

the reduction in the number of groups, but on the other hand the calculations required to solve the problem

become less expensive.

Remark 2: The above derivation of Eqn.(15) is exact if

a) ∂φ∂t = 0 and

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Page 122: Larsen Lecture Notes

b) D(E),Σt(E),Σs(E′ → E), χ(E), and νΣf (E) are constants for E and E

′in each energy group,

or

φ(r, E, t) = ψ(r, t) ϕ(E)

Structure of Σsg′g: In scattering process, high-energy neutrons lose energy and thermal neutrons gain

or lose energy. Therefore, if we have only one thermal energy group, then since

Σsg′g is the rate at which neutrons scatter from group g′to g.

Then

Σsg′g =

⎧⎨⎩

0 g′> g

≤ 0 g′ ≤ g

(16)

In addition, if the group structure is chosen carefully, it is sometimes possible to have

Σsg′g =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

0 g′> g

≤ 0 g′

= g and g′ − 1 = g

0 g′ − 1 < g

In this case, the groups are said to be directly coupled; neutrons can scatter downward at most one group

at a time. Henceforth, we will assume that Eqn.(16) holds, but we will not assume that the groups are

directly coupled. Thus, if we define

Σrg = Σtg − Σsg′g = removal cross section for group g (17)

then Eqn.(15) can be written

1vg

∂tφg − ∇ ·Dg ∇φg + Σrg φg =

g−1∑g′=1

Σsg′g φg′ + χg

G∑g′=1

(νΣf )g′ φg′ + Qg , 1 ≤ g ≤ G (18)

These are the multigroup diffusion equations with fission and downscattering (but no upscattering). The

appropriate initial and boundary conditions are:

φg(r, 0) = φig(r) , r ∈ D (19)

φg(r, t) + 2 D(r) n(r) · ∇ φg(r, t) = 0 , r ∈ ∂D , t > 0 (20)

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Page 123: Larsen Lecture Notes

These equations can be discretized in time and space exactly as in the previous chapter. Assuming that

this is done, one than has, even for steady-state problems, a much more complicated set of equations to

solve than in the one-group case that we considered in Chapter 4. Therefore, it is not feasible to solve these

equations by a direct method such as Gaussian elimination, and an iteration scheme must be used.

Power Iteration for Fixed Source Problems: For steady-state problems, Eqn.(18) can be written

−∇ ·D1 ∇φ1 + Σr1 φ1 = (χ1 F + Q1)

−∇ ·D2 ∇φ2 + Σr2 φ2 = (Σs12 φ1) + (χ1 F + Q2)...

−∇ ·DG ∇ φG + ΣrG φG = (∑G−1

g′=1Σsg′G φg′ ) + (χG F + QG)

where

F (r, t) =G−1∑g′=1

ν Σfg′ φg′ = fission source

This suggests the iteration scheme:

−∇ ·D1 ∇φ(n+1)1 + Σr1 φ

(n+1)1 = (χ1 F

(n) + Q1)

−∇ ·D2 ∇φ(n+1)2 + Σr2 φ

(n+1)2 = (Σs12 φ1) + (χ1 F

(n) + Q2)...

−∇ ·DG ∇ φ(n+1)G + ΣrG φ

(n+1)G = (

∑G−1g′=1

Σsg′G φ(n+1)

g′ ) + (χG F (n) + QG)

(21)

f (n+1)(r) =G∑

g′=1

ν Σfg′ φ(n+1)

g′ (r) (22)

The boundary condition for each of these equations is

φg(r) + 2 D(r) n(r) · ∇ φg(r) = 0 , r ∈ ∂D (23)

We stop iterating when the fission source is sufficiently converged, i.e., when

maxr∈D |f (n+1)(r) − f (n+1)(r)| < ε , (24)

where ε is a preassigned number, typically, ε = 10−5.

Remarks:

1. At each step in this scheme, one is only required to solve a one-group diffusion equation.

6

Page 124: Larsen Lecture Notes

2. In one-dimensional problems, each one-group diffusion problem can be solved directly by Gauss elimina-

tion. However, in two or three dimensional problems, one must iterate to solve the equations that old within

each group. These iterations are called inner iterations .

3. A single sweep through all of Eqns.(21),(22),(23)) constitutes one outer iteration. Thus, a single outer

iteration consists of possibly many inner iterations in each group.

Inverse Power Iteration for Eigenvalue Problems: Multigroup k-eigenvalue problems typically have the

form

−∇ ·Dg ∇ φg + Σrg φg = (g−1∑g′=1

Σsg′g φg′ ) +1kχg

G∑g′=1

ν Σfg′ φg′ 1 ≤ g ≤ G, r ∈ D (25)

φg + 2 Dg n · ∇ φg = 0 , r ∈ ∂D (26)

We note that if the functions φg(r) satisfy this problem, then Aφg(r) also satisfies it for any constant A.

Therefore, we can make the solution functions φg(r) unique by specifying a normalization, such as:

∫D

(G∑

g′=1

ν Σfg′ φg′ (r)) d3r =

∫D

F (r) d3r =∫D

d3r = V

Physically, this states that the average value of the fission source F (r) across the system is unity.

The inverse power iteration method is defined as follows. We begin the (n+1)-st outer iteration with an

eigenvalue estimate k(n) and a fission source estimate F (n)(r) satisfying

1V

∫D

F (n)(r) d3r = 1 (27)

The first step is to determine φ(n+1/2)g by solving the following fixed-source problem:

−∇ ·Dg ∇ φ(n+1/2)g + Σrg φ

(n+1/2)g = (

∑g−1

g′=1Σsg′g φ

(n+1/2)

g′) + χg

k(n) F(n) , r ∈ D

φg + 2 D n · ∇ φg = 0 , r ∈ ∂D(28)

Next, we define

f (n+1/2)(r) =G∑

g′=1

ν Σfg′ φ(n+1/2)

g′ (r) , (29)

and then sum up Eqns.(21),(22),(23) over g and use∑χg = 1:

G∑g=1

(−∇ ·Dg ∇ φ(n+1/2)g + Σrg φ(n+1/2)

g −g−1∑g′=1

Σsg′g φ(n+1/2)

g′) =

1k(n)

F (n)

7

Page 125: Larsen Lecture Notes

Now we integrate over the spatial domain D and solve for k(n):

k(n) =

∫DF (n)(r) d3r∫

D

∑Gg=1 [−∇ ·Dg ∇ φ

(n+1/2)g + Σrg φ

(n+1/2)g − ∑g−1

g′=1Σsgg′ φg′ ] d3r

Using the most recent estimate for thr scalar fluxes in this formula, we now define

k(n+1) =

∫DF (n+1/2)(r) d3r∫

D

∑Gg=1 [−∇ ·Dg ∇ φ

(n+1/2)g + Σrg φ

(n+1/2)g − ∑g−1

g′=1Σsgg′ φg′ ] d3r

Dividing these last two equations, we obtain the simpler result

k(n+1)

k(n)=

∫D F (n)(r) d3r∫

D F (n+1/2)(r) d3r

or

k(n+1) = k(n) [1V

∫D

F (n)(r) d3r] (30)

Finally, we normalize F (n+1/2)(r) and obtain

F (n+1)(r) =F (n+1/2)(r)

1V

∫D F (n+1/2)(r′) d3r

(31)

The inverse power iteration method is now specified by Eqns.(28)-(31).We stop iterating when

| k(n+1) − k(n) | < ε

and

maxr∈D | F (n+1)(r) − F (n)(r) | < δ

where ε and δ are preassigned (again, typically, ε = 10−5 and δ = 10−4).

Now we shall discuss some applications of Multigroup Diffusion Theory.

One-Group Diffusion Theory : Here one performs a group-collapse down to single energy group that

spans all neutron energies, from thermal to fast.

Eqn.(15) and χ1=1 reduce to tha familiar (from Chapter 4) result

1v1

∂tφ1 − ∇ ·D1 ∇φ1 + Σa1 φ1 = νΣf1 φ1 + Q1

8

Page 126: Larsen Lecture Notes

E = EE = E1 omin max

Clearly, this one-group model can only contain a limited amount of physics.

Two-Group Diffusion Theory : Here one performs a collapse down to two groups, one describing fast

neutrons and the other describing thermal neutrons.

E = EE = E E = 1 ev1 omin max

thermalgroup

fast group

The dividing energy E1 between the two groups is chosen so that

1) Σs21 = 0 (no neutrons scatter up from the thremal group to fast group)

2) χ2 = 0 (⇒ χ1 = 1) (all fission neutrons are fast)

Then, the general two-group k-eigenvalue problem

− ∇ ·D1 ∇φ1 + Σr1 φ1 = Σs21 φ2 + χ1k (νΣf1 φ1 + νΣf2 φ2)

− ∇ ·D2 ∇φ2 + Σr2 φ2 = Σs12 φ1 + χ2k (νΣf1 φ1 + νΣf2 φ2)

becomes

− ∇ ·D1 ∇φ1 + Σr1 φ1 =1k(νΣf1 φ1 + νΣf2 φ2) (32)

− ∇ ·D2 ∇φ2 + Σr2 φ2 = Σs12 φ1 (33)

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Page 127: Larsen Lecture Notes

with boundary conditions

φg + 2 Dg n · ∇ φg = 0 r ∈ ∂D g = 1, 2 (34)

To proceed with problem analytically, we shall replace the boundary conditions Eqn.(34) by the simpler

(and cruder) conditions

φg = 0 r ∈ ∂D g = 1, 2 (35)

Now we can solve the problem Eqn.(32), Eqn.(35) by taking

φg(r) = ϕg ψg(r) g = 1, 2 (36)

where ψg(r) satisfies the eigenvalue problem

2 ψ + B2g ψ = 0 r ∈ D

ψ = 0 r ∈ D

with B2g the geometric buckling. Then the boundary conditions Eqn.(35) are satisfied and Eqns.(32)-(34)

give (for constant cross sections)

(D1 B2g + Σr1)ϕ1 = 1

k ((νΣf )2 ϕ1 + (νΣf )2 ϕ2)

(D2 B2g + Σr2)ϕ2 = Σs12 ϕ1

This is an algebric eigenvalue problem for k. The solution is:

k =νΣf1

D1 B2g + Σr1

+νΣf2

D2 B2g + Σr2

− Σs12D1 B2

g + Σr1(37)

ϕ1 = A (D2 B2g + Σr2)

ϕ2 = A (Σs12)(38)

where A is a constant. By setting k=1, we obtain a criticality condition relating all the cross sections in the

problem to the geometric buckling.

Remark 1: This procedure will not woek with the original group-dependent boundary conditions Eqn.(34).

Remark 2: In Duderstadt Hamilton, pp. 297-298, it is shown that the formula Eqn.(37) can be identified

with the six-factor formula.

10

Page 128: Larsen Lecture Notes

One and One-Half Group Diffusion Theory: In Eqns.(32)-(22), it is sometimes permissible to set D2 =

0, when the thermal neutron diffusion is unimportant. We then obtain from Eqn.(33),

φ2 =Σs12Σr2

φ1

and introducing this into Eqn.(32), we obtain the modified one-group or the one and one-half group equation

− ∇ ·D1 ∇φ1 + Σr1 φ1 =1k

(νΣf1 φ1 + νΣf2Σs12Σr2

) φ2

This single equation will be much easier to solve than the coupled system of two equations (32)-(22).

