Intro to Neutron Transport

Introduction to neutron transport
Fausto Malvagi
To cite this version:
Fausto Malvagi. Introduction to neutron transport. Master. Neutronique I, INSTN, Saclay, France. 2024.
⟨hal-05007454v1⟩
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Submitted on 26 Mar 2025 (v1), last revised 6 Mar 2026 (v4)
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Neutron-Matter Interactions
The transport equation
Analytical solutions
The Transport Equation (1/6)
Fausto Malvagi
CEA - IRSN - INSTN
2024
1 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
1
Neutron-Matter Interactions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
2
The transport equation
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
3
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
2 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Introduction (1)
When I look at the behavior of a single neutron interacting
with a nuclide, this is described by nuclear physics
cross sections
kinematic parameters
When I look at the behavior of a neutron population moving
in a fixed medium, this is described by transport theory
neutron density (or flux)
chain reactions
When I look at the behavior of a system (a reactor)
interacting with a neutron population, this is described by
reactor physics
heat production (energy deposition) from fission and
absorption changes the temperature
transmutation from neutron reactions and radioactive decay
changes the composition
3 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
Outline
1
Neutron-Matter Interactions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
2
The transport equation
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
3
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
4 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
Neutron-Matter Interactions (1)
Potential Scattering
(n, n) elastic scattering
Compound nucleus formation
(n, n) elastic scattering
(n, n′ ) inelastic scattering
(n, γ) radiative capture
(n, f ) fission
(n, 2n) · · · others
5 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
Neutron-Matter Interactions(2)
In transport theory we want to know:
How many neutrons exit from an interaction
none disappearance
one elastic or inelastic scattering
a few (n, 2n), (n, 3n), . . .
0 − 7 fission −→ possibility of a chain reaction!
Their kinematic parameters
energy E = 1/2mn vn2
Eout ≤ Ein for scattering ?
angle anisotropy of deflection
For some reactions there is a correlation between outgoing
energy and outgoing angle
6 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
Microscopic Cross Sections (1)
Let’s consider a one-speed,
mono-directional neutron beam
impinging normally on a thin target:
NA atoms/cm2
Reaction rate
The neutron-matter
reaction rate is
proportional to the
intensity of the beam I
and to the atom density
NA :
I neutrons/cm2/s
R
#
cm2 · sec
=
σ
2
= cm
I
N
A
#
#
cm2 · sec cm2
7 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
Microscopic Cross Sections (2)
We can define the microscopic cross section σ as an
interaction probability:
σ=
R/NA
nb reactions/nucleus/s
=
I
nb neutrons/cm2 /s
The microscopic cross sections for different reactions are
additive:
total σt = σa + σs
scattering σs = σel + σinel
absorption σa = σc + σf + · · ·
The ratio σr /σt is the probability that, if there is an
interaction, the process chosen is r .
8 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
All the physics for transport theory
Cross Section Regions
Evaluated Nuclear Data Files
All cross sections for ≈ 400
nuclides (with uncertainties) and
two dozens materials
ENDF/B USA V-8.0
JEFF Europe V-3.3
JENDL Japan V-5.0
TENDL Thalis group (NRJ)
Thermal region (E < 5eV )
Resolved Resonances Range
FENDL Specific fusion
BROND Russia
Unresolved Resonances
Range
Continuum
9 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
Macroscopic cross sections (1)
Let us consider a one-speed neutron beam of intensity I
impinging normally on a thick slab:
Reaction rate R in [x, x + dx]:
I0
dR = σt INdx = Σt Idx
-
Macroscopic cross section
x = 0 x x + dx
Σt ≡ σt N
We can compute the uncollided intensity according to
-
dI = −dR →
dI
= −Σt I → I (x) = I0 e −Σt x
dx
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
Macroscopic cross sections (2)
Mean free path
Z ∞
λ = ⟨x⟩ =
xp(x)dx
0
p(x)dx probability of first collision between x and x + dx
e −Σt x probability of traveling the distance x without
collisions
Σt dx probability of having a collision in dx
Z ∞
λ=
xe −Σt x Σdx = 1/Σ
0
The macroscopic cross section is the inverse of a mean free
path.
11 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
Macroscopic cross sections (3)
We can define partial and total macroscopic cross sections
Σr = Nσr
Σt = Nσt = Σr 1 + Σr 2 + · · ·
.
We can define macroscopic cross sections for a mixture of
nuclides X , Y , . . .
Y
Σt = ΣX
t + Σt + · · ·
The ratio ΣX
t /Σt is the probability that, if a collision occurs in
the mixture, it concerns the component X .
Collision frequency
The probability of collision per unit volume per unit time is
Σt dx/dt = v Σt where v is the speed of the neutron.
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
The phase space
A ”classical” neutron is characterized by
its position ⃗r and velocity ⃗v .
θ
Velocity is frequently replaced by kinetic
energy E = 1/2mv 2 and direction of
propagation Ω̂ = ⃗v /∥⃗v ∥.
ϕ
Spherical coordinates
Ω̂ = sinθ cosϕ êx + sinθ sinϕ êy + cosθ êz
Z
d2 Ω ≡ sinθ dθ dϕ
Z 2π Z π
d2 Ω f (Ω̂) =
dϕ
dθ sinθf (θ, ϕ)
4π
0
0
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
Differential cross sections
’
P(E → E ′ , Ω̂ → Ω̂′ )dE ′ d2 Ω′ : probability that a neutron with
(E , Ω̂) before the collision will be in the volume
dE ′ d2 Ω′ around (E ′ , Ω̂′ ) after the collision.
(E ′ , Ω̂′ )
(E , Ω̂)
θ: deflection angle
differential cross section
Σs (⃗r , E → E ′ , Ω̂ → Ω̂′ ) ≡ Σs (⃗r , E )P(E → E ′ , Ω̂ → Ω̂′ )
normalization condition:
Z
Z ∞
′
d2 Ω
dE ′ Σs (⃗r , E → E ′ , Ω̂ → Ω̂′ ) = Σs (⃗r , E )
4π
0
14 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
Outline
1
Neutron-Matter Interactions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
2
The transport equation
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
3
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
15 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The neutron population
In a reactor the number of neutrons is large enough
(≈ 108 /cm3 ) to justify a statistical treatment
→ the neutron density is the average number of neutrons per
unit of volume of the phase space.
The number of neutrons is very small compared to the
number of atoms of the underlying medium (ratio ≈ 10−15 )
→ we can neglect neutron-neutron interactions.
We treat the neutrons as classical neutral particles
→ they are completely described by (⃗r , ⃗v )
→ they travel in straight lines with exponential flights
between collisions
We neglect neutron decay, relativity, gravity, spin effects, etc.
16 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The (angular) neutron density
n(⃗r , ⃗v )d3 r d3 v number of neutrons in the volume d3 r around ⃗r
and with velocity in the volume d3 v around ⃗v .
n(⃗r , E , Ω̂)d3 r dE d2 Ω number of neutrons in the volume d3 r
around ⃗r and with energy between E and E + dE and
direction of propagation in the volume d2 Ω around Ω̂.
Change of variables
n(⃗r , E , Ω̂)d3 r dE d2 Ω = n(⃗r , ⃗v )d3 r d3 v
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The neutron flux
angular flux the number of neutrons per second which cross a
unit surface perpendicular to the direction of
propagation [#/cm2 /s].
ψ(⃗r , E , Ω̂, t) = vn(⃗r , E , Ω̂, t)
scalar flux number of neutrons in the volume d3 r around ⃗r and
with energy between E and E + dE .
Z
ϕ(⃗r , E , t) =
d2 Ω ψ(⃗r , E , Ω̂, t)
4π
Reaction rates
R(⃗r , E , t) = Σ(⃗r , E ) ϕ(⃗r , E , t) number of reactions per second
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The neutron current (1)
current density ⃗j(⃗r , E , Ω̂, t) = Ω̂ψ(⃗r , E , Ω̂, t) = ⃗v n(⃗r , E , Ω̂, t)
neutron current number of neutrons which cross the surface d2 S
per unit time
n̂
Ω̂
n̂S · ⃗j(⃗r , E , Ω̂, t) d2 S dE d2 Ω
d2 S
19 / 47
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
Neutron-Matter Interactions
The transport equation
Analytical solutions
The neutron current (2)
partial current number of neutrons which cross the surface d2 S
per unit time in a given sense
J+
J−
+
Z
d2 Ω n̂ · ⃗j(⃗r , E , Ω̂, t)
J (⃗r , E , t) =
n̂·Ω̂≥0
n̂
d2 S
J − (⃗r , E , t) =
Z
d2 Ω n̂ · ⃗j(⃗r , E , Ω̂, t)
n̂·Ω̂≤0
net current: the net number of neutrons which cross a surface in
the direction of the normal n̂ · J⃗ = J + − J − .
Exercise
Show that for an isotropic flux and a unit surface with arbitrary
orientation we have that the partial current is J + = ϕ/4 and the net
current is zero.
20 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The transport equation(1)
Linearized form of the Boltzmann equation from gas kinetics.
We write an equation for the neutron balance in an arbitrary
volume V enclosed by the surface ∂V , for neutrons of velocity
⃗v .