Multigroup Perturbation Theory: Let us now consider the multigroup k-eigenvalue problem

L φg = −∇ ·Dg ∇ φg + Σrg φg − (∑g−1

g′=1Σsg′g φg′ ) − 1

k χg∑G

g′=1 ν Σfg′ φg′ r ∈ D

= A φg − 1k B g

(39)

B(L) : φg + 2 Dg n · ∇ φg = 0 , r ∈ ∂D (40)

We suppose that we have consructed a (numerical or analytical) solution, which we denote simply by

φ0g and k, and we must ask: what effect will the small perturbation of a cross section in Eqn.(39) have on

the reactivity ρ = 1 − (1/k)? To answer this question, we must use perturbation theory. However, there

is a difficulty: in general, there is no way to define an inner product for which the multigroup operator L

is self-adjoint. Therefore, the perturbation theory used in Chapter 4 is not of direct use here. However, we

shall see that a rather simpler generilization of this theory is of use.

Let φ(r) denote a vector with G components, whose g-th component is φg(r). Then, for any two such vector

functions φ(r) and ζ(r), let us define the inner product

(φ, ζ) =G∑g=1

∫D

φg(r) ζg(r) d3r (41)

With this inner product, one can show that

L∗ φg = −∇ ·Dg ∇ φg + Σrg φg − (G∑

g′=g+1

Σsg′g φg′ ) − ν Σfgk

G∑g′=1

χg φg r ∈ D (42)

B(L∗) : φg + 2 Dg n · ∇ φg = 0 , r ∈ ∂D (43)

and therefore, L is not self-adjoint, and we cannot use the theorem on page 4-19. However, we can use the

following generalization of this theorem:

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Page 129: Larsen Lecture Notes

(Fredholm Alternative) Theorem: Let there exists a nonzero function φ0 such that

L φ0 = 0 and φ0 satisfiesB(L) (44)

Then, there exists a nonzero function φ∗0 such that

L0 φ∗0 = 0 and φ∗0 satisfiesB(L∗) (45)

and the problem

L g = Q (46)

has a solution if and only if

( φ∗0, Q) = 0

Reasonableness proof (one way only): If Eqns.(44),(45),(46) have solution, then

( φ∗0, Q ) = ( φ∗0, Lg ) = ( L∗φ∗0, g ) = ( 0, g ) = 0

Application: Letus return to the multigroup eigenvalue problem

L φ0 = (A − 1kB) φ0 (47)

where the operators A and B are defined in Eqn.(39). We wish to consider the effect on k of perturbations

in the operators A and B. Thus, we let

A → A + ΔA

B → B + ΔB

φ0 → φ0 + Δφ

and

1k = 1 − ρ → 1 − (ρ + Δρ) = 1

k − Δρ

Introducing these perturbations into Eqn.(47), we obtain

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Page 130: Larsen Lecture Notes

0 = [A → A + ΔA − ( 1k − Δρ)(B → B + ΔB)] (φ0 + Δφ)

= [(A − 1kB) + (ΔA − 1

kΔB + ΔρB)] (φ0 + Δφ) + O(Δ)

= [L + (ΔA − 1kΔB + ΔρB)] (φ0 + Δφ) + O(Δ2)

= L φ0 + L Δφ + (ΔA − 1kΔB + ΔρB) φ0 + O(Δ2)

= L Δφ + (ΔA − 1kΔB + ΔρB) φ0 + O(Δ2) for L φ0 = 0

or, deleting the O(Δ2) terms,

L Δφ = (−ΔA +1kΔB − ΔρB) φ0

Now, according to the Fredholm Alternative Theorem, a solution of this equation exists if and only if

0 = (φ∗0, (−ΔA +1k

ΔB − ΔρB) φ0)

or, rearranging,

Δρ =(φ0, (−ΔA + 1

kΔB) φ0)(φ∗0, B φ0)

This formula gives (to first order) the small change in ρ due to any small change in the operators A and B.

For example, if we alter the fission cross section

ν Σfg → ν ΣfgΔν Σfg

then

ΔA = = 0

ΔB φ0 = χg∑Gg′=1 Δν Σfg′ φ0g′

so

Δρ = 1k

(φ∗0 , ΔB φ0)

(φ∗0 , B φ0)

= 1k

∑G

g=1

∫D

[φ∗0g(r) χg

∑G

g′=1

Δν Σfg

′ φ0g′ (r)] d3r∑G

g=1

∫D

[φ∗0g(r) χg

∑G

g′=1

ν Σfg

′ φ0g′ (r)] d3r

= 1k

∫D

[∑G

g=1φ∗

0g(r) χg ] [∑G

g′=1

Δν Σfg

′ φ0g′ (r)] d3r∫

D[∑G

g=1φ∗

0g(r) χg

∑G

g′=1

ν Σfg

′ φ0g′ (r)] d3r

Remark 1: To use multigroup perturbation theory, one must compute two unperturbed eigen functions;

one corresponding to the forward operator L, the other to the adjoint operator L∗.

Remark 2: The Fredholm Alternative Theorem described here reduces to the problem on p. 4-19 when

L is self-adjoint.

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Page 131: Larsen Lecture Notes

Remark 3: The solution φ∗0 of the adjoint problem is called the ’importance function’, and in later chap-

ters we will see that it has other applications.

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CHAPTER 6 : FAST SPECTRUM CALCULATIONS (Chapter 8, Duderstadt and Hamil-

ton)

- Introduction

- Elastic ’S-Wave’ or ’Potential’ Scattering

- Infinite Medium Spectrum Equation

- Slowing down Equation

- Hydrogen (A=1) With No Absorption

- Hydrogen (A=1) With an Infinitely Massive Absorber

Resonance Eacape Probability

- Slowing Down Density q(E)

- Neutron Lethargy

Scalar Flux, Collision Density

Elastic Scattering Probability Function

Slowing Down in Hydrogen Without Absorption

Slowing Down Density in Hydrogen

Mean Lethargy Gain per Collision ζ

Number of Collisions to Thermalize

Moderating Power and Moderating Ratio

Multigroup Structure

- Slowing Down in Nonabsorbing Media (Σa = 0) for A > 1

- Slowing Down in Absorbing Media (Σa > 0) for A > 1

- Infinite Medium Resonance Absorption

- A Close Look at Resonance Escape Probabilities

- Approximate Evaluations of Resonance Integrals

Narrow Resonance (NR) Approximation

Narrow Resonance Infinite Mass (NRIM) Approximation

- Slowing Down in a Finite Medium

Slowing Down in Hydrogen (A=1)

Age Approximation (A >> 1)

Continuous Slowing Down Approximation

Consistent Age Approximation (A >> 1)

Grueling-Goertzel Approximation (A >> 1 and A=1)

- Treatment of Spatial Dependence

Age-Diffusion (or ’Age’, ’Fermi Age’, ’Inconsistent Age’, or ’Continuous Slowing Down’) Theory

Fermi Age Equation

Muft-Gram Method, The B1 Method

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In this chapter, we will consider the problem of computing energy spectra φ(E) for use in multigroup

diffusion calculations. There are three basic types of spectral approximations in common use:

1) Zero-th order approximation: The equations for φ(E) strictly apply only in infinite homogenous media

and contain no spatial dependence at all. [D and H, Chapters 8,9]

2) First-order approximation: ’Buckling’ terms are included in the equations for φ(E). [D and H, Chap-

ters 8,9]

3) Second-order approximation: Cross sections are averaged over a spatially heterogeneous unit cell (e.g.,

a fuel cell). [D and H, Chapter 10]

In this chapter, we will just consider the zero-th order and first-order approximations.

Physics important in various energy ranges:

105ev − 107ev (fast) : No upscattering

Doenscattering (elastic and inelastic)

Resonance absorption

Fission sources

1ev − 105ev (fast) : No upscattering

Doenscattering (elastic)

Resonance absorption

0 − 1ev (thermal) : Upscattering and downscattering

We will only treat elastic downscattering and certain types of resonancs absorption. Our analysis will be

most applicable in the 1ev − 105ev energy range, where inelastic scattering and upscattering are absent.

Elastic ’S-Wave’ or ’Potential’ Scattering: This is defined by

Σs(E′ → E) = Σs(E

′) P (E

′ → E) (1)

where

P (E′ → E) =

⎧⎨⎩

1(1 − α) E′ α E

′< E < E

0 otherwise(2)

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Page 134: Larsen Lecture Notes

E

E )

E’

P(E’

1

E’(1− α)

α = (A − 1A + 1

)2 , A = mass number of target nucleus (3)

We note that

∫ ∞0

P (E′ → E) dE =

∫ E′

αE′ dE′

(1 − α) E′ = 1∫ ∞0

Σs(E′ → E) dE = Σs(E

′)

(4)

Also,

αE′< E < E

E < E′

and αE′< E

E < E′

and E′< E / α

E < E′< E / α

Therefore, we can also write

P (E′ → E) =

⎧⎨⎩

1(1 − α) E′ E < E

′< E/α

0 otherwise(5)

The infinite medium spectrum equation is now defined as:

Σt(E) φ(E) =∫ ∞

0

Σs(E′ → E) φ(E

′) dE

′+ S(E) (6)

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Page 135: Larsen Lecture Notes

This is just the diffusion equation for an infinite medium, with no spatial dependence and the known

fission source χ(E) contained in S(E). Also, it is the transport equation for an infinite medium, with no

spatial dependence, an isotropic source S(E) containing χ(E), and written in terms of the scalar flux rather

than the angular flux. For 0 < E < 105ev, χ(E) = S(E) = 0. Also,for elastic scattering,

Σs(E′ → E) = 0 for E

′< E ,

and we obtain the slowing down equation

Σt(E) φ(E) =∫ E/α

E

Σs(E′)

(1 − α) E′ φ(E′) dE

′+ S(E) (7)

Hydrogen (A=1, α = 0) with no absorption (Σa = 0,Σt = Σs): For A=1 and α = 0, Eqn.(7) reduces to:

Σs(E) φ(E) =∫ ∞

0

Σs(E′)

E′ φ(E′) dE

′+ S(E) (8)

This equation can be solved analytically, by converting it to a first-order differential equation and then

solving this differential equation. We set

F (E) = Σt(E) φ(E) = Σs(E) φ(E) (9)

which is the collision rate density.

Then Eqn.(8) can be written

F (E) =∫ ∞

E

F (E′)

E′ dE′

+ S(E) (10)

To convert this into a differential equation, we need the following general formula:

d

dx

∫ a(x)

b(x)

f(x′, x) dx

′=

∫ b(x)

a(x)

∂f

∂x(x

′, x) dx

′+ f [b(x), x] b

′(x) − f [a(x), x] a

′(x) (11)

Differentiating Eqn.(10) with respect to E and using this formula, we get

F′(E) = − F (E) / E + S

′(E)

E F′(E) + F (E) = E S

′(E) + S(E) − S(E)

d

dEE F (E) =

d

dEE S(E) − S(E) (12)

To solve this differential equation, we need a boundary condition. Let us assume that

S(E) = 0 for E ≥ E0 (13)

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Page 136: Larsen Lecture Notes

Then, for E ≥ E0, Eqn.(12) gives

ddEE F (E) = 0 so,

E F (E) = C = constant or

F (E) = CE , E ≥ E0

What is the value of the constant C? If there is no source of neutrons for E ≥ E0, then there should be

no neutrons for E ≥ E0, so C=0. Hence,

F (E) = 0 , E ≥ E0, (14)

This is the proper boundary condition.

Now, integrating Eqn.(12) over E0 < E′< E, we obtain

E′F (E

′) |E0E = E

′S(E

′) |E0E − ∫ E0

E S(E′) dE

−E F (E) = − E S(E) − ∫ E0

E S(E′) dE

F (E) = S(E) + 1E

∫ E0

ES(E

′) dE

φ(E) =1

Σs(E)[S(E) +

1E

∫ E0

E

S(E′) dE

′] (15)

If we assume S(E) = 0 for E < E∗

then, for E < E∗, Eqn.(15) reduces to

φ(E) =1

Σs(E)1E

∫ E0

E∗S(E

′) dE

′] (16)

Thus, since Σs(E) = constant, we get

φ(E) � constant

E, E < E∗ (17)

In deriving this result, we assumed S(E) = 0 for E < E∗ and E0 < E.

If S(E) represents a fission source χ(E), then E0 � 108 ev and E∗ � 105 ev.

Hydrogen (A=1, α = 0) with an infinitely massive absorber: Eqn.(7) now becomes:

Σt(E) φ(E) =∫ ∞

E

Σs(E′) φ(E

′)

E′ dE′

+ S(E) (18)

(In a collision with an absorber nucleus, a neutron is either absorbed, or is scattered with the same energy.)

We define

Σt(E) φ(E) = F (E) = collision rate density (19)

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Page 137: Larsen Lecture Notes

EEoE*

S(E)

1

Ε Σ (Ε)sφ(Ε)= ( Σ (Ε )

Eo

E*

dE’ )s ’

and

A(E) =Σs(E)E Σt(E)

(20)

Then,

F (E) =∫ ∞E

Σs(E′) Σt(E

′) φ(E

′)

E′ Σt(E′ ) dE

′+ S(E)

=∫ ∞E

A(E′) F (E

′) dE

′+ S(E)

(21)

Hence,

F′(E) = − A(E) F (E) + S

′(E)

ddE (F − S) = − A F = − A (F − S) − A S

ddE (F − S) + A (F − S) = − A S

ddE (F − S) exp(− ∫ E0

EA(E

′′) dE

′′) = − A S

′exp(− ∫ E0

EA(E

′′) dE

′′)

Taking S(E0) = 0, so F (E0) = 0, we get

− (F (E) − S(E)) exp(− ∫ E0

E A(E′′) dE

′′) = − ∫ E0

E A(E′) S(E

′) exp(− ∫ E0

E A(E′′))

or

F (E) = S(E) +∫ E0

E S(E′) A(E

′) e

∫E

EA(E

′′) dE

′′dE

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Page 138: Larsen Lecture Notes

However,

A(E′) exp(

∫ E′

EA(E

′′) dE

′′)

= Σs(E′)

E′ Σt(E′) exp(

∫ E′ Σs(E′′

)

E′′ Σt(E′′ )

dE′′)

E

= Σs(E′)

E′ Σt(E′) exp(

∫ E′ 1E

′′ [1 − Σa(E′′

)

Σt(E′′ )] dE

′′)

E

= Σs(E′)

E′ Σt(E′) exp(ln(E

E ) − ∫ E′ Σa(E′′

)

Σt(E′′ )dE

′′)

E

= 1E [Σs(E

′)

Σt(E′ ) ] [exp( − ∫ E′ Σa(E

′′)

Σt(E′′ )dE

′′)]

E

= 1E u(E

′) p(E

′, E)

Therefore,

F (E) = S(E) +1E

∫ E0

E

S(E′) u(E

′) p(E

′, E) dE

or

φ(E) =1

Σt(E)[S(E) +

1E

∫ E0

E

S(E′) u(E

′) p(E

′, E) dE

′] (22)

In the absence of absorption, Σa = 0,Σs = Σt, u = p = 1, and Eqn.(22) reduces to Eqn.(15). Therefore,

u(E′) p(E

′, E) = the fraction of (source) neutrons with energy E

′that slow down to energy E

= the probability that a (source) neutron at energy E′will not be absorbed while slowing down to energy E.

(23)

But,

u(E′) =

Σs(E′)

Σt(E′)

= the probability that a neutron with energy E′is scattered (not absorbed) in its first collision

(24)

so,

p(E′, E) = exp( − ∫ E′ Σa(E

′′)

Σt(E′′ )dE

′′)

E

= probability that a neutron, which has initial energy E′and undergoes a scattering event, is not absorbed while slo

= resonance escape probability

(25)

Slowing Down Density: We define the slowing down density, q(r, E), by

q(r, E) d3r = the number of neutrons that slow down past E per second in d3r about r.

We have

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Page 139: Larsen Lecture Notes

energyE’EE’’(fixed) (E’>E)(E’’<E)

φ(r, E′) d3r dE

′= the flux of neutrons in d3r about r having energy indE

′about E

′,

so

Σs(E′ → E

′′) [φ(r, E

′) d3r dE

′] dE

′′= the number of neutrons within d3r about r that scatter from

within dE′about E

′to within dE

′′about E

′′per second.

Therefore,

q(r, E) =∫ E

E′′=0

[∫ E0

E′=EΣs(E

′ → E′′) φ(r, E

′) dE

′] dE

′′(26)

Using Eqn.(11), we obtain

∂∂E q(r, E) = [

∫ E0

E′=E Σs(E′ → E) φ(r, E

′) dE

′] + [

∫ EE′′=0 [−Σs(E→E

′′) φ(r, E)] dE

′′

However,

Σs(E′ → E) = 0 for E

′< E

Σs(E → E′′) = 0 for E

′′> E

so

∂∂E q(r, E) =

∫ E0

E′=E Σs(E′ → E) φ(r, E

′) dE

′ − ∫ E0

E′=0 Σs(E′ → E) dE

′φ(r, E)

= [Σt(E) φ − S] − [Σs(E)] φ

= − S(E) + Σa φ(r, E)

Again, for E > E0, we have S(E)=0, so φ(E)=0, so q(E)=0. Thus,

q(r, E′) |E0E = − ∫ E0

E S(E′) dE

′+

∫ E0

E Σa(E′) φ(E

′) dE

or

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Page 140: Larsen Lecture Notes

q(r, E) =∫ E0

E S(E′) dE

′ − ∫ E0

E Σa(E′) φ(E

′) dE

= the rate at which source neutrons are produced with ranges > E

= the rate at which neutrons with energies > E are absorbed.

(27)

This is a ’balance’ or ’conversation’ law, which must hold in a steady-state situation:

energy

(source neutrons with energy > E)

(neutrons with energy >E that are absorbed)

slow down pastenergy E

(neutrons that

)

The slowing-down density for hydrogen is explicitly given by:

q(E) =∫ EE′′=0 [

∫ E0

E′=E Σs(E′ → E

′′) φ(E

′) dE

′] dE

′′

=∫ EE′′=0 [

∫ E0

E′=EΣs(E

′)

E′ φ(E′) dE

′] dE

′′

= E∫ E0

E′=EΣs(E

′)

E′ Σt(E′) Σt(E

′) φ(E

′) dE

= E∫ E0

E′=E A(E′) F (E

′) dE

= E [F (E) − S(E)]

=∫ E0

E

S(E′) x(E

′) p(E

′, E) dE

′(28)

where u(E′) and p(E,E

′) are given by Eqns.(24)-(25).

Neutron Lethargy: Lethargy is a variable that is often used in slowing down calculations in place of the

energy variable; this is done because the slowing down equations become mathematically somewhat simplier.

To define lethargy, let E0 be the maximum neutron energy in a system. Then

u = ln(E0E ) = lethargy of a neutron with energy E

E = E0 e−u

(29)

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Page 141: Larsen Lecture Notes

E

u

Eo

We note that

1) u ¿ 0 ,

2) as E → 0, u → ∞.

Also,

u = ln(E0) − ln(E) , du = − dE

E(30)

If a neutron has initial energy Ei and (after one or more collisions) a final energy Ef < Ei, then

uf − ui = [ln(E0) − ln(Ef )] − [ln(E0) − ln(Ei)] = ln(EiEf

) (31)

We also note that:

3) uf − ui > 0 (as a neutron loses energy, it gains lethargy)

4) uf − ui is independent of E0 (the change in lethargy is independent of E0).

In many applications, u is a more convenient variable than E because the mathematical form of the slowing

down equation becomes simpler. We will now convert the slowing down equation to u.

Scalar Flux, Collision Density: We have

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Page 142: Larsen Lecture Notes

φ(E) dE = the flux of neutrons in dE about E

F(E) dE = Σt(E) φ(E) dE = the number of neutrons in dE about E that undergo collisions per sec. per

cm3.

We wish to define φ(u), f(u) such that

φ(u) |du| = the flux of neutrons in du about u

f(u) |du| = the number of neutrons in du about u that undergo collisions per sec. per cm3.

Thus, we want

φ(u) |du| = φ(E) dE

f(u) |du| = F (E) dE where du = − dEE

(32)

so

φ(u) = φ(E) |dEdu | = E φ(E)

f(u) = F (E) |dEdu | = E F (E)

= E Σt(E) φ(E) = Σt(E) φ(u) = Σt(u) φ(u)

(33)

where

Σt(u) = Σt(E) = Σt(E0 e−u)

From Eqn.(15), we have for the case of hydrogen with no absorption,

E F (E) = constant , E < E∗ ,

so

f(u) = E F (E) = constant , u < u∗

In more general situations, we have f(u) � constant. Hence, f(u) is easier to compute than F(E).