V
∂V
Changes in time of n
= changes due to streaming
+ changes due to collisions
+ changes due to sources.
21 / 47
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
Neutron-Matter Interactions
The transport equation
Analytical solutions
The transport equation(2)
time change term change of the number of neutrons of velocity
⃗v ± d3 v in the volume V per unit time
Z
∂
d3 r d3 v n(⃗r , ⃗v , t)
∂t V
external source term number of neutrons born in volume V with
velocity ⃗v ± d3 v from external sources per unit time
Z
d3 r d3 v s(⃗r , ⃗v , t)
V
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Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
Neutron-Matter Interactions
The transport equation
Analytical solutions
The transport equation(3)
streaming term number of neutrons of velocity ⃗v ± d3 v which
enter the volume V minus the number of neutrons
which exit, per unit time
Z
−
d2 S d3 v n̂S · ⃗j(⃗r , ⃗v , t)
∂V
Z
⃗ · ⃗j(⃗r , ⃗v , t)
d3 r d3 v ∇
(∗)
⃗ r , ⃗v , t)
d3 r d3 v ⃗v · ∇n(⃗
(∗∗)
=
V
V
∂V
Z
=
V
(*): divergence theorem
(**): ⃗v and ⃗r are independent variables
23 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The transport equation(4)
collision term minus the number of neutrons with velocity ⃗v ± d3 v
which collide (they are either absorbed or scattered)
per unit time
Z
−
d3 r d3 v Σ(⃗r , v )vn(⃗r , ⃗v , t)
V
plus the number of neutrons which scatter from ⃗v ′
into ⃗v ± d3 v , for any ⃗v ′ , per unit time
Z
Z
+
d3 r d3 v d3 v ′ Σs (⃗r , ⃗v ′ → ⃗v )v ′ n(⃗r , ⃗v ′ , t)
V
24 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The transport equation(5)
Putting everything together we obtain:
Z
Z
∂n
′
′
′
⃗
d3 r
+ ⃗v · ∇n + Σvn − d3 v Σs (⃗v → ⃗v )v n − s = 0
∂t
V
Recalling that the volume of integration V was arbitrary, we
obtain the transport equation
Z
∂n
⃗ + Σvn = d3 v ′ Σs (⃗v ′ → ⃗v )v ′ n + s
+ ⃗v · ∇n
∂t
Changing the independent variable from ⃗v to E , Ω̂ and the
dependent variable from n to ψ
Z ∞ Z
1 ∂ψ
⃗ + Σψ =
+ Ω̂ · ∇ψ
dE ′ d2 Ω′ Σs (E ′ , Ω̂′ → E , Ω̂)ψ + s
v ∂t
0
4π
25 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The canonical form
The transport equation
1 ∂ψ
⃗ + Σ(⃗r , E )ψ(⃗r , E , Ω̂, t) = s(⃗r , E , Ω̂, t)
+ Ω̂ · ∇ψ
v ∂t Z
Z
∞
dE ′
+
0
d2 Ω′ Σs (⃗r , E ′ , Ω̂′ → E , Ω̂)ψ(⃗r , E ′ , Ω̂′ , t)
4π
Linear equation (if the cross sections are independent of the
flux!) −→ Possible use of Green’s functions
Integro-differential equation. Differential in time and space,
integral in energy and angle −→ Continuity of the solution?
Both hyperbolic (Σs → 0) and elliptic (Σs → Σ) behaviors
⃗ term
Boundary layers are possible ←− Ω̂ · ∇ψ
26 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The limit conditions
Differential equations need initial and boundary conditions
Initial condition
The flux is known everywhere at some initial time
ψ(⃗r , E , Ω̂, t = t0 ) = Λ(⃗r , E , Ω̂)
Boundary condition
The incoming flux is known at all times
V
∂V
ψ(⃗rs ∈ ∂V , E , Ω̂, t) = Γ(⃗rs , E , Ω̂, t)
n̂out · Ω̂ < 0
The outgoing flux is part of the solution!
27 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
Neutron sources
Spontaneous fission is an inhomogeneous source
(α, n) reactions is another inhomogeneous source
Neutron-induced fission is a homogeneous source
The (induced) fission term
Fψ ≡
Z
X χi (E ) Z ∞
dE ′ ν i (E ′ )Σif (⃗r , E ′ ) d2 Ω′ ψ(⃗r , E ′ , Ω̂′ )
4π
0
4π
i
ν i (E ) average number of neutrons per fission; it depends on
the incident neutron energy and on the target isotope
χi (E ) fission spectrum: energy distribution of neutrons
from fission; it depends on the target isotope
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
A Cauchy problem
The full transport equation:
1 ∂ψ
+ Lψ = Hψ + F ψ + s(⃗r , E , Ω̂, t)
v ∂t
⃗ + Σ(⃗r , E )ψ(⃗r , E , Ω̂, t)
Lψ ≡ Ω̂ · ∇ψ
Z ∞ Z
Hψ ≡
dE ′ d2 Ω′ Σs (⃗r , E ′ → E , Ω̂′ → Ω̂)ψ(⃗r , E ′ , Ω̂′ , t)
0
4π
Initial value problem whose solution exists for t ∈ [t0 , T ) and
is unique
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The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
Criticality calculations (k-eigenvalue problem)
In reactor physics we seek a steady state solution with neutron
induced fissions only
Lψ = Hψ + F ψ
We consider the associated eigenvalue problem
Lψ = Hψ +
1
Fψ
k
(∗)
We can always find (many) couples (k, ψ)i which satisfy the
critical equation (*). The highest k is called the keff .
The multiplication factor keff
keff is the number by which we need to divide the fission term in
order to obtain a steady state solution.
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The power iteration
The power iteration is the numerical algorithm usually used to
solve criticality problems. We start from an initial guess ψ 0
for the flux. We then compute by iteration:
Lψ n = Hψ n +
kn =
1
k n−1
n
Fψ
F ψ n−1
F ψ n−1 /k n−1
It can be shown that (k n , ψ n ) −−−→ (keff , ψ) where keff is the
n→∞
highest eigenvalue
The solution of the eigenvalue problem is thus reduced to the
iterative solution of many fixed source problems
31 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The α-eigenvalue problem
We look for the time behavior by separation of variables
Time-dependent solution
ψ(⃗r , E , Ω̂, t) = ψ̃(⃗r , E , Ω̂)e −αt
⃗ ψ̃ + Σ + α ψ̃ = H ψ̃ + F ψ̃
Ω̂ · ∇
v
(∗)
We have a new eigenvalue problem. We now have to look for
the couples (α, ψ̃)i such that equation (∗) is satisfied.
The infinite number of couples are the time-dependent modes
of the transport equation. The mode corresponding to the
smallest α is the asymptotic mode
In general the α-eigenvalue modes are different from the
k-eigenvalue modes. They are equal if k = 1 (α = 0)
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
Brute force discretization (1)
We need to partition the space (⃗r , E , Ω̂) in ”cells” where the
coefficients Σ are constant and the solution ψ is linear.
Spatial discretization
The system is of the order of the meter (reactor core), the
heterogeneities are of the order of the millimeter (fuel pin rings)
−→ (103 )3 ≈ 109 meshes
Angular discretization
102 − 103 angles are needed to describe the angular flux
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
Brute force discretization (2)
Energy discretization
For a linear reconstruction of the
cross section it takes between 600
(1 H) and 150000 points (238 U). A
unified energy grid demands a few
million points.
Discretization of the transport equation
Total number of meshes for a complete discretization of a
criticality (steady state) calculation:
109 × 103 × 106 = 1018 meshes
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
Industrial calculations constraints
Precision required for neutronic calculations:
keff ∼ 100 pcm (0.1 %)
pin power ∼ 1 %
Computing time required for neutronic calculations:
3D core at steady state a few minutes
cycle lenght a few hours
Simplified (but complex!) computational schemes are
developed in order to satisfy those requirements. They need
to be validated and their domain of validity established.
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
The Monte-Carlo method
Particle simulation (no PDF equation solved!)
Neutrons are followed from beginning (birth) to end
(absorption or leakage) throughout their random walk.
Almost no approximations are made in space and energy
Complex and detailed geometries can be treated
Pointwise cross sections are used
Convergence is slow:
√ statistical uncertainty on macroscopic
quantities is ∼ 1/ N with N the number of histories
simulated
Running time: ∼ 1 million histories per hour of simulation per
cpu.
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Neutron-Matter Interactions
The transport equation
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
Outline
1
Neutron-Matter Interactions
Microscopic Cross Sections
Macroscopic Cross Sections
The phase space
2
The transport equation
Neutron Population
Derivation of the transport equation
Initial and boundary conditions
Criticality calculations
3
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
37 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
Purely absorbing media (1)
No scattering
If we set Σs = 0, then neutrons never change their speed nor their
direction of propagation. Energy is then just a parameter and we
can write
⃗ + Σψ(⃗r , Ω̂) = s(⃗r , Ω̂)
Ω̂ · ∇ψ
Ω̂
⃗rs
u
⃗r
38 / 47
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
Neutron-Matter Interactions
The transport equation
Analytical solutions
Purely absorbing media (2)
We separate the two cases of homogeneous and heterogeneous
media
Homogeneous medium Σ ̸= Σ(⃗r )
ψ(⃗r , Ω̂) = Γ(⃗rs , Ω̂)e
−Σ|⃗
r −⃗
rs |
Z |⃗r −⃗rs |
+
du s(⃗r − u Ω̂, Ω̂)e −Σ|⃗r −uΩ̂|
0
Heterogeneous medium Σ = Σ(⃗r )
ψ(⃗r , Ω̂) = Γ(⃗rs , Ω̂)e
−τ (⃗
r ,⃗
rs )
Z |⃗r −⃗rs |
+
du s(⃗r − u Ω̂, Ω̂)e −τ (⃗r ,⃗r −uΩ̂)
0
39 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
Optical distance
We can measure the distance between two points not in cm, but in
mean free paths
Optical distance τ (⃗r0 , ⃗r1 )
The distance between two points ⃗r0 and ⃗r1 measured in units of
mean free paths
τ (⃗r1 , ⃗r0 ) =
Z |⃗r1 −⃗r0 |
du Σ(⃗r0 + u Ω̂)
Ω̂ = ⃗r0 − ⃗r1
0
Note the symmetry relationship: τ (⃗r0 , ⃗r1 ) = τ (⃗r1 , ⃗r0 )
40 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
The point kernel (1)
Point kernel
We consider the transport equation with an isotropic point source:
⃗ + Σψ(⃗r , Ω̂) =
Ω̂ · ∇ψ
S0
δ(⃗r )
4π
Angular flux solution
ψ(⃗r , Ω̂) =
S0 e −τ (r )
δ(Ω̂ − Ω̂r )
4πr 2
Ω̂r = ⃗r /r
Scalar flux solution
ϕ(⃗r ) =
S0 e −τ (r )
4πr 2
41 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
The point kernel (2)
Point kernel with spatially distributed source
We consider the transport equation with an isotropic distributed
source:
⃗ + Σψ(⃗r , Ω̂) = S0 (⃗r )
Ω̂ · ∇ψ
4π
We can use the previous solution for the point source as a Green’s
function
Scalar flux solution for distributed source
Z
′
e −τ (⃗r ,⃗r )
S (⃗r ′ )
ϕ(⃗r ) = d3 r ′
2 0
′
4π |⃗r − ⃗r |
42 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
The integral form (1)
We rewrite the transport equation:
⃗ + Σψ(⃗r , E , Ω̂) = Q(⃗r , E , Ω̂)
Ω̂ · ∇ψ
where
Q(⃗r , E , Ω̂) ≡
Z ∞ Z
dE ′ d2 Ω′ Σs (⃗r , E ′ → E , Ω̂′ → Ω̂)ψ(⃗r , E ′ , Ω̂′ ) + s
0
4π
We assume that the sources and the scattering are isotropic:
s(⃗r , E , Ω̂) =
S0 (⃗r )
4π
Σs (⃗r , E ′ → E , Ω̂′ → Ω̂) =
1
Σs (⃗r , E ′ → E )
4π
43 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
The integral form (2)
In this case the term Q doesn’t depend on angle
Q(⃗r , E , Ω̂) =
Q0
4π
Z ∞
dE ′ Σs (⃗r , E ′ → E )ψ(⃗r , E ′ ) + S0 (⃗r )
Q0 (⃗r , E ) ≡
0
and thus we can formally solve as in the Point Kernel example
Peierls’ equation
Z
ϕ(⃗r ) =
′
d3 r ′
e −τ (⃗r ,⃗r ;E )
Q (⃗r ′ , E )
2 0
′
4π |⃗r − ⃗r |
This is an integral equation for the scalar flux ϕ(⃗r , E ).
44 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
The one-speed equation in 1D (1)
µ
∂ψ
Σs
+ Σ(x)ψ(x, µ) =
ϕ(x) + s(x, µ)
∂x
2
Separation of variables
1
We look for a separable solution ψ(x, µ) = χ(x)φ(µ)
2
We end up with an eigenvalue problem Aφλ = λφλ where λ is
called the separation constant
3
The full solution is of the form:
X
ψ(x, µ) =
aλ (x)φλ (µ)
λ
45 / 47
Neutron-Matter Interactions
The transport equation
Analytical solutions
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
The one-speed equation in 1D (2)
−ν0
-1
1
ν0
ν
Solution
ψ(x, µ) = a0+ φ0+ (µ)e −x/ν0 + a0− φ0− (µ)e x/ν0
Z 1
+
dνA(ν)φν (µ)e −x/ν
−1
a0+ , a0− , A(ν) constants of integration
φ0+ , φ0+ , φ0 eigenfunctions
ν0 , ν eigenvalues
46 / 47
Purely absorbing media
The point kernel
Integral form
One-speed, one-dimension
Neutron-Matter Interactions
The transport equation
Analytical solutions
The one-speed equation in 1D (3)
Dispersion relation for eigenvalues
λ(ν) ≡ 1 −
cν 1 + ν
ln
=0
2
1−ν
Discrete eigenfunctions
φ0± =
cν0
2(ν0 ∓ ν)
Singular eigenfunctions
φν (µ) =
cν
1
P
+ λ(ν)δ(ν − µ)
2 ν−µ
−1≤ν ≤1
47 / 47
The diffusion equation
Examples
One Dimensional Problems
The Diffusion Equation
Space Problem
Fausto Malvagi
IRSN - CEA - INSTN
2024
1 / 34
The diffusion equation
Examples
One Dimensional Problems
1
The diffusion equation
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
2
Examples
The naked slab with sources
The critical slab
The reflected critical slab
3
One Dimensional Problems
Helmoltz’s Equation
The albedo
2 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
Outline
1
The diffusion equation
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
2
Examples
The naked slab with sources
The critical slab
The reflected critical slab
3
One Dimensional Problems
Helmoltz’s Equation
The albedo
3 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
The one-speed transport equation
We assume that all neutrons have the same speed, and this
speed is not affected by scattering. The steady-state transport
equation can be written as:
⃗ + Σ(⃗r )ψ(⃗r , Ω̂) =
Ω̂ · ∇ψ
Z
=
d2 Ω′ Σs (⃗r , Ω̂′ → Ω̂)ψ(⃗r , Ω̂′ ) + s(⃗r , Ω̂)
4π
This assumption is non-physical, but we will see that the
energy dependent problem is usually solved as a system of
one-speed equations (the multi-group method).
4 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
From transport to diffusion
The transport equation is exact but difficult to solve because
of its 6-dimensional phase space.
The unknown of the transport equation is the angular flux ψ,
but we really only need the scalar flux ϕ in order to compute
the reaction rates Σr ϕ (like the fission rates Σf ϕ).
We would like to have an equation for the scalar flux directly.
There are several different ways to derive the diffusion equation for
the scalar flux starting from the transport equation using different
sets of assumptions
5 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
How to derive the diffusion equation (1)
I. Taylor expansion
We start from the integral form of the transport equation with
isotropic scattering and we assume that the scalar flux has a weak
dependence on space
⃗ + Higher Order Terms (in space)
ϕ(⃗r ) ≈ ϕ(⃗r0 ) + (⃗r − ⃗r0 ) · ∇ϕ
II. Spherical harmonics expansion
We start from the integro-differential form and we assume that the
angular flux has a weak dependence on angle
⃗ r ) · Ω̂ + Higher Order Terms (in angle)
ψ(⃗r , Ω̂) ≈ A(⃗r ) + B(⃗
6 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
How to derive the diffusion equation (2)
III. Asymptotic expansion
We start from the integro-differential form and we assume that
absorption is small and the gradients are also somewhat small but
not as much
Z
⃗ + Σ(⃗r )ψ(⃗r , Ω̂) =
ϵΩ̂ · ∇ψ
d2 Ω′ (Σ − ϵ2 Σa )ψ(⃗r , Ω̂′ ) + s
4π
This is an assumption on the equation rather than on the solution!
Different assumptions lead to the same result: the diffusion
equation. What does this tell us?
7 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
Integrating the transport equation
We want to derive an equation for the scalar flux, which is the
integral over angle of the angular flux
Z
ϕ(⃗r ) =
d2 Ω ψ(⃗r , Ω̂)
4π
We can integrate over angle the transport equation for ψ and
see if we get an equation for ϕ:
Z
d2 Ω (Bψ = s)
4π
where
⃗ + Σψ −
Bψ ≡ Ω̂ · ∇ψ
Z
d2 Ω′ Σs (Ω̂′ → Ω̂)ψ(Ω̂′ )
4π
8 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
The first angular moment (1)
We can integrate the terms one by one and obtain
1
Z
⃗r)
⃗ =∇
⃗ · J(⃗
d2 Ω Ω̂ · ∇ψ
4π
2
Z
d2 Ω Σψ = Σ(⃗r )ϕ(⃗r )
4π
3
Z
Z
d2 Ω
4π
d2 Ω′ Σs (Ω̂′ → Ω̂)ψ(Ω̂′ ) = Σs (⃗r )ϕ(⃗r )
4π
4
Z
d2 Ω s = S0 (⃗r )
4π
9 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
The first angular moment (2)
We obtain the first (zeroth order) angular moment of the
transport equation
⃗ r ) + Σa (⃗r )ϕ(⃗r ) = S0 (⃗r )
⃗ · J(⃗
∇
This equation is exact , but it contains two unknowns: ϕ and
⃗
J.