Elastic Scattering Probability Function: We have

P (E′ → E) dE = probability that a neutron undergoing a scattering at energy E

′, has final energy

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Page 143: Larsen Lecture Notes

within dE about E

P (E′ → E) =

⎧⎨⎩

1(1 − α) E′ α E

′< E < E

0 otherwise

We wish to define p(u′ → u)

p(u′ → u) |du| = P (E

′ → E) dE (34)

Thus, for αE′< E < E

′,

p(u′ → u) = P (E

′ → E) |dEdu | = E(1 − α) E′

= E0 e−u

(1 − α) E0 e−u′ = eu

′−u

1 − α

This holds forαE

′< E < E

or

αE0e−u′

< E0 e−u < E0 e

−u′

or

αe−u′< e−u < e−u

or

e−u′+lnα < e−u < e−u

or

−u′+ lnα < − u < − u

or

u′+ ln 1

α > u > u′

Thus,

αE′< E < E

′ → u′+ ln

> u > u′

(35)

These inequalities can be rewritten as

E < E′< E/α → u − ln 1

α < u′< u

However, we cannot have E′> E0 or u

′< 0, so

E < E′< min(E0, E/α) → max(0, u − ln

) < u′< u (36)

Therefore,

P (u′ → u) =

⎧⎨⎩

eu′−u

1 − α u′< u < u

′+ ln 1

α

0 otherwise(37)

12

Page 144: Larsen Lecture Notes

and for the case of hydrogen (A=1, α=0), we have the simpler result

P (u′ → u) =

⎧⎨⎩

eu′−u u

′< u

0 otherwise(38)

Slowing Down in Hydrogen Without Absorption: From Eqn.(10),

E F (E) =∫ ∞

E

EF (E

′)

E′ dE′

+ E S(E)

or

f(u) =∫ 0

u′=u

E0e−u

E0e−uf(u

′) (−du′

) + s(u) (39)

or

f(u) =∫ u

0

eu′−u f(u

′) du

′+ s(u) (40)

Slowing Down Density in Hydrogen: We have

q(u) = q(E) = E [F (E) − S(E)]

= f(u) − s(u) =∫ u0eu

′−u f(u′) du

Therefore,

d

duq(u) = eu·u f(u) +

∫ u

0

[− eu′−u f(u

′)] du

′= f(u) − q(u)

so

ddu q(u) + q(u) = f(u)

q(0) = 0 [fromEqn.(40)](41)

This result, for A=1, is exact. Approximate results for A > 1 having a similar form will be derived later.

Mean Lethargy Gain per Collision: Let a neutron with initial energy E′

scatter off an atom with mass

number A, and let the neutron have final energy E. Then, the change in the lethargy of the neutron is:

u − u′

= lnE

E= Δu

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Page 145: Larsen Lecture Notes

Let us now define

ζ = mean lethargy gain

= < Δu >

=∫ ∞u′ (u − u

′) p(u

′ → u) du

=∫ u′

+ln 1α

u′ (u − u′) e−(u − u

′)

1 − α du

= 11 − α

∫ ln 1α

0t e−t dt

= 11 − α [−(1 + t) e−t]ln

0

ζ = 1 + α (ln α

1 − α) α = (

A − 1A + 1

)2 (42)

We note that ζ depends only on A, the mass number of the nucleus, and is independent of the initial energy

E′. One can show

ζ � 1 , α � 0 (Asmall)

ζ � 1 − α2 � 1

A , α � 1 (Alarge)

and

α

ζζα

d

d

d ζd α

=

=−1/2

− ❏❏

Number of Collisions to Thermalize: If a neutron has initial energy Ei and (after one or more collisions),

a final energy Ef , than by Eqn.(31)

lnEiEf

= uf − ui = N < Δu > = N ζ

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Page 146: Larsen Lecture Notes

Hence,

N =1ζlnEiEf

(43)

where ζ is a function only of A, the mass number of the nucleus.

Moderating Power and Moderating Ratio: A goog moderator has:

large ζ, large Σs, small Σa

Hence, two figures of merit for moderators are:

moderating power = ζ Σs

moderating ratio = ζ Σs

Σa

(44)

(For a good moderator, these are >> 1.)

Multigroup Structure: Multigroup structures are typically chosen so that the gain in lethargy Δu is the

same for each group. Therefore, by Eqn.(31),

Δu = constant = lnEg−1Eg

, all g

so

E0

E1=

E1

E2= . . . . . =

Eg−1

Eg= . . . . =

EG−1

EG= eΔu (45)

Multiplying all these equations, we get

E0

EG= eG Δu so eΔu = (

E0

EG)1/G

Now, multiplying the first of g of Eqn.(45), we obtain

E0

Eg= (eΔu)G = (

E0

EG)g/G

Hence,

Eg = E0 (EGE0

)g/G = E1−g/G0 E

g/GG 0 ≤ g ≤ G

This formula explicitly gives the multigroup boundaries as a function of E0 = Emax, EG = Emin, and G

= the number of groups. This formula is identical to that given by Eqn.(5.2), so these multigroup structures

are logarithmic.

15

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Slowing Down in Nonabsorbing Media(Σa = 0withA > 1) : ForΣa = 0,Σs = Σt, and therefore

Eqn.(7) becomes

F (E) =1

1 − α

∫ E/α

E

F (E′)

E′ dE′

+ S(E) ,

where, as usual,

F (E) = Σt(E) φ(E)

Equivalently,

E F (E) = 11 − α

∫ E/αE

E F (E′)

E′ dE′

+ E S(E)

f(u) = 11 − α

∫max(0,u−ln 1α )

uE0e

−u

E0e−u f(u′) (−du′

) + s(u)

or

f(u) =1

1 − α

∫ u

max(0,u−ln 1α )

eu′−u f(u

′) du

′+ s(u) (46)

This equation can be solved explicitly, but to do this we must consider succesive intervals of width Δu = ln(1/α).

First, we shall consider the interval 0 < u < ln(1/α):

f(u) =∫ u

0

eu′−u f(u

′) du

′+ s(u) (47)

f(0) = s(0) (48)

Then

f′(u) = 1

1 − α [f(u) − ∫ u0 eu

′−u f(u′) du

′] + s

′(u)

= 11 − α f(u) − [f(u) − s(u)] + s

′(u)

= α1 − α f(u) + s(u) + s

′(u)

= α1 − α [f(u) − s(u)] + 1

1 − α s(u) + s′(u)

so

(f − s)′ − α

1 − α (f − s) = 31 − α

dds(f − s) e−

α1 − αu = 1

1 − α e−α

1 − αu s(u)

Now, integrating from 0 to u, we get

[f(u) − s(u)] e−α

1 − αu =α

1 − α

∫ u

0

e−α

1 − αu′s(u

′) du

or

f(u) = s(u) +α

1 − α

∫ u

0

1 − α (u−u′) s(u

′) du

′0 < u < ln 1/α (49)

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Now, let us consider in general

ln 1α < u

Then, Eqn.(46) becomes

f(u) = s(u) +α

1 − α

∫ u

u−ln 1α

eu−u′s(u

′) du

′(50)

Differentiating with respect to u and using Eqn.(11) we obtain

f′(u) =

α

1 − α[f(u) − α f(u − ln

)] − [f(u) − s(u)] + s′(u) (51)

This is a differential difference equation, which must be solved recursively, on succesive intervals. To show

this, let us define

fn(u) = f(u) for n ln1α

< u < (n+ 1) ln1α

(52)

Then, we have already determine f0(u) [it is given by Eqn.(49)]. Setting u = n ln 1/α, Eqn.(50) becomes

fn(n ln1α

) − s(n ln1α

) =αn

1 − α

∫ n ln 1α

(n−1) ln 1α

eu′fn−1(u

′) du

′(53)

Now, rewriting Eqn.(51), we get

(f − s)′

= 11 − α [f(u) − α f(u − ln 1

α )] − f + s

= α1 − α f(u) − α

1 − α f(u − ln 1α ) + s

= α1 − α (f − s) + 1

1 − α s − α1 − α f(u − ln 1

α )

so

(f − s)′ − α

1 − α (f − s) = 11 − α [s − α f(u − ln 1

α )]ddu (f − s) e−

α1 − α u = e

− α1 − α

u

1 − α [s − α f(u − ln 1α )]

Integrating from n ln 1α to u [with n ln 1

α < u < (n + 1) ln 1α ], we obtain

[fn(u) − s(u)] e−α

1 − α u − [ αn

1 − α

∫ n ln 1α

(n−1) ln 1α

eu′fn−1(u

′) du

′] e−

α1 − α u

= 11 − α

∫n ln 1

αu e

− α1 − α

u′

[s(u′ ) − α fn−1(u′ − ln 1

α )] du′

= 11 − α

∫n ln 1

αu e

− α1 − α

u′s(u′ ) du′ − α

11−α

1 − α

∫ u− ln 1α

(n−1) ln 1α

e− α

1 − αu′fn−1(u

′ ) du′

Hence,

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Page 149: Larsen Lecture Notes

fn(u) = s(u) + 11 − α

∫n ln 1

αu e

− α1 − α

(u−u′ )s(u′ ) du′

+ αn

1−α

1 − α (eα

1 − α u)∫ n ln 1

α

(n−1) ln 1α

eu′fn−1(u

′) du

− αn

1−α

1 − α (eα

1 − α u)∫ u− ln 1

α

(n−1) ln 1α

e−α

1 − α u′fn−1(u

′) du

(54)

This explicitly gives fn in terms of fn−1. Since we have already determined f0, then we can (in principle)

determine fn for all n ≥ 0. [In practice, it is best to use this relationship numerically.]

Now, in regions where S(u)=0, one can show that

fn(u) = fn−1(u) = C = constant

is a solution of Eqn.(54). Numerical calculations show that if S(u)=0, then we in fact have

fn(u) = � C , nsufficientlylarge

What is the value of the constant C? If we have exactly

fn(u) = C

then, because there is no absorption,

C = f(u) = E F (E) = E Σs(E) φ(E)

Now, by Eqns.(27) and (26), we have for E < E∗

∫ E0

E∗s(E) dE = q(E) =

∫ E

E′′=0

Σs(E′ → E

′′) φ(E

′) dE

′dE

′′

However, for a neutron with energy E′slowing down past E to E

′′,

we get

E < E′< E/α , α E

′< E

′′< E

which means that there are two possible ranges of E′and E

′′

so, using Eqns.(1) and (2),

18

Page 150: Larsen Lecture Notes

α E’ E’’ E E’

∫ E0

E∗ s(E) dE =∫ E/αE′=E

∫ EE′=αE′

Σs(E′) φ(E

′)

(1 − α)E′ dE′′dE

=∫ E/αE′=E [

∫ EE′=αE′ C

(1 − α)(E′)2 dE′′]

= C1 − α

∫ E/αE′=E

E − α E′

(E′)2 dE′

= . . . . C [1 + α ln α1 − α ]

= C ζ

Hence,

f(u) � C =1ζ

∫ E0

E∗s(E) dE

or

Σt(E) φ(E) � 1E

∫ E0

E∗s(E) dE (55)

This generalizes Eqn.(16) to the case A > 1.

Slowing Down with Absorption for A > 1: For an infinitely massive absorber, we can rewrite Eqn.(7) as

F (E) =1

1 − α

∫ E/α

E′=E

Σs(E′)

E′ Σt(E′)F (E

′) dE

′+ S(E) (56)

We can now proceed, as before, to obtain recursion relations for

fn(u) = f(u) n ln 1α < u < (n+ 1) ln 1

α

The resulting equations are useful in numerical work, but not in analytical work, so we will not derive

them here.

Infinite Medium Resonance Absorption: Heavy nuclei hacve sharp capture resonances that produce res-

onance absorption of neutrons as they slow down. High Energy resonances are not solved abd are difficult

to treat analytically. Low energy resonances are narrow and resolved, and can be treated analytically; we

19

Page 151: Larsen Lecture Notes

shall do this here.

For hydrogen, with an infinitely massive absorber, we introduce S(E) = S0 δ(E − E0) into Eqn.(22)

and obtain [exactly, for E < E0]

Σt(E) φ(E) = S0 δ(E − E0) + 1E

∫ E0

E S0 δ(E′ − E0) u(E

′) p(E,E

′) dE

= 1E S0 u(E0) p(E,E0) + S0 δ(E − E0)

=1E

[S0 Σs(E0)

Σt(E0)] e

−∫

E0

E

Σa(E′)

E′ Σt(E

′ ) dE′

(57)

20

Page 152: Larsen Lecture Notes

energyEoEi−1EiEi+1E

for a heavy nucleusσγ (Ε)

Εi +Ei−Δ Δ

EE3 E2

E Eo

1/E

1/EΣ (Ε)Σ (Ε)

a

t

1

21

Page 153: Larsen Lecture Notes

EEEE E o123

E

Eo

a

tE’dE’

Σ (Ε )Σ (Ε )

’’

EE E E E3 2 1 o

1.0So u(E )o

E

E’dE’

Eo

E

a

t

exp [ − ]Σ (Ε )

Σ (Ε )

22

Page 154: Larsen Lecture Notes

EoE1E2E3

E

o u(E )S

E(no absorption)

Σ (Ε) φ(Ε) = tSo u(E )

E E’dE’

E

Eo

exp[− ]Σ (Ε )’

Σ (Ε )’

From Eqn.(25),

pi = exp (− ∫ Ei+Δ

Ei−ΔΣa(E

′)

E′ Σt(E′)dE

′)

= probability that a neutron is not absorbed while slowing down past the i− th resonance

= i− th resonance escape probability

(58)

[Remark: In typical applications, we have for a single resonance,

pi � 1εi =∫ Ei+Δ

Ei−ΔΣa(E

′)

E′ Σt(E′) dE

′<< 1(59)These inequalities will be very important when we

later treat the case A > 1.] Now, normalizing F(E) so that F(E)=1/E for E1 < E < E0, we get (for

hydrogen) the following figure:

The phenomenon of resonance absorption is important for the following reasons:

1. It is a physical process that must be accounted for in the calculation of multigroup constants.

2. Reonance absorption in fertile material can lead to production of fissile material, and thus is impor-

tant for the accurate prediction of breeding ratios.

3. Resonance absorption plays a role in reactor kinetics. [As the temperature of a reactor increases,

the resonances broaden, the resonance escape probabilities pi decrease, and therefore the reactivity decreases.]

23

Page 155: Larsen Lecture Notes

EEEEE3 2 1 o

F=1 / E

F (E)= 1 / E (no absorption)

F(E) for narrow, resolved resonance absorption

Note : the dips in this curve are exxagrated

F= P1 / EF = P1 P2 / E

F=P1 P2 P3 /E

A Closer Look at Resonance escape Probabilities: (Hydrogen, A=1) We shall now consider resonance

absorption in an infinite homogenous mixture of:

a) an absorber (denoted by A): a heavy isotope with well-resolved absorption resonances. WE require

this isotope to be heavy enough that the scattering of neutrons off it can be ignored (the neutron energy

loss is negligible).

b) a moderator consisting of hydrogen (denoted by H), with a constant Σt over a resonance and a neg-

ligible Σa = Σγ .

Also, we shall ignore inelastic scattering. Then, from Eqn.(25) and Fig.1,

p(E0→E) = exp (− ∫ Ei+Δ

Ei−ΔΣa(E

′)

E′ Σt(E′) dE

′)

= probability that a neutron at E0 which is scattered is not absorbed before slowing down to E

= exp [− ∑E<Ei<E0

∫ Ei+Δ

Ei−ΔΣa(E

′)

E′ Σt(E′ ) dE

′]

= exp [− ∑E<Ei<E0

εi]

= E<Ei<E0 e− εi

= E<Ei<E0 Pi (60)

We now consider a single Pi:

Pi = exp [−∫ Ei+Δ

Ei−Δ

NA σAγ (E′) + NH σHγ (E

′)

NA [σAγ (E′) + σAs (E′)] + NH [σHγ (E′) + σHs (E′)]dE

E′ ]

24

Page 156: Larsen Lecture Notes

Then, near Ei,

σAs << σAγ [absorber scattering is small]

σHγ << σHs � constant [absorbercaptureissmall]

Thus,

Pi � exp [− ∫ Ei+Δ

Ei−Δ

NA σAγ (E

′)

NA σAγ (E′ ) + NH σH

s (E′)dE

E′ ](61)

Using the Breit-Wigner formula [D H, p.50], we obtain

σγ(E) = σ0ΓγΓ

ψ(g, x)

ψ(g, x) =g

2√π

∫ ∞

−∞

e−14 (x−y)2g2

1 + g2dg (62)

where

g = Γ( A4 Ei k T

) (A = mass number)

x = 2 (E − Ei

Γ )

and [D H, pp.26-27]:

Γγ = radioactive line width

Γ = total line width

Ei = energy at which resonance occurs

σ0 = value of the total cross section σt at Ei

k = Boltzman’s constant

T = temperature

Then, Eqn.(61) gives:

25

Page 157: Larsen Lecture Notes

Pi = exp [− ∫ Ei+Δ

Ei−Δ

NA σ0ΓγΓ ψ

NA σ0ΓγΓ ψ + NH σH

s

dE′

E′ ]

= exp [− ∫ ∞−∞

ψ

ψ + (NH σH

s ΓNA σ0 Γγ

)

Γ2dx

Ei]

= exp [− ΓEi

∫ ∞0

ψ(g,x′)

ψ(g,x′ ) + βdx

′]

= exp [− ΓEi

J(g, β)] [This holds only for hydrogen]

In these equations, J is a decreasing function of β and a weakly decreasing function of ζ. Therefore,

1. As T increases, J increases, and Pi decreases. [This is the Doppler Effect: Resonances broaden, and

more neutrons are absorbed.]

2. As NH / NA increases, β decreases, so J decreases, so Pi increases. [Adding moderator material or

deleting absorber material decreases the number of absorptions.]

3. As Ei (the resonance energy) decreases, ζ increases, so J decreases weakly, so J/Ei increases weakly,

so Pi decreases. [As Ei decreases, more neutrons are absorbed.]

For T1 < T2 , Pi |T1 > Pi |T2

E E

T T1 2

φ

φ

E iEi

26

Page 158: Larsen Lecture Notes

Resonance Escape Probabilities (and Integrals) for Non-Hydrogenous (A> 1) Moderators: We recall that

for a problem with no absorption,

φ(E) =S0

ζ E Σs(E)

To treat problems with resonance integrals, we shall assume that for a well-resolved resonance, F takes the

above form between resonances, i.e.,

EEEE2 1 o

(source energy)

Σ φ = ζ Ε

Σ φ = = ζ Ε ζ Ε

Σ φ = = ζ Ε ζ Ε

So

S P Po 1 2 Seff(3)

S Po 1 Seff(2)

s

s

s (no absorption)

Thus, the effective source driving the i-th resonance is

S(i)eff = S0 ( i−1

j=1 Pj )

and

Σs(E) φ(E) =S

(i)eff

ζϕi(E) Ei − Δ < E < Ei + Δ (63)

where

ϕi(E) � 1E

Ei − Δ < E < Ei−1 − Δ (64)

27

Page 159: Larsen Lecture Notes

We now define

Pi = resonance escape probability for the i− th resonance

= effective source rate to resonance − absorption rate in resonanceeffective source rate to resonance

=S

(i)eff

−∫

Ei+Δ

Ei−ΔΣa(E) φ(E) dE

S(i)eff

= 1 − ∫ Ei+Δ

Ei−Δ Σa(E) φ(E)

S(i)eff

dE

= 1 − ∫ Ei+Δ

Ei−ΔΣa(E)Σs(E)

ϕ(E)ζ dE

= 1 − NA

Σs(Ei) ζ

∫ Ei+Δ

Ei−Δσa(E) ϕ(E) dE

= 1 − NAζ

IiΣs(Ei)

(65)

= e− NA

ζ

IiΣs(Ei) (66)

Then, the total resonance escape probability over all the resonances Ei between E < Ei < E0 is

P (E) = E<Ei<E0 Pi = exp [− NAζ

∑E<Ei<E0

IiΣs(Ei)

] (67)

where

Ii =∫ Ei+Δ

Ei−Δ

σa(E) ϕ(E) dE = resonance integral (68)

and where ϕi(E) is the flux, normalized to satisfy

ϕi(E) � 1E, Ei − Δ < E < Ei−1 − Δ (69)

(just above the resonance).

Example Problem 1: We wish to compute a multigroup cross section

Σag =

∫ Eg−1

Egσa(E) ϕ(E) dE∫ Eg−1

Egϕ(E) dE

We assume that the resonances Ei, 1 ≤ i ≤ I,

Eg < EI < . . . . < Ei < . . . < E1 < Eg−1

lie in this group, are well-resolved, and that all absorptions occur in these resonances.

Solution: We shall assume all Pi are very close to one [see Eqn.(59)]. Then, we have approximately [with

Σs(E) � constant]

28

Page 160: Larsen Lecture Notes

ϕ(E) = 1E [ + O(ε)] so

Σag =NA

∫ Eg−1Eg

σa(E) 1E dE∫

Eg−1Eg

1E dE

[ + O(ε2) ]

=NA

∑I

i=1

∫iσa(E) ϕ(E) dE

lnEg−1

Eg

[ + O(ε2) ]

= NA

Δug

∑Ii=1 [ + O(ε2) ]

Example Problem 2: Show that the formulas Eqn.(58) [exact, for A=1] and Eqn.(66) [approximate, for

A ≥ 1] agree for A=1 and a narrow resonance:

εi =∫ Ei+Δ

Ei−Δ

Σa(E′)

E′ Σt(E′)dE

′<< 1 (70)

Solution: Let A=1. Then, normalizing ϕ(e) so that φ(E) = 1/E for E > Ei+Δ, Eqn.(57) exactly gives:

Σt(E) φ(E) = constantE exp [− ∫ E0

EΣa(E

′)

E′ Σt(E′) dE

′]

= constantE [ 1 + O(εi) ]

= Σt(Ei+Δ)E + O(εi) ]

= Σs(Ei+Δ)E + O(εi) ]

= Σs(Ei)E + O(εi) ]

so

1E Σt(E)

=φ(E)

Σs(Ei)+ O(εi) (71)

Now, by Eqn.(58),

Pi = e−εi = exp [− ∫i

Σa(E′)

E′ Σt(E′ ) dE

′]

= exp [− ∫i

Σa(E′) φ(E

′)

Σs(Ei)dE

′(1 + O(εi))]

= exp [− NA

Σs(Ei)

∫i

Σa(E′) φ(E

′) σs(Ei) dE

′(1 + O(εi))]

= exp [− NA

Σs(Ei)Ii + O(ε2i )]

For A=1, ζ = 1, and Eqn.(66) gives

Pi = exp [− NAΣs(Ei)

Ii]

Thus, the two formulas agree with a second-order O(ε2i ) error.