We can try to obtain an equation for J⃗ by multiplying the
transport equation by Ω̂ and integrating again:
Z
d2 Ω Ω̂(Bψ = s)
4π
10 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
The second angular moment (1)
We can integrate the terms one by one and obtain
1
Z
⃗ =∇
⃗ · Π(⃗r )
d2 Ω Ω̂Ω̂ · ∇ψ
4π
2
Z
⃗r)
d2 Ω Ω̂Σψ = Σ(⃗r )J(⃗
4π
3
Z
Z
d2 Ω Ω̂
4π
⃗r)
d2 Ω′ Σs (Ω̂′ → Ω̂)ψ = µ̄0 Σs (⃗r )J(⃗
4π
4
Z
d2 Ω Ω̂s = S⃗1 (⃗r )
4π
11 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
The second angular moment (2)
We obtain the second (first order) angular moment of the
transport equation
⃗ r ) = S⃗1 (⃗r )
⃗ · Π(⃗r ) + Σtr (⃗r )J(⃗
∇
where
Σtr = Σ − µ̄0 Σs
Z
Π(⃗r ) =
d2 Ω Ω̂Ω̂ψ(⃗r , Ω̂)
4π
This equation is still exact, but it contains the new unknown
Π.
12 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
The closure relationship
For each new moment we take of the transport equation, a
new unknown (of higher dimensionality) is introduced.
We need an approximation to close the system of equations.
We assume the dependence of the angular flux is linear wrt Ω̂:
⃗ r ) · Ω̂
ψ(⃗r , Ω̂) ≈ A(⃗r ) + B(⃗
We obtain:
ψ(⃗r , Ω̂) ≈
1
3
⃗r)
ϕ(⃗r ) +
Ω̂ · J(⃗
4π
4π
1
Π(⃗r ) ≈ ϕ(⃗r ) Id3
3
13 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
The P1 equations
We arrive at the following system of equations
⃗ · J⃗ + Σa ϕ = S0
∇
1⃗
∇ϕ + Σtr J⃗ = S⃗1
3
If the sources are isotropic (and they often are) then S⃗1 = 0
and we can ”solve” the second equation for J⃗ to obtain:
Fick’s Law for neutrons
⃗
J⃗ = −D ∇ϕ
where
D=
1
;
3Σtr
Σtr = Σ − µ̄0 Σs
14 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
Validity of the diffusion equation
The diffusion equation
An approximate second order differential equation for the scalar
flux ϕ(⃗r )
⃗ · D(⃗r )∇ϕ(⃗
⃗ r ) + Σa (⃗r )ϕ(⃗r ) = S0 (⃗r )
−∇
Neutrons must make a lot of scattering collisions (diffusions)
in order for the flux to be smooth in angle and space
Need to have little absorption and little leakage
Must be far from boundaries: the diffusion equation
approximates the asymptotic solution of the transport
equation but it neglects the boundary layers
The diffusion equation is a robust approximation
15 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
Void boundary condition
In transport the void boundary condition corresponds to zero
angular flux for all incoming directions:
ψ(⃗rs , Ω̂) = 0;
∀ Ω̂ · n̂s ≤ 0
In diffusion theory we can ask that the incoming partial
current vanishes at the boundary:
xs
Z
J − (xs ) =
d2 Ω n̂ · Ω̂ ψ(xs , Ω̂)
n̂·Ω̂<0
ϕ(x)
Medium
Void
ϕ(xs ) D dϕ
+
=0
4
2 dx xs
2D
This condition can be approximated by asking that the flux
vanishes at the extrapolated distance
≈
ϕ(xs + 2D) = 0
16 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
Interface conditions
At the interface between two materials the angular flux is
continuous
Mat I Mat II
ψ + (⃗rs , Ω̂) = ψ − (⃗rs , Ω̂)
xs
In diffusion we demand that the scalar flux and the current be
continuous:
ϕ+ (⃗rs ) = ϕ− (⃗rs );
J⃗+ (⃗rs ) = J⃗− (⃗rs )
This implies that the gradient of the scalar flux is
discontinuous at material interfaces.
⃗ ϕ| = D − ∇
⃗ ϕ|
D +∇
⃗
⃗
r+
r−
s
s
17 / 34
The diffusion equation
Examples
One Dimensional Problems
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
Three kinds of problems
Source problems
⃗ · D ∇ϕ
⃗ + Σa ϕ = S
−∇
A steady state solution always exists.
Criticality problems
⃗ · D ∇ϕ
⃗ + Σa ϕ = νΣf ϕ
−∇
Criticality problems are eigenvalue problems: a steady state
solution exists only if a certain criticality condition is met.
Subcritical problems with sources
⃗ · D ∇ϕ
⃗ + Σa ϕ = νΣf ϕ + S
−∇
A steady state solution exists only if the system is subcritical
(keff < 1)
18 / 34
The diffusion equation
Examples
One Dimensional Problems
The naked slab with sources
The critical slab
The reflected critical slab
Outline
1
The diffusion equation
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
2
Examples
The naked slab with sources
The critical slab
The reflected critical slab
3
One Dimensional Problems
Helmoltz’s Equation
The albedo
19 / 34
The diffusion equation
Examples
One Dimensional Problems
The naked slab with sources
The critical slab
The reflected critical slab
Slab with uniform source (1)
Problem
S
−a
−D
a
x
d 2ϕ
+ Σa ϕ = S
dx 2
20 / 34
The diffusion equation
Examples
One Dimensional Problems
The naked slab with sources
The critical slab
The reflected critical slab
Slab with uniform source (2)
Solution
1.0
/(S/ a)
0.8
0.6
0.4
0.2
0.0
a=1*L
a=2*L
a=4*L
a=8*L
1.00 0.75 0.50 0.25 0.00 0.25 0.50 0.75 1.00
x/a
S
cosh x/L
ϕ(x) =
1−
Σa
cosh a/L
21 / 34
The diffusion equation
Examples
One Dimensional Problems
The naked slab with sources
The critical slab
The reflected critical slab
Critical slab (1)
Problem
−a
D
a
x
d 2ϕ
+ (νΣf − Σa )ϕ = 0
dx 2
22 / 34
The diffusion equation
Examples
One Dimensional Problems
The naked slab with sources
The critical slab
The reflected critical slab
Critical slab (2)
Solution
1.0
0.8
0.6
0.4
0.2
0.0
1.00 0.75 0.50 0.25 0.00 0.25 0.50 0.75 1.00
x/a
ϕ(x) = A cos
χ2 ≡ νΣfD−Σa
πx 2a
χa = π/2
23 / 34
The diffusion equation
Examples
One Dimensional Problems
The naked slab with sources
The critical slab
The reflected critical slab
Reflected critical slab (1)
Problem
I
−a − b
−a
II
a
a+b
x
d 2 ϕI
+ χ2 ϕ I = 0
dx 2
d 2 ϕII
− 1/L2 ϕII = 0
dx 2
24 / 34
The diffusion equation
Examples
One Dimensional Problems
The naked slab with sources
The critical slab
The reflected critical slab
Reflected critical slab (2)
Solution
1.0
0.8
0.6
0.4
0.2
0.0
2.0
1.5
1.0
0.5
0.0
x/a
0.5
1.0
1.5
2.0
ϕI (x) = A cos(χx)
a+b−x
cos(χa)
ϕII (x) = A
sinh
sinh(b/L)
L
tan(χa) =
DII coth(b/L)
DI
χL
25 / 34
The diffusion equation
Examples
One Dimensional Problems
Helmoltz’s Equation
The albedo
Outline
1
The diffusion equation
The one-speed equation
The diffusion equation from the PN method
The boundary conditions
2
Examples
The naked slab with sources
The critical slab
The reflected critical slab
3
One Dimensional Problems
Helmoltz’s Equation
The albedo
26 / 34
The diffusion equation
Examples
One Dimensional Problems
Helmoltz’s Equation
The albedo
The Laplacian
Cartesian coordinates
∇2 f =
∂2f
∂2f
∂2f
+
+
∂x 2 ∂y 2 ∂z 2
Cylindrical coordinates
∇2 f =
∂2f
1 ∂f
1 ∂2f
∂2f
+
+
+
∂r 2
r ∂r
r 2 ∂ϕ2 ∂z 2
Spherical coordinates
∂2f
2 ∂f
1
∂
∇ f = 2+
+ 2
∂ρ
ρ ∂ρ ρ sin θ ∂θ
2
∂f
1
∂2f
sin θ
+ 2 2
∂θ
ρ sin θ ∂ϕ2
27 / 34
The diffusion equation
Examples
One Dimensional Problems
Helmoltz’s Equation
The albedo
The Helmholtz equation
∇2 f + B 2 f = 0
Solutions for one-dimensional geometries
Geometry
B2 < 0
1/L2 = −B 2
B2 > 0
χ2 = B 2
Cartesian
sinh(x/L), cosh(x/L)
exp(x/L), exp(−x/L)
I0 (r /L), K0 (r /L)
sinh(ρ/L)/ρ, cosh(ρ/L)/ρ
exp(ρ/L)/ρ, exp(−ρ/L)/ρ
sin(χx), cos(χx)
Cylindrical
Spherical
J0 (χr ), Y0 (χr )
sin(χρ)/ρ, cos(χρ)/ρ
28 / 34
The diffusion equation
Examples
One Dimensional Problems
Helmoltz’s Equation
The albedo
Solutions to the Helmholtz equation
Case B 2 > 0: oscillatory
1.00
jo( )
yo( )
1.0
0.75
0.50
cos(r)/r
sin(r)/r
1.0
0.5
0.5
0.0
0.0
0.5
0.5
0.25
0.00
0.25
0.50
0.75
cos(x)
sin(x)
1.00
0
2
4
6
x
8
10
12
1.0
14
1.0
0
2
4
6
8
10
12
14
0
2
4
6
r
8
10
12
14
Case B 2 < 0: exponential
10
4
3
1
10
io( )
ko ( )
8
2
cosh(r)/r
sinh(r)/r
8
6
6
4
4
0
1
2
cosh(x)
sinh(x)
3
4
2.0
1.5
1.0
0.5
0.0
x
0.5
1.0
1.5
(a) Cartesian
2.0
2
2
0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
(b) Cylindrical
4.0
0
0.0
0.5
1.0
1.5
2.0
r
2.5
3.0
3.5
4.0
(c) Spherical
29 / 34
The diffusion equation
Examples
One Dimensional Problems
Helmoltz’s Equation
The albedo
Homogeneous systems (1)
For homogeneous systems we can write the criticality
condition as:
material buckling = geometric buckling
k∞ − 1
2
Bm
≡
= Bg2 ; L2 = D/Σa
L2
For one group homogeneous systems we can write the
multiplication factor keff as:
keff =
νΣf
k∞
=
2
Σa + DBg
1 + L2 Bg2
The term DBg2 represents the leakage of the system.
30 / 34
The diffusion equation
Examples
One Dimensional Problems
Helmoltz’s Equation
The albedo
Homogeneous systems (2)
Homogeneous simple shapes have simple solutions for the
geometric buckling Bg2 and the flux ϕ:
Shape
Slab
Cylinder
Sphere
ρ ∈ [0, R]
Cube
x, y , z ∈ [−a, a]
Finite
Cylinder
r ∈ [0, R]
z ∈ [−H, H]
x ∈ [−a, a]
r ∈ [0, R]
Bg2
π 2
2a ι0 2
R
π 2
R
π 2
3 2a
ι0 2
π 2
+ 2H
R
ϕ
cos πx
2a
J0 ιR0 r
sin πρ
R /ρ
πy πz
cos πx
2a cos 2a cos 2a
πz
J0 ιR0 r cos 2H
31 / 34
The diffusion equation
Examples
One Dimensional Problems
Helmoltz’s Equation
The albedo
Homogeneous systems (3)
Homogeneous 1D systems: solutions rescaled from center to
boundary
1.0
cartesian
cylindrical
spherical
0.8
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
32 / 34
The diffusion equation
Examples
One Dimensional Problems
Helmoltz’s Equation
The albedo
The reflector albedo (1)
The reflector albedo is the number of neutrons coming back
to the system after having wandered into the reflector.
β = J ref→core /J core→ref
In diffusion theory and plane geometry we can find
β=
ϕII (a) + 2DII dϕ
dx
a
ϕII (a) − 2DII dϕ
dx
a
=
1−γ
1+γ
I
II
a
b
2DII
b→∞ 2DII Σa →0
γ=
coth(b/LII ) −−−→
−−−→ 0
LII
LII
33 / 34
The diffusion equation
Examples
One Dimensional Problems
Helmoltz’s Equation
The albedo
The reflector albedo (2)
Numerical values for the asymptotic (infinitely thick) reflector
albedo for some materials:
Material
Graphite
Heavy water
Light water
D(cm)
L(cm)
βasym
0.8
0.8
0.2
55
130
2.8
0.94
0.97
0.80
For a sphere the limiting value of the albedo is < 1:
γ
=
2D
R
1 + coth(δ/L)
R
L
R Σa →0 2D
δ→∞ 2D
1+
−−−→
−−−→
R
L
R
R
δ
34 / 34
Introduction
Point kinetics equations
Simplified models
Point Kinetics
The Time-dependent Problem
Fausto Malvagi
IRSN - INSTN
2024
1 / 26
Introduction
Point kinetics equations
Simplified models
1
Introduction
2
Point kinetics equations
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
3
Simplified models
One delayed family
The prompt jump approximation
Reactivity window insertion
2 / 26
Introduction
Point kinetics equations
Simplified models
Outline
1
Introduction
2
Point kinetics equations
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
3
Simplified models
One delayed family
The prompt jump approximation
Reactivity window insertion
3 / 26
Introduction
Point kinetics equations
Simplified models
Time scales
In a reactor we have four time scales:
Life of fission prompt neutrons [10−7 − 10−3 secs]
From birth to absorption or leakage: slowing down and
thermalization.
Emission of delayed neutrons [0.1 - 60 secs]
Evolution of fission products [hours - days]
They are usually strong neutron absorbers in the thermal
spectrum. Xenon, Samarium ...
Evolution of actinides [weeks - months]
Fissiles (235 U, 239 Pu, 241 Pu) and absorbers (240 Pu, 242 Pu)
4 / 26
Introduction
Point kinetics equations
Simplified models
Prompt neutrons
Neutrons
thermal
from fission
E
0.025 eV
2 MeV
v
2.103 m/s
2.107 m/s
The lifetime of a prompt neutron ℓ depends on the type of reactor
Reactor Type
Fast
LWR
Graphite
ℓ [s]
10−7
2.510−5
10−3
nb of generations in one sec
10,000,000
40,000
1,000
For a PWR with keff = 1.0001 the power increases of a factor
1.000140000 = 55 in a second: we cannot control the reactor!
5 / 26
Introduction
Point kinetics equations
Simplified models
Fission fragments (1)
Figure: Fission yields as a function of product mass A. [Wikipedia]
At each fission we produce some prompt neutrons (0 − 7), two
(or three) fission fragments and gamma rays
6 / 26
Introduction
Point kinetics equations
Simplified models
Fission fragments (2)
The fission fragments are highly unstable because of neutron
excess (they are far from the stability valley)
Figure: Fission fragments distance from stability valley [Wikipedia]
7 / 26
Introduction
Point kinetics equations
Simplified models
Fission products (1)
Figure: Double cascade of fission products [Wikipedia]
The fission fragments/products mostly decay by β − emission
A small number of fission products decay by emitting a
neutron. Those emissions are delayed wrt the fission event,
because of all the (beta) decays that must happen before the
final neutron emission.
8 / 26
Introduction
Point kinetics equations
Simplified models
Fission products (2)
Only two precursors of delayed neutrons are individualized:
87 Br et 137 I
87
β−
87
Br −−→
137
β
n 86
Kr* −
→
− 137
I −−→
Kr
n 136
Xe* −
→
Xe
The other 100 are grouped in 4 (or 6) families with average
decay constants.
Delayed neutron fraction β
Fraction of fission neutrons which are delayed
β=
νd
νp + νd
9 / 26
Introduction
Point kinetics equations
Simplified models
Delayed neutrons (1)
Only a small fraction of neutrons are delayed.
Fissionable isotopes
Isotope
232 Th
233 U
235 U
238 U
239 Pu
241 Pu
Fission
fast
thermal
thermal
fast
thermal
thermal
β [pcm]
2433
296
679
1828
224
535
10 / 26
Introduction
Point kinetics equations
Simplified models
Delayed neutrons (2)
Delayed neutron families βi and λi depend on the fissioning
nuclide.
Delayed families for 235 U
Group
1
2
3
4
5
6
Total
βi [pcm]
24
123
117
262
108
45
679
λi [s−1 ]
0.0127
0.032
0.116
0.311
1.40
3.88
0.088
τ [s]
78.74
31.25
8.62
3.22
0.71
0.26
11.31
11 / 26
Introduction
Point kinetics equations
Simplified models
Delayed neutrons (3)
Average neutron lifetime
It is now computed taking into account the emission time
⟨ℓ⟩ = (1 − β)ℓ +
6
X
βi (1/λi + ℓ)
i=1
For a PWR we have ⟨ℓ⟩ ≈ 0.08, which corresponds to 13
generations in one second. If we consider again the previous
example of keff = 1.0001, we now have (keff )13 = 1.0013: the
power increases of about 0.1% in one second. The reactor is now
controllable!