Approximate Evaluations of Resonance Integrals: Now we wish to compute the resonance integral

Ii =∫ Ei+Δ

Ei−Δ

σa ϕ(E) dE �∫i

σa ϕ(E) dE , (72)

29

Page 161: Larsen Lecture Notes

where:

1. We have a uniform mixture of an absorber A and a moderator M, both with mass numbers ≥ 1,

2. φ(E) � 1/E for E above the resonance,

3. ΣMγ << ΣMs � constant across the resonance,

4. The sources introduce neutrons into the system with energies higher than the resonances.

The slowing down equation nera the resonance is then

Σt(E) ϕ(E) =∫ E0

E Σs(E′ → E) ϕ(E

′) dE

=∫ E0

E[ΣMs (E

′) PM (E

′ → E) + ΣAs (E′) PA(E

′ → E)] ϕ(E′) dE

� ∫ E/αM

EΣM

s ϕ(E′)

(1 − αM ) E′ dE′

+∫ E/αA

EΣA

s ϕ(E′)

(1 − αA) E′ dE′

(73)

For many resonances of interest in reactor analysis, the width of the resonance is much smaller than the

typical energy loss in collision with a moderator nucleus.

EEi Ei

EiEi

α M/

−Δ−Δ

2 Δ << Ε − Εα M

ii

30

Page 162: Larsen Lecture Notes

Then, since φ(E′) � 1/E in almost all of the interval E < E

′< E/αM ,

∫ E/αM

EΣM

s ϕ(E′)

(1 − αM ) E′ dE′ � ΣM

s

(1 − αM )

∫ E/αM

E1

(E′)2dE

= ΣMs

(1 − αM ) (− 1E′ )E/αM

E

= ΣMs

(1 − αM ) (− αM

E − 1E′ )

= ΣMs

E ,

so Eqn.(74) becomes

Σt(E) ϕ(E) � ΣMsE

+∫ E/αA

E

ΣAs ϕ(E′)

(1 − αA) E′ dE′

(74)

Now we shall consider two cases:

Narrow Resonance (NR) Approximation: We assume that the resonance width is very small compared

to the average energy a neutron loses in a collision with an absorber nucleus. Then, as above,

∫ E/αA

E

ΣAs ϕ(E′)

(1 − αA) E′ dE′ � ΣAs

E

Thus, Eqn.(75) gives

Σt(E) ϕNR(E) =ΣMs + ΣAs

E=

Σs(Ei)E

so Eqn.(71) yields

Ii =∫i σa(E

′) ϕNR(E

′) dE

= Σs(Ei)∫i

σa(E′)

E′ Σt(E′ ) dE

= Σs(Ei)∫i

σa(E′)

E′ [ΣMt (E′ ) + ΣA

s (E′) + ΣAa (E′)] dE

= Σs(Ei)∫i

σa(E′)

E′ [ΣMs (Ei) + ΣA

s (Ei) + NA σAa (E′)] dE

=∫i

σa(E′)

E′ [1 + (NA / Σs) σa(E′)] dE

′(75)

Now we can use the Breit-Wigner formulas [Eqn.(62)].

Narrow Resonance Infinite Mass (NRIM) Absorber, or, Wide Resonance (WR) Approximation: Here we

assume that the mass number of the absorber is very large. Then αA � 1, and a resonance width is much

larger than the energy a neutron loses in a collision with an absorber nucleus. Because αA � 1,

∫ E/αA

E

ΣAs ϕ(E′)

(1 − αA) E′ dE′ � ΣAs (E) ϕ(E)

(1 − αA) E(E

αA− E) � ΣAs (E) ϕ(E)

31

Page 163: Larsen Lecture Notes

so Eqn.(75) gives

Σt(E) ϕ(E) = ΣMs

E + ΣAs (E) ϕ(E) so

ϕNRIM (E) = ΣMs

E [Σt(E) − ΣAs (E)]

and then Eqn.(73) yields

Ii =∫i

σa(E) ΣMs

E [Σt(E) − ΣAs (E)]

dE

=∫i

σa(E) ΣMs

E [Σs(E) + ΣAs (E)] dE

=∫i

σa(E)E

[1 + (NA / ΣMs ) σa(E)] dE (76)

Again, we can use the Breit-Wigner formulas [Eqn.(62)] to obtain an explicit expression for these integrals.

Slowing Down in a Finite Medium: The basis of much of our analysis will be the lethargy-dependent P1

equations. To derive these, we start with the lethargy-dependent transport equation

μ ∂∂xψ(x, μ, u) + Σt(u) ψ(x, μ, u)

= 12

∫ u0

∫ 1

−1[Σso(u

′ → u) + 3 μ μ′Σs1(u

′ → u)] ψ(x, μ′, u

′) dμ

′du

′+ 1

2 S0(x, μ)

Setting

φ(x, μ) =∫ 1

−1ψ(x, μ

′, u) dμ

′= scalar flux ,

φ(x, μ) =∫ 1

−1 μ′ψ(x, μ

′, u) dμ

′= current ,

we get

μ ∂∂xψ + Σt ψ = 1

2

∫ u0 Σso(u

′ → u) φ(x, u′) du

′+ 3μ

2

∫ u0 Σs1(u

′ → u) j(x, u′) du

′S0(x, u)

Now, we set

ψ � 12

[ φ(x, u) + 3 μ j(x, u) ]

and operate by∫ 1

−1(·) dμ and

∫ 1

−1μ(·) dμ to obtain

d

dxj(x, u) + Σt(u) φ(x, u) =

∫ u

0

Σso(u′ → u) φ(x, u

′) du

′+ S0(x, u) (78)

13

d

dxφ(x, u) + Σt(u) j(x, u) =

∫ u

0

Σs1(u′ → u) j(x, u

′) du

′(79)

32

Page 164: Larsen Lecture Notes

These are the lethargy-dependent P1 equations, with:

Σso(u′ → u) = Σs(u

′) eu

′−u

1 − α for u′< u < ln 1/α

= 0 otherwise(80)(80)

Σso(u′ → u) = Σso(u

′ → u) [A+ 1

2e

u′−u2 − A− 1

2e

u′−u2 ] (81)

Eqns.(79) and (80) form the basis of the most of the remaining work in this chapter. They are approximate

only because of the angular approximation made in the application of the P1 method. Because the scatter-

ing integrals in these equations are somewhat complicated, a rewriting of these integrals (for A=1) or an

approximation to these integrals (for A¿¿1) is usually also performed. (This only affects temrs that operate

on the u variable.)

To do this, we define the slowing down densities

qo(x, u) =∫ u

u′=0

∫ ∞

u′′=uΣso(u

′ → u′′) φ(x, u

′) du

′′du

′, (82)

q1(x, u) =∫ u

u′=0

∫ ∞

u′′=uΣs1(u

′ → u′′) j(x, u

′) du

′′du

′(83)

Differentiating Eqn.(83), we get

d

duqo =

∫ ∞

u′′=uΣso(u→ u

′′) φ(x, u) du

′′ −∫ u

u′=0

Σso(u′ → u) φ(x, u

′) du

′(84)

But, using Eqn.(81), we obtain

∫ ∞u′′=u Σso(u→ u

′′) du

′′=

∫ u+ln 1/α

uΣs(u)1 − α eu−u

′′du

′′

= Σs(u)1 − α eu (− e−u

′′)u+ln 1/αu

= Σs(u)1 − α eu [− e−u+lnα + e−u]

= = Σs(u)1 − α [− α + 1] = Σs(u) ,

so Eqn.(85) yields

d

duqo = Σs(u) φ(x, u) − intu

u′=0Σso(u

′ → u) φ(x, u′) du

′(85)

In a similar manner, Eqn.(84) yields

d

duq1 = [

∫ ∞

u′′=uΣso(u→ u

′′) ] du

′′j(x, u) −

∫ u

u′=0

Σs1(u′ → u) j(x, u

′) du

′(86)

However, using Eqn.(82), we obtain

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Page 165: Larsen Lecture Notes

∫ ∞u′′=u Σs1(u→ u

′′) du

′′=

∫ u+ln 1/α

u′′=uΣs(u)1 − α eu−u

′′[A+1

2 eu−u

′′2 − A−1

2 eu′′−u2 ] du

′′

= Σs(u)1 − α [e

32u A+1

2

∫ u+ln 1/α

u e−32u

′′du

′′ − e12u A−1

2

∫ u+ln 1/α

u e−12u

′′du

′′]

= Σs(u)1 − α [− A+1

3 e32u (e−

32u+ 3

2 lnα − e−32u) + (A − 1) e

12u (e−

12u+ 1

2 lnα − e−12u)]

= Σs(u)1 − α [− A+1

3 (α3/2 − 1) + (A − 1)(α1/2 − 1)]

= Σs(u) [A + 13

1 − α3/2

1 − α − (A − 1) 1 − α1/2

1 − α ]

= Σs(u) [A + 13

1 + α1/2 + α1 + α1/2 − (A − 1) 1

1 + α1/2 ]

= Σs(u) A + 11 + α1/2 [1 + α1/2 + α

3 − α1/2]

= Σs(u) A + 11 + α1/2

1 − 2 α1/2 + α3

= Σs(u) A + 11 + α1/2

(1 − α1/2)2

3

= 13 Σs(u) A + 1

1 + ( A − 1A + 1 )

[1 − A − 1A + 1 ]2

= 13 Σs(u) (A + 1)2

2 A [ 2A + 1 ]2

= 46 A Σs(u) = 2

3 A Σs(u)

Hence, if we define

μ0 =2

3 A(87)

then Eqn.(87) becomes

∂q1∂u

= μ0 Σso(u) j(x, u) −∫ u

u′=0

Σs1(u′ → u) j(x, u

′) du

′(88)

Inroducing Eqn.(86) into (79) and Eqn.(89) into (80), we get

∂j∂x(x, u) + Σt(u) φ(u) = Σs(u) φ(u) − ∂qo

∂x (x, u) + So(x, u)13∂φ∂x (x, u) + Σt(u) j(u) = μo Σs(u) j(u) − ∂q1

∂x (x, u)

or, defining

Σne(u) = Σt(u) − Σelastics (u) = nonelastic cross section

Σtr(u) = Σt(u) − μo Σs(u) = transport cross section(89)

we get

∂j

∂x(x, u) + Σne(u) φ(u) = − ∂qo

∂x(x, u) + So(x, u) (90)

13∂φ

∂x(x, u) + Σtr(u) j(u) = − ∂q1

∂x(x, u) (91)

This gives two equations in four unknowns. Eqns.(83) and (84) yield two other equations:

qo(x, u) =∫ u

u′=u−ln 1/α

∫ u+ln 1/α

u′′=u

Σs(u′)

1 − αeu

′−u′′φ(x, u

′) du

′′du

′(92)

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Page 166: Larsen Lecture Notes

q1(x, u) =∫ u

u′u−ln 1/α

∫ u+ln 1/α

u′′=u

Σs(u′)

1 − α[A + 1

2e

32 (u

′−u′′) − A − 1

2e

12 (u

′′−u′)] j(x, u

′) du

′′du

′(93)

However, the integral terms in these equations are too complicated to be of analytical use. To proced, we

shall consider special cases (for A=1) or approximations (for A >> 1).