12 / 26
Introduction
Point kinetics equations
Simplified models
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
Outline
1
Introduction
2
Point kinetics equations
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
3
Simplified models
One delayed family
The prompt jump approximation
Reactivity window insertion
13 / 26
Introduction
Point kinetics equations
Simplified models
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
Time dependent one-speed diffusion equation (1)
We start from the time-dependent one-speed diffusion equation
1 ∂ϕ
⃗ · D ∇ϕ
⃗
= (νΣf − Σa )ϕ(⃗r , t) + ∇
v ∂t
Separation of variables
Assume
X
An exp(−λn t)ψn (⃗r )
ϕ(⃗r , t) =
n
Then
∇2 ψn + Bn2 ψn (⃗r ) = 0
λn = vDBn2 + v Σa − v νΣf
14 / 26
Introduction
Point kinetics equations
Simplified models
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
Time dependent one-speed diffusion equation (2)
Keeping only the first term
ϕ(⃗r , t) ∼ A1 exp
Mean neutron lifetime
−1
ℓ ≡ v Σa (1 + L2 Bg2 )
k −1
ℓ
t ψ1 (⃗r )
Multiplication factor
k≡
νΣf /Σa
1 + L2 Bg2
Reactor period
T ≡
ℓ
k −1
15 / 26
Introduction
Point kinetics equations
Simplified models
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
Kinetics with precursors
Introducing the precursors concentrations Ci (⃗r , t):
X
1 ∂ϕ
− D∇2 ϕ + Σa ϕ = (1 − β)νΣf ϕ +
λi Ci
v ∂t
i
∂Ci
= −λi Ci + βi νΣf ϕ
∂t
Separation of variables
ϕ(⃗r , t) = vn(t)ψ1 (⃗r )
Ci (⃗r , t) = Ci (t)ψ1 (⃗r )
16 / 26
Introduction
Point kinetics equations
Simplified models
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
Point kinetics: first form
We obtain the Point kinetics equations:
6
X
dn
k(1 − β) − 1
=
n(t) +
λi Ci (t)
dt
ℓ
i=1
k
dCi
= βi n(t) − λi Ci (t)
i = 1, . . . , 6
dt
ℓ
NB: The reactor is not a point, but its spatial shape does not
change in time.
Transients usually start from a steady-state state, in which the
neutron and precursors populations are constant in time. From the
second equation we obtain:
Ci0 =
βi
n0
ℓλi
−→
Ci0 ∼ 104 n0
17 / 26
Introduction
Point kinetics equations
Simplified models
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
Point kinetics: second form
6
X
ρ(t) − β
dn
=
n(t) +
λi Ci (t)
dt
Λ
i=1
dCi
βi
= n(t) − λi Ci (t)
dt
Λ
Mean generation time
Λ≡
i = 1, . . . , 6
Reactivity
ℓ
k
ρ(t) =
k(t) − 1
k(t)
We see the importance of having ρ < β !
18 / 26
Introduction
Point kinetics equations
Simplified models
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
The in-hour equation (1)
Assuming the reactivity is a constant ρ0 in time, we can look for
exponential solutions: n(t) = n0 e st and Ci (t) = Ci0 e st . The
characteristic equation (or in-hour equation) for s is:
6
ρ0 =
1 X sβi
sℓ
+
sℓ + 1 sℓ + 1
s + λi
i=1
3
2
0
1
0
1
2
3
20
15
10
5
s
0
5
10
19 / 26
Introduction
Point kinetics equations
Simplified models
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
The in-hour equation (2)
s1 , the ”biggest” root of the in-hour equation, is the asymptotic
value [n ∼ exp(st)]. τ0 ≡ 1/s1 is the reactor period.
Notable limits
ρ = 0 −→ s1 = 0
ρ → 1 −→ s1 = ∞
ρ → −∞ −→ s1 = −λ1
ρ ≪ β −→ s1 =
k −1
⟨ℓ⟩
ρ ≫ β −→ s1 =
k −1
ℓ
20 / 26
Introduction
Point kinetics equations
Simplified models
One delayed family
The prompt jump approximation
Reactivity window insertion
Outline
1
Introduction
2
Point kinetics equations
Kinetics with no delayed neutrons
Kinetics with delayed neutrons
The in-hour equation
3
Simplified models
One delayed family
The prompt jump approximation
Reactivity window insertion
21 / 26
Introduction
Point kinetics equations
Simplified models
One delayed family
The prompt jump approximation
Reactivity window insertion
One delayed family (1)
Equation
dn
ρ0 − β
=
n(t) + λC (t)
dt
Λ
dC
β
= n(t) − λC (t)
dt
Λ
"
β=
X
i
βi ;
1 X βi
λ=
β
λi
#−1
i
The solution to this system (for constant reactivity) is a sum of
two exponentials: one is the asymptotic solution and the other one
is a rapid transient.
22 / 26
Introduction
Point kinetics equations
Simplified models
One delayed family
The prompt jump approximation
Reactivity window insertion
One delayed family (2)
Solution
n(t) = p1 e s1 t + p2 e s2 t
2.0
λρ0
β − ρ0
s2 ≈ −
p1 ≈
β
β − ρ0
p2 ≈ −
$rho>0
$rho<0
1.8
100
1.4
C(t)
n(t)
ρ0
β − ρ0
$rho>0
$rho<0
110
1.6
1.2
90
80
1.0
0.8
0.6
β − ρ0
Λ
s1 ≈
70
0
2
4
time [s]
6
8
10
0
2
4
time [s]
6
8
10
23 / 26
Introduction
Point kinetics equations
Simplified models
One delayed family
The prompt jump approximation
Reactivity window insertion
Prompt jump approximation (1)
We want to replace the fast transient by a discontinuity. Starting
from the one family model we neglect dn/dt :
ρ(t) − β
n(t) + λC (t)
0=
Λ
dC
β
= n(t) − λC (t)
dt
Λ
We can ”solve” for n from the first equation and substitute in the
second:
dρ
dn
+ λρ(t) n(t) = 0
[ρ(t) − β] +
dt
dt
24 / 26
Introduction
Point kinetics equations
Simplified models
One delayed family
The prompt jump approximation
Reactivity window insertion
Prompt jump approximation (2)
We see that a jump in reactivity ρ1 → ρ2 will cause a jump in
neutron density n1 → n2 according to
n2
β − ρ1
=
n1
β − ρ2
25 / 26
Introduction
Point kinetics equations
Simplified models
One delayed family
The prompt jump approximation
Reactivity window insertion
Reactivity insertion
2.2
0.025
1.10
0.020
1.08
0.015
1.6
0.010
1.4
0.005
1.2
0.000
1.02
1.0
0.005
1.00
0.5
1.0
t [s]
1.5
2.0
2.5
C(t)/C0
1.8
0.0
=1e-03
=5e-04
=1e-05
1.12
0.030
[pcm]
N(t)/N0
2.0
0.035
(t)
0.035
0.030
0.025
0.020
0.015
1.06
[pcm]
=1e-03
=5e-04
=1e-05
(t)
2.4
0.010
1.04
0.005
0.000
0.005
0.0
0.5
1.0
t [s]
1.5
2.0
2.5
26 / 26
The energy discretization
Condensation and homogenisation
Problems
Multigroup Method
Fausto Malvagi
IRSN - INSTN
2024
1 / 27
The energy discretization
Condensation and homogenisation
Problems
1
The energy discretization
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
2
Condensation and homogenisation
3
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
2 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
Outline
1
The energy discretization
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
2
Condensation and homogenisation
3
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
3 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
The energy discretisation
We start from the steady-state transport equation
⃗ + Σ(⃗r , E )ψ(⃗r , E , Ω̂) = s(⃗r , E , Ω̂)
Ω̂ · ∇ψ
Z ∞ Z
+
dE ′ d2 Ω′ Σs (⃗r , E ′ → E , Ω̂ · Ω̂′ )ψ(⃗r , E ′ , Ω̂′ )
0
4π
We want to discretize the energy variable E.
Group g
Eg
Eg −1
E
Multigroup flux
Z Eg −1
dE ψ(⃗r , E , Ω̂)
ψg (⃗r , Ω̂) ≡
Eg
4 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
Integrating the transport equation (1)
We want to derive an equation for the multigroup flux, which
is the integral over a group of the angular flux
We thus try to integrate the whole transport equation over a
group g :
Z Eg −1
dE (Bψ = s)
Eg
where
⃗ + Σψ −
Bψ ≡ Ω̂ · ∇ψ
Z ∞
dE ′ Σs (E ′ → E , Ω̂′ · Ω̂)ψ(⃗r , E ′ , Ω̂′ )
0
we note that
Z ∞
dE f (E ) =
0
Ng Z E
g −1
X
g =1 Eg
dE f (E ) =
Ng
X
fg
g =1
5 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
Integrating the transport equation (2)
We can integrate the terms one by one and obtain
1
Z Eg −1
⃗ = Ω̂ · ∇ψ
⃗ g
dE Ω̂ · ∇ψ
Eg
2
Z Eg −1
dE Σψ = Σg ψg
Eg
3
Z Eg −1
Z ∞
Z
dE ′
d2 Ω Σs (E ′ → E , Ω̂′ · Ω̂)ψ(⃗r , E ′ , Ω̂′ ) =
dE
Eg
0
=
4
4π
XZ
g′
d2 Ω Σs,g ′ →g (Ω̂′ · Ω̂)ψg ′ (⃗r , Ω̂′ )
4π
Z Eg −1
dE s = Sg (⃗r , Ω̂)
Eg
6 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
Multigroup cross sections
Multigroup cross sections
Z Eg −1
dE Σ(E )ψ(⃗r , E , Ω̂)/ψg (⃗r , Ω̂)
Σg (⃗r ) ≡
Eg
Multigroup transfer matrix
Σs,g ′ →g (⃗r , Ω̂′ · Ω̂) ≡
Z Eg −1 Z Eg ′ −1
dE ′ Σs (E ′ → E )ψ(E ′ , Ω̂′ )/ψg ′ (⃗r , Ω̂)
≡
dE
Eg
Eg′
7 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
Multigroup transport equations
Multigroup transport equations
⃗ g + Σg ψg (⃗r , Ω̂) =
Ω̂ · ∇ψ
Ng Z
X
d2 Ω′ Σs,g ′ →g (⃗r , Ω̂′ · Ω̂)ψg ′ (⃗r , Ω̂′ )
g ′ =1 4π
+Sg (⃗r , Ω̂)
g = 1, . . . , Ng
System of Ng equations for the multigroup flux unknowns ψg .
In the solwing-down region (1eV ≤ E ≤ 20MeV ) the neutrons
can only lose energy. The system is thus lower triangular.
8 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
Iterative solution of the multigroup equations
Multigroup transport equations
⃗ gn + Σg (⃗r )ψgn (⃗r , Ω̂) =
Ω̂ · ∇ψ
Z
d2 Ω′ Σs,g →g (⃗r , Ω̂′ · Ω̂)ψgn (⃗r , Ω̂′ ) + qgn
4π
qgn (⃗r , Ω̂)
=
g
−1 Z
X
g ′ =1
+
d2 Ω′ Σs,g ′ →g (⃗r , Ω̂′ · Ω̂)ψgn ′ (⃗r , Ω̂′ )
4π
Z
Ng
X
d2 Ω′ Σs,g ′ →g (⃗r , Ω̂′ · Ω̂)ψgn−1
r , Ω̂′ )
′ (⃗
g ′ =g +1 4π
The first equation looks just like a one-speed equation
9 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
A second look at the multigroup cross sections (1)
Multigroup cross sections
Z Eg −1
dE Σ(E )ψ(⃗r , E , Ω̂)/ψg (⃗r , Ω̂)
Σg (⃗r , Ω̂) ≡
Eg
Σg depends on the unknown ψg
Σg depends on the (even more) unknown ψ(E )
Σg depends on the propagation angle Ω̂
For a same material, Σg depends on the spectrum at the point
In order to compute Σg , the coefficient for the equation for
ψg , we need a knowledge of the solution ψ(⃗r , E , Ω̂)
10 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
Solving a problem via multigroup transport
This is a two-step process:
1 First we need to compute the appropriate multigroup cross
sections
They depend on the problem at hand
They require the solution of some local problem
Difficult! Very few codes that can do this: CASMO (USA),
APOLLO2/3 (F), WIMS (UK), DRAGON (Can), SERPENT
(Fin), OpenMC (USA)
2
Then we solve the multigroup transport equations
We need to discretize space. Plenty of methods: finite
differences, finite elements, methods of characteristics, etc.
We need to discretize angle. Two methods: SN (collocation
method) and PN (functional expansion in Spherical Harmonics)
Plenty of codes that do this.
11 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
Multigroup meshes (1)
Meshes used by APOLLO-2/3 codes
Meshes
1.0
0.8
0.6
0.4
0.2
0.0
10 6
10 5
10 4
10 3
E [MeV]
10 2
10 1
100
101
Figure: Multigroup meshes: AP2-99 (blue), XMAS-172 (red), SHEM-281
(green)
12 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
Multigroup meshes (2)
Discretisation of U238 resonances
SHEM-281
U238
U238
(n,tot)
104
103
[b]
[b]
103
102
101
100
0.5
(n,tot)
104
102
101
1.0
1.5
2.0
2.5
E [MeV]
3.0
3.5
4.0
4.5
1e 5
100 4
10
10 3
E [MeV]
10 2
13 / 27
The energy discretization
Condensation and homogenisation
Problems
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
Multigroup diffusion
Without derivation:
⃗ · Dg ∇ϕ
⃗ g + Σa,g (⃗r )ϕg (⃗r ) =
−∇
X
Σs,g ′ →g (⃗r )ϕg ′ (⃗r ) + qg (⃗r )
g ′ ̸=g
qg (⃗r ) =
X
i
χig
X
νΣif ,g ′ (⃗r )ϕg ′ (⃗r ) + sg (⃗r )
g′
Multigroup diffusion should be used with only a few groups!
3D reactor cores are mostly simulated in 2-group diffusion
with homogenized assemblies.
14 / 27
The energy discretization
Condensation and homogenisation
Problems
Outline
1
The energy discretization
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
2
Condensation and homogenisation
3
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
15 / 27
The energy discretization
Condensation and homogenisation
Problems
Cross sections condensation
Starting from a cross section set defined on a fine energy mesh
{Σg } for which we have the corresponding fluxes {ϕg }, we can
define a new cross section set defined on a coarse mesh ΣG by flux
weighting the fine cross sections as:
Condensed cross sections
ΣG =
X
ϕg Σg /
g ∈G
X
ϕg
g ∈G
This corresponds to:
Reaction rates conservation
ΣG ϕG =
X
ϕg Σg
g ∈G
16 / 27
The energy discretization
Condensation and homogenisation
Problems
Cross section homogeneisation
We can use the same reasoning in space as well.
If we want to replace a fine spatial mesh region {h} by a
coarser spatial mesh {H} we average the cross sections not
only by volume but by flux also:
X
X
X
ΣH =
Vh Σh /
Vh −→ ΣH ϕH ≡
Vh ϕh Σh
(∗)
h∈H
h∈H
h∈H
A ”better” method is used: Super Homogenisation (SPH). It
iterates the second of Eq. (*) until reaction rates are
conserved between configuration H and h.
17 / 27
The energy discretization
Condensation and homogenisation
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
Outline
1
The energy discretization
Integration of the Boltzmann equation
Multigroup transport
Multigroup diffusion
2
Condensation and homogenisation
3
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
18 / 27
The energy discretization
Condensation and homogenisation
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
The two group diffusion equation
General formulation of 2G diffusion
⃗ · D1 ∇ϕ
⃗ 1 + Σr ,1 ϕ1 = Σs,2→1 ϕ2 + χ1 (νΣf ,1 ϕ1 + νΣf ,2 ϕ2 )
−∇
⃗ · D2 ∇ϕ
⃗ 2 + Σr ,2 ϕ2 = Σs,1→2 ϕ1 + χ2 (νΣf ,1 ϕ1 + νΣf ,2 ϕ2 )
−∇
Removal cross section
Σr ,i = Σa,i + Σs,i→j
19 / 27
The energy discretization
Condensation and homogenisation
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
Simplified two-group diffusion
Simplify the 2G equations when
All fissions are thermal
All neutrons from fission are fast
no upscattering
⃗ · D1 ∇ϕ
⃗ 1 + Σr ,1 ϕ1 = νΣf ,2 ϕ2
−∇
⃗ · D2 ∇ϕ
⃗ 2 + Σa,2 ϕ2 = Σs,1→2 ϕ1
−∇
20 / 27
The energy discretization
Condensation and homogenisation
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
From 2G diffusion to 4 (3!) factors formula
Starting from the simplified 2G diffusion, show that for an
homogeneous system we have
k∞ = pf η
where:
resonance escape probability: p = Σs,1→2 /Σr ,1
thermal utilization factor f = ΣFuel
a,2 /Σa,2
reproduction factor η = νΣf ,2 /ΣFuel
a,2
21 / 27
The energy discretization
Condensation and homogenisation
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
Thermal column (1)
Problem
We consider a semi-infinite system with incoming fast neutrons
S
x
0
ϕ1 (0) = 1;
ϕ2 (0) = 0
22 / 27
The energy discretization
Condensation and homogenisation
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
Thermal column (2)
Solution
H1 column
100
H2 column
100
thermal
fast
10 1
10 1
10 3
Flux
Flux
10 2
10 4
10 2
10 3
10 5
10 6
0
20
40
x [cm]
60
80
Figure: Light water
100
0
thermal
fast
20
40
x [cm]
60
80
100
Figure: Heavy water
23 / 27
The energy discretization
Condensation and homogenisation
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
Critical slab (1)
Problem
−a
D1
D2
a
x
d 2 ϕ1
+ Σr ,1 ϕ1 = νΣf ,2 ϕ2
dx 2
d 2 ϕ2
+ Σa,1 ϕ2 = Σs,1→2 ϕ1
dx 2
24 / 27
The energy discretization
Condensation and homogenisation
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
Critical slab (2)
MC simulation solution
thermal
fast
0.4
[a.u.]