Slowing Down in Hydrogen (A=1, α=0): Eqns.(93) and (94) become:

qo(x, u) =∫ uu′=0

∫ ∞u′′=u Σs(u

′) eu

′−u′′φ(x, u

′) du

′′du

=∫ uu′=0

Σs(u′) eu

′−u φ(x, u′) du

′ (94)

q1(x, u) =∫ uu′=0

∫ ∞u′′=u Σs(u

′) e3/2 (u

′−u′′) j(x, u

′) du

′′du

= 23

∫ uu′=0

Σs(u′) e3/2 (u

′−u) j(x, u′) du

′ (95)

These satisfy

∂qo

∂u = Σs(u) φ − qo∂q1∂u = 2/3 Σs(u) j − 3/2 q1

or

∂qo)(x,u) + qo(x,u) = Σs(u) φ(x,u)(96)

23∂q1)(x,u) + q1(x,u) = 4

9 Σs(u) j(x,u)(97)

These two equations are exact only for hydrogen (A=1). However, approximations that we will consider for

A >> 1 result in equations of this form, i.e.,

λo∂qo

∂u + qo = βo Σs φ ,

λ1∂q1∂u + q1 = β1 Σs j ,

(98)

with siutably defined constants λo, βo, λ1, andβ1.

Age Approximation: If A >> 1, then α � 1 and ln(1/α) � 0, so the interbals of integration in Eqns.(93)

and (94) are small. Hence, in the integrands we can use the following approximations:

Σs(u′) φ(x, u

′) � Σs(u) φ(x, u) + (u

′ − u) ∂∂uΣs(u) φ(x, u)

Σs(u′) j(x, u

′) � Σs(u) j(x, u) + (u

′ − u) ∂∂uΣs(u) j(x, u)

and get

qo = βo Σs(u) φ(x, u) − λo βo∂

∂uΣs(u) φ(x, u) (99)

q1 = β1 Σs(u) j(x, u) − λ1 β1∂

∂uΣs(u) j(x, u) (100)

35

Page 167: Larsen Lecture Notes

where

βo(α) =∫ u

u′=u−ln 1/α

∫ u+ln 1/α

u′′=u

eu′−u′′

1 − αdu

′′du

′= . . . = 1 + α

ln α

1 − α= ζ , (101)

λo(α) βo(α) =∫ u

u′=u−ln 1/α

∫ u+ln 1/α

u′′=u

eu′−u′′

1 − α(u − u

′) du

′′du

′= . . . = ζ − α

2 (1 − α)ln2 ζ , (102)

β1(α) =∫ u

u′u−ln 1/α

∫ u+ln 1/α

u′′=u

11 − α

[A + 1

2e

32 (u

′−u′′) − A − 1

2e

12 (u

′′−u′)] du

′′du

′(103)

λ1(α) β1(α) =∫ u

u′u−ln 1/α

∫ u+ln 1/α

u′′=u

11 − α

[A + 1

2e

32 (u

′−u′′) − A − 1

2e

12 (u

′′−u′)] (u − u

′) du

′′du

(104)

We set

βo = ζ (correct)

β1 = 0 (inconsistent)

λo = λ1 = 0 [low-order; first derivative terms in Eqn.(100) and (101)]

and get

qo(x, u) = ζ Σs(u) φ(x, u)

q1(x, u) = 0

Taking q1 = 0 in Eqn.() and eliminating j from Eqn.(92), we obtain

− ∂∂x

13 Σtr(u)

∂∂xφ(x, u) + Σne(u) φ(x, u) = − ∂

∂uqo(x, u) + So(x, u)

qo(x, u) = ζ Σs(u) φ(x, u)(105)

which is called the inconsistent age equation, or the inconsistent P1 equations, or the continuous slowing

down approximation. [The term ’inconsistent’ is used because q1 has been simply set to zero.] We will return

to these equations later.

Consistent Age Approximation: Here, we set

βo = ζ (consistent)

β1 = Eqn.() (consistent)

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Page 168: Larsen Lecture Notes

λo = λ1 = 0 (low-order)

Then, Eqns.(100) and (101) reduce to

qo = ζ Σs(u) φ(x, u)

qo = β1 Σs(u) j(x, u)

and Eqns.(91) and (92) yield

∂∂x j(u, x) + Σne(u) φ(x, u) = − ζ ∂

∂uΣs(u) φ(x, u) So13

∂∂xφ(u, x) + Σtr(u) j(x, u) = − β1

∂∂uΣs(u) j(x, u)

(106)

Grueling-Goertzel Approximation: In Eqns.(100) and (101),

q0 = ζ Σs φ − λo∂∂uζ Σs φ

q1 = β1 Σs j − λ1∂∂uβ1 Σs j

we use

q0 = ζ Σs φ

q1 = β1 Σs j

to write

q0 = ζ Σs φ − λo∂∂uqo

q1 = β1 Σs j − λ1∂∂uq1

or

λo∂∂uqo + q0 = ζ Σs φ

λ1∂∂uq1 + q1 = β1 Σs j

(107)

We note that these equations are of the desired form [i.e., of Eqn.(99)] and that they actuall reduce to the

correct equations for hydrogen (A=1). Thus, they are correct for both large and small A, and they turn out

to be reasonably accurate for immediate A. Eqn.(108) are used in conjunction with Eqns.(91) and (92) to

yield four equations in four unknowns.

37

Page 169: Larsen Lecture Notes

SUMMARY

∂∂x j(u, x) + Σne(u) φ(x, u) = − ∂

∂uqo(x, u) So(x, u)13

∂∂xφ(u, x) + Σtr(u) j(x, u) = − ∂

∂uq1(x, u)

λo∂∂uqo(x, u) + q0(x, u) = ζ Σs(u) φ(x, u)

λ1∂∂uq1(x, u) + q1(x, u) = β1 Σs(u) j(x, u)

(108)

λ0 β0 λ1 β1

Hydrogen (A=1) 1 1 2/3 4/9

Inconsistent Age (Continuous Slowing Down) 0 ζ 0 0

Consistent Age 0 ζ 0 Eqn.()

Grueling-Goertzel Eqn.() ζ Eqn.() Eqn.()

Treatment of Spatial Dependence (Three Ways): To continue this chapter, we shall describe three ways

to include the effects of spatial dependence in fat spectrum calculations. (All three of these methods assume

a homogenous spatial medium.)

1. Age diffusion theory, which combines diffusion theory in space with age theory in u,

2. Muft-Gam calculations, which use a buckling approximation in space applied to the lethargy-dependent

P1 equations,

3. The B1 method, which uses a buckling approximation in space applied to the lethargy-dependent trans-

port equation.

Age Diffusion [or Age, Fermi Age, Inconsistent Age, or Continous Slowing Down] Theory: The starting

point for this method is the lethargy-dependent diffusion Eqn.(106) with a delta function (in u) source:

− D(u) ∂2

∂x2φ(x, u) + Σne(u) φ(x, u) = − ∂∂uqo(x, u) + So(x) δ(u)

qo(x, u) = ζ Σs(u) φ(x, u)(109)

Eliminating φ, we obtain

∂uqo(x, u) +

Σne(u)ζ Σs(u)

qo(x, u) =D(u)ζ Σs(u)

∂2

∂x2qo(x, u) + So(x) δ(u)

Now, defining

qo(x, u) = q(x, u) exp [−∫ u

0

Σne(u′)

ζ Σs(u′)du

′] (110)

we obtain

38

Page 170: Larsen Lecture Notes

∂q∂u exp [− ∫ u

0Σne(u

′)

ζ Σs(u′ )du

′] + q exp [− ∫ u

0Σne(u

′)

ζ Σs(u′ )du

′] [− Σne

ζ Σs] + Σne

ζ Σsq exp [− ∫ u

0Σne(u

′)

ζ Σs(u′ )du

′]

= Dζ Σs

∂2

∂x2 q exp [− ∫ u0

Σne(u′)

ζ Σs(u′ ) du′] + So(x) δ(u)

or

ζ Σs(u)D(u)

∂q∂u = ∂2

∂x2 q + ζ Σs(u)D(u) exp [

∫ u0

Σne(u′)

ζ Σs(u′) du′] + So(x) δ(u)

= ∂2

∂x2 q + ζ Σs(0)D(0) So(x) δ(u)

(111)

Finally, we define the Fermi Age as

=∫ u

0

D(u′)

ζ Σs(u′)du

′[d

du=

D(u)ζ Σs(u)

] (112)

and we also define

q(x, u) = q(x, ) (113)

Then,

∂q

∂=

∂q

∂u

∂u

∂=

∂q

∂u

ζ Σs(0)D(0)

so Eqn.(112) becomes

∂q

∂=

∂2

∂x2q(x, ) + [

ζ Σs(0)D(0)

So(x)] δ(u) (114)

However, for � 0 and u � 0, Eqn.(113) gives

� uζ Σs(0)D(0)

= a u (115)

so u � 1a .

We have

δ(u) = δ(a

= a δ() =ζ Σs(0)D(0)

δ()) (116)

[ Proof:

∫ ∞−∞ δ(a ) f() d =

∫ ∞−∞ δ(t) f(at) a dt

= a∫ ∞−∞ δ(t) f(at) dt = a f(0) =

∫ ∞−∞ a δ() f() d ]

Therefore, Eqns.(115) and (116) yield

∂q

∂=

∂2

∂x2q(x, ) + So(x) δ() (117)

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Page 171: Larsen Lecture Notes

By Eqn.(113), u < 0 implies t < 0, and since q0(u) = 0 for u < 0, q = 0 for t < 0.

Operating on Eqn.(117) by∫ ε−ε(·) du, we obtain

q(x, ε) =∫ ε

0

∂2

∂x2q(x, ) d + So(x)

and letting ε→ 0, we get

q(x, 0+) = So(x)

Therefore, Eqn.(12) can be replaced by

∂q

∂=

∂2

∂x2q(x, ) > 0 (118)

q(x, 0) = So(x) ′initial condition′ (119)

Eqn.(118) is the Fermi Age Equation. It has the form of a standart time-dependent diffusion equation, with

playing the role of time. [Hence the term =’age’.] We note that is an increasing function of u. Hence,

E decreases → u increases → increases

Example Problem 1: Solve Eqn.(13) for an infinite medium with a planar source

So(x) = So δ(x − xo)

Solution:

q(x, ) = So1√4 π

e−(x − xo)2 / 4 (120)

[This is a special solution which cannot be derived by any of the methods we have described in these notes.]

Thus, as neutrons lose energy, they diffuse away from the source, and Eqn.(120) describes this quantita-

tively. The scalar flux is explicitly given by

φ(x, u) = qo(x,u)ζ Σs(u)

= 1ζ Σs(u) exp [− ∫ u

0Σne

ζ Σsdu

′] q

= 1ζ Σs(u) exp [− ∫ u

0Σne

ζ Σsdu

′] So 1√

4 πe−(x − xo)2 / 4

= 1ζ Σs(u) exp [− ∫ u

0Σne

ζ Σsdu

′] So 1√

4 π∫ 0

u Dζ Σs

e−(x − xo)2 / 4∫ 0 u D

ζ Σs

40

Page 172: Larsen Lecture Notes

x

q o (x , )

xo

τ

τ

τ

τ

1

2

3

0 < τ < τ < τ1 2 3

Ε > Ε > Ε > Εο 1 2 3

Example Problem 2: Now let us consider a point source in an infinite, homogenous, three-dimensional

medium. Problem (118) generalizes to:

∂q∂ (x, y, z, ) = ( ∂

2

∂x2 + ∂2

∂y2 + ∂2

∂z2 )q(x, y, z, )

q(x, y, z, 0) = So δ(x − xo) δ(y − yo) δ(z − zo)

Solution:

q(r, ) = Soe−|r − ro|2 / 4

(4 π )3/2

[This is another special solution which cannot be derived by any of the methods we have described in these

notes.] From this result, we obtain

< r2 > =∫ ∫ ∫ |r − ro|2 e−|r − ro|2 / 4

(4 π )3/2 dx dy dz

= 4 π∫ ∞r=0 r2 e−r2 / 4

(4 π )3/2 r2 dr

= 1√4 π

(4 )5/2

3/2

∫ ∞0

r2

4 π e−r2 / 4 r2

4 π d r√4π

= 16√π

∫ ∞0

t4 e−t2dt

= 8√π

∫ ∞0

y3/2 e−y dy

= 8√π

Γ(5/2) = 8√π

32

12 Γ(1/2) = 6

Therefore,

41

Page 173: Larsen Lecture Notes

< r2 > = 6 = the mean−square distance from the source that a neutron travels in slowing down to age

(121)

Muft-Gram Method: Here, one consistently solves Eqn.(109), but with an assumed aeparable form for

the spatial dependence:

φ(x, u) = φ(u) ei B x

j(x, u) = i j(u) ei B x

qo(x, u) = qo(u) ei B x

q1(x, u) = q1(u) ei B x

So(x, u) = So(u) ei B x

(122)

where B is the geometric buckling. [The idea is that f(x) = ei B x satisfies the equation

d2

dx2f(x) + B2 f(x) = 0 ,

which approximately describes the spatial variation of the angular flux.] Eqn.(109) reduce to

− B j(u) + Σne(u) φ(u) = − dduqo(u) So(u)

B3 φ(u) + Σtr(u) j(u) = − d

duq1(u)

λodduqo(u) + q0(u) = ζ Σs(u) φ(u)

λ1dduq1(u) + q1(u) = β1 Σs(u) j(u)

(123)

and we now have a system of equations with only one dependent variable u. This system is solved numeri-

cally. To show how this is done, we first introduce a group structure

u

u u

u

uu

1 n

1/2 3/2 n−1/2 n+1/20 =

(Usually, one takes Δu = constant). Then, integrating Eqn.(123) over the n-th group, we get

− B jn + Σne,n φn = − qo,n+1/2 − qo,n−1/2

Δu+ So,n

13B φn + Σtr,n jn = − q1,n+1/2 − q1,n−1/2

Δu

λoqo,n+1/2 − qo,n−1/2

Δu+ qo,n = ζ Σsn φn

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Page 174: Larsen Lecture Notes

λ1

q1,n+1/2 − q1,n−1/2

Δu+ q1,n = β1 Σsn jn

Now, we make the approximation

qk,n+1/2 − qk,n−1/2 � qk,n − qk,n−1 (k = 0, 1)

with

qk,−1 = 0 (k = 0, 1)

and we obtain

− B jn + Σne,n φn = − qo,n − qo,n−1

Δu+ So,n (124)

13B φn + Σtr,n jn = − q1,n − q1,n−1

Δu(125)

λoqo,n − qo,n−1

Δu+ qo,n = ζ Σsn φn (126)

λ1q1,n − q1,n−1

Δu+ q1,n = β1 Σsn jn (127)

These equations are now rearranged, Eqn.(126) and (127) yield

qo,n =λo qo,n−1 + Δu ζ Σsn φn

λo + Δu(128)

q1,n =λ1 q1,n−1 + Δu β1 Σsn jn

λ1 + Δu(129)

These are used to eliminate qon and q1n from Eqns.(124) and (125):

− B jn + Σne,n φn = qo,n−1 − ζ Σsn φn

λo + Δu + Son

13 B φn + Σtr,n jn = q1,n−1 − β1 Σsn jn

λ1 + Δu

The second of these equations is now solved for jn:

jn =q1,n−1 − B/3 (λ1 + Δu) φnβ1 Σsn + (λ1 + Δu) Σtr,n

(130)

and this is used to eliminate jn from the first:

φn =( qo,n−1λo + Δu ) + ( B q1,n−1

β1 Σsn + (λ1 + Δu) Σtr,n) + Son

( ζ Σsn

λo + Δu

) + (B2 (λ1 + Δu) / ζ

β1 Σsn + (λ1 + Δu) Σtr,n) (131)

Now, for n=1, q00 = q10 = 0, and

Eqn.(27) → φ1 → Eqn.(128) → q01

φ1 → Eqn.(130) → j1 → Eqn.(129) → q11

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Page 175: Larsen Lecture Notes

Having q01 and q11, we set n=2, and then

Eqn.(27) → φ2 → Eqn.(128) → q02

φ2 → Eqn.(130) → j2 → Eqn.(129) → q12

Eqns.(128)-(131) are solved recursively, in this manner, to get φn, jn, andq1n for all n ≥ 1.

B1 Method: Here we use the same separable spatial approximations as in the Muft-Gam method, but

the starting point is the lethargy-dependent transport equation (78) written below, rather than the lethargy-

dependent P1 equations:

μ ∂∂xψ(x, μ, u) + Σt(u) ψ(x, μ, u) = 1

2

∫ u0

Σso(u′ → u) φ(x, u

′) du

+ 3μ2

∫ u0

Σs1(u′ → u) j(x, u

′) du

′+ 1

2 So(x, u) ,(132)

where

φ(x, u) =∫ 1

−1

ψ(x, μ′, u) dμ

′, (133)

j(x, u) =∫ 1

−1

μ′ψ(x, μ

′, u) dμ

′, (134)

Using Eqns.(86) and (89), Eqn.(132) becomes:

μ ∂∂xψ(x, μ, u) + Σt(u) ψ(x, μ, u) = 1

2 [Σs(u) φ(x, u) − ∂∂uqo(x, u)]

+ 3μ2 [μo Σso(u) j(x, u) − ∂

∂uq1(x, u)] + 12 So(x, u)

(135)

Also, q0 and q1 are now approximated by Eqn.(109):

λo∂

∂uqo(x, u) + qo(x, u) = βo Σs(u) φ(x, u) (136)

λ1∂

∂uq1(x, u) + q1(x, u) = β1 Σs(u) j(x, u) (137)

Introducing the same assumed separable form for the spatial variation as before,

So(x, u) = So(u) ei B x

ψ(x, μ, u) = ψ(μ, u) ei B x

φ(x, u) = φ(u) ei B x

j(x, u) = i j(u) ei B x

qo(x, u) = qo(u) ei B x

q1(x, u) = q1(u) ei B x

into Eqn.(135), we get

Eqn.(135) → [i B u + Σt(u)] ψ =12

[Σs φ − d

duqo] +

3μ2

[μo Σso j − d

duq1] +

So2

(138)

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Page 176: Larsen Lecture Notes

Eqn.(133) → φ =∫ 1

−1

ψ dμ′

(139)

Eqn.(134) → i j =∫ 1

−1

μ′ψ dμ

′(140)

Eqn.(136) → λo∂

∂uqo(u) + qo(u) = βo Σs(u) φ(u) (141)

Eqn.(137) → λ1∂

∂uq1(u) + q1(u) = β1 Σs(u) j(u) (142)

Integrating Eqn.(138) over μ, we obtain

(i B) (i j) + Σt φ = Σs φ − d

duqo (143)

or

− B j(u) + Σne(u) φ(u) = − d

duqo(u) + So(u) (144)

Using Eqn.(144), we write Eqn.(138) as

ψ(μ, u) = 1i B μ − Σt(u)

12 [− B j(u) + Σt(u) φ(u)]

+ i μi B μ + Σt(u)

32 [μo Σs(u) j(u) − d

duq1(u)]

Integrating this equation over μ, we obtain (exactly)

φ(u) = (I1) [− B j(u) + Σt(u) φ(u)] + (I2) [μo Σs(u) j(u) − d

duq1(u)] (145)

where

I1 =12

∫ 1

−1

du

i B μ + Σt(u)= . . . =

1Btan−1 B

Σt(u)(146)

I2 =32

∫ 1

−1

i μ du

i B μ + Σt(u)= . . . =

3B

[1 − Σt(u)B

tan−1 B

Σt(u)] (147)

Eqns.(145)-(147) now give, after some rearranging,

B

3φ(u) + γ(u) Σt(u) j(u) = − d

duq1(u) (148)

γ(u) =B

3 Σt(u) tan−1 B

Σt(u)

1 − Σt(u)B tan−1 B

Σt(u)

− μoΣso(u)Σt(u)

(149)

Eqns.(141),(142),(144), and (148) are the basis of the B1 method. They are identical to Eqn.(123) if we

identify

Σtr(u) = γ(u) Σt(u)

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Page 177: Larsen Lecture Notes

Thus, numerical methods for solving the Muft-Gam equations can be used to numerically solve the B1 equa-

tions.

Remark 1: Because no approximation is made in the angular variable, the B1 method is more accurate

than the Muft-Gam method. [Eqn.(123)].

Remark 2: To give a feeling for the effect of the buckling terms in the Muft-Gam and B1 methods, let

us consider an ’inconsistent’ approximation in which we set q1 = 0 in the second of Eqn.(123) and ignore

the fourth of these equations. We then have the system:

− B j(u) + Σne(u) φ(u) = − dduqo(u) + So(u)

B3 φ(u) + Σtr(u) j(u) = 0

λodduqo(u) + qo(u) = ζ Σs(u) φ(u)

(150)

The second of these equations can be written

j(u) = − B

3 Σtr(u)φ(u)

and thus, the equation can be written

[B2

3 Σtr(u)φ(u) + Σne(u)] φ(u) =

qoλo

− ζ

λoΣs(u) φ(u) + S (151)

However, the third of Eqn.(150) gives

qo =βoλo

∫ u

0

e−(u−u′)/λo Σs(u

′) φ(u

′) du

so Eqn.(151) can be written

[B2

3 Σtr(u)φ(u) + Σne(u) +

ζ

λo] φ(u) =

βoλ2o

∫ u

0

e−(u−u′)/λo Σs(u

′) φ(u

′) du

′+ S(u) (152)

This is an equation only involving φ. The terms on the left side of this equation represent losses due to

collisions, while the terms on the right side represent gains due to downscattering into lethergyu and sources.

The spatial dependence here is represented by the term B2, which produces an effective increase in the col-

lision rate on the left side of the equation without corrsponding increase in the source on the right side.

Therefore, the space dependence here acts like an extra removal term; this is correct because the net effect

of the spatial variation is to allow neutrons to leak out of the system, and this is a loss mevhanism. The

larger the system the smaller is B, and hence the smaller is the leakage term in Eqn.(151).

46