0.3
0.2
0.1
0.0
40
20
0
x [cm]
20
40
The diffusion solution is separable in space and energy
25 / 27
The energy discretization
Condensation and homogenisation
Problems
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
Critical reflected slab (1)
Problem
I
−a − b −a
II
a
a+bx
Write the 2G diffusion equation for each region I and II
Determine the general solution for each region
Use boundary and interface conditions to determine the
constants of integration
You will end up with an homogeneous system of four linear
equations in four unknowns
26 / 27
The two group diffusion equation
Thermal columns
Critical slab
Critical reflected slab
The energy discretization
Condensation and homogenisation
Problems
The critical reflected slab (2)
MC simulation solution
0.30
thermal
fast
0.25
0.20
0.15
0.10
0.05
0.00
60
40
20
0
20
40
60
27 / 27
Introduction
Scattering
Slowing down in infinite media
Neutron Slowing Down
Fausto Malvagi
IRSN - INSTN
2024
1 / 32
Introduction
Scattering
Slowing down in infinite media
1
Introduction
Flux spectrum in thermal reactors
2
Scattering
Elastic scattering
Inelastic scattering
Lethargy
3
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
2 / 32
Introduction
Scattering
Slowing down in infinite media
Flux spectrum in thermal reactors
Outline
1
Introduction
Flux spectrum in thermal reactors
2
Scattering
Elastic scattering
Inelastic scattering
Lethargy
3
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
3 / 32
Introduction
Scattering
Slowing down in infinite media
Flux spectrum in thermal reactors
Flux spectrum in a thermal reactor
If we look at the flux spectrum in a thermal reactor we can clearly
identify three regions:
The fission spectrum region
The slowing-down region
The thermal region spectrum
[a.u.]
10 1
10 2
10 8
10 6
10 4
E [MeV]
10 2
100
Figure: Flux average spectrum in a thermal reactor
4 / 32
Introduction
Scattering
Slowing down in infinite media
Flux spectrum in thermal reactors
Flux spectrum (2)
Fission spectrum region
Flux roughly proportional to fission spectrum (T ∼ 1 MeV)
Mostly inelastic scattering, URR and continuum, anisotropic
scattering (in CM)
Slowing down region
Flux roughly proportional to 1/E
Mostly elastic scattering, RRR, isotropic scattering (in CM)
Thermal region
Flux roughly Maxwellian (T ∼ background)
Neutron energy comparable with background thermal motion:
upscattering possible
Neutron energy comparable with chemical bonds: excitation of
molecular vibrational modes (inelastic scattering)
5 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Outline
1
Introduction
Flux spectrum in thermal reactors
2
Scattering
Elastic scattering
Inelastic scattering
Lethargy
3
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
6 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Elastic scattering (1)
vout
vin
ψ
Vout
Figure: Scattering in the Laboratory system
The target nuclide is at rest and ψ is the deflection angle
We have rotational symmetry around the axis defined by ⃗vin .
Conservation of total kinetic energy and momentum hold: 6
unknowns and 4 conditions
A scattering collision always happens on a plane!
7 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Elastic scattering (2)
θ
wout
win
Wout
Win
Figure: Scattering in the Center of Mass system
In the CM system, the total momentum is zero. The neutron
and nuclide speeds don’t change, only their directions (which
stay opposite).
The result of the collision is just a rotation of an angle θ.
8 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Elastic scattering (3)
Going back to the Laboratory system, it can be shown that there is
a correlation between the outgoing (kinetic) energy of the neutron
and the deflection angle in the CM system θ. A is the nuclide
atomic mass (in neutron mass units).
Energy-angle correlation
Eout
A2 + 1 + 2A cos θ
=
Ein
(A + 1)2
The azimuth angle ϕ is arbitrary
The deflection angle θ is a random variable ”sampled” from
the anisotropy in nuclear data
Everything else is determined by the conservation equations
9 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Elastic scattering (4)
Eout
A2 + 1 + 2A cos θ
=
Ein
(A + 1)2
Limits:
θ = 0: straight-ahead collision Eout = Ein (minimum energy
loss)
θ = 180: back-scattering Eout = αEin ; (maximum energy loss)
α=
A−1
A+1
2
A = 1 (H1): −→ α = 0. In water the neutron can lose all its
energy in just one collision
10 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Elastic scattering (5)
If the scattering is isotropic in the CM system, we have
P(µ)dµ = dµ/2 where µ = cos θ. We also have
η = cos ψ = √
P( )
1 + A cos θ
A2 + 1 + 2A cos θ
P( )
2.00
0.52
1.75
0.51
1.50
0.50
1.00
A=1
A=2
A=5
A=10
A=50
1.25
0.75
0.49
0.50
0.48
0.25
1.00
0.75
0.50
0.25 0.00
= cos
0.25
0.50
Figure: CM system
0.75
1.00
0.00
1.00
0.75
0.50
0.25 0.00
= cos
0.25
0.50
0.75
1.00
Figure: Lab system
11 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Elastic scattering (6)
A2 + 1 + 2A cos θ
Eout
=
Ein
(A + 1)2
When the scattering collision is isotropic in the CM system, then
the outgoing energy distribution is uniform in the Lab system
(
1
, if αEin ≤ Eout ≤ Ein
P(Eout ) = (1−α)Ein
0,
otherwise
P(Eout )
1
(1−α)Ein
αEin
Ein
Eout
12 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Inelastic scattering
The kinematics can also be solved for the inelastic scattering. If Q
is the excitation energy of the target nuclide after the collision, we
have:
γ 2 + 1 + 2γ cos θ
Eout
=
Ein
(A + 1)2
1 + γ cos θ
cos ψ = p
γ 2 + 1 + 2γ cos θ
p
γ = A 1 − [(A + 1)/A]Q/Ein
Threshold energy
Eth =
A+1
Q
A
13 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Lethargy (1)
Upon an (elastic) scattering, a neutron loses a fraction of its
energy.
This suggests a change of variable from energy to lethargy
u ≡ ln(E0 /E ). E0 is some reference energy. E0 = 10 MeV is
usually used for reactor applications, so as to always have
positive lethargies.
We now have du = −dE /E , and also ϕ(E )dE = −ϕ(u)du.
In the slowing-down process the neutron lethargy is increasing
with the ”age” of the neutron.
Lethargy gain per collision:
w ≡ ∆u = − ln {0.5[1 + α + (1 − α) cos θ]}
14 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Lethargy (2)
Energy variable
Lethargy variable
Max energy loss:
max lethargy gain
∆max E = (1 − α)Ein
energy loss distribution
P(Eout )dEout =
dEout
(1 − α)Ein
average energy loss
⟨Eout ⟩ =
1−α
Ein
2
wmax = ϵ = − ln α
lethargy gain distribution
P(w )dw =
e −w
dw
1−α
average lethargy gain
⟨w ⟩ = ξ = 1 − αϵ/(1 − α)
ξ: slowing-down power
15 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Lethargy (3)
How many collisions to thermal energy?
From 2 MeV to 1 eV we have ∆u = ln(2.E6/1.) = 14.5 units of
lethargy. Since for Hydrogen the average gain in lethargy per
collision is ξ = 1, the number of collisions needed to slow-down a
fission neutron in water is around 15.
16 / 32
Introduction
Scattering
Slowing down in infinite media
Elastic scattering
Inelastic scattering
Lethargy
Lethargy (4)
Slowing-down quantities
Slowing-down power ξ
Moderating power ξΣs
Moderating ratio ξΣs /Σa
Moderator
ξ
nb colls
ξΣs
ξΣs /Σa
H2 O
D2 O
He
Be
C
U238
.920
.509
.425
.209
.158
.008
16
29
43
69
91
1730
1.35
0.176
1.6E-5
0.158
0.060
0.003
71
5670
83
143
192
0.0092
17 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Outline
1
Introduction
Flux spectrum in thermal reactors
2
Scattering
Elastic scattering
Inelastic scattering
Lethargy
3
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
18 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
The transport equation in an infinite medium (1)
The slowing-down equation
We write down the steady-state transport equation for an infinite
medium:
Z ∞
Σ(E )ϕ(E ) =
dE ′ Σs (E ′ → E )ϕ(E ′ ) + S(E )
0
Neutron Spectrum
Cross Sections
103
H1
H2
He4
C12
O16
H1
H2
He4
C12
O16
102
[b]
[a.u.]
102
101
101
100
10 1
10 4
10 3
E [MeV]
10 2
10 10 10 8 10 6 10 4 10 2 100
E [MeV]
102
19 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Arrival density
E
E + dE
E
Figure: Arrival density ρ(E )
The arrival density ρ(E )dE is the number of neutrons which
scatter into the interval [E , E + dE ] per unit time and unit volume,
coming from some energy E ′ > E .
The arrival density ρ(E )
Z ∞
ρ(E ) =
dE ′ Σs (E ′ → E )ϕ(E ′ )
E
20 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Slowing-down current (1)
E
E
Figure: Slowing-down current q(E )
The slowing-down current q(E ) is the number of neutrons which
scatter, per unit time and unit volume, from an energy E ′ > E to
an energy E ′′ < E .
The slowing-down current q(E )
Z E
Z ∞
′′
q(E ) =
dE
dE ′ Σs (E ′ → E ′′ )ϕ(E ′ )
0
E
21 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Slowing-down current (2)
It can be shown that the slowing-down current q(E ) obeys the
balance equation:
Second form of the slowing-down equation
dq
= Σa (E )ϕ(E ) − S(E )
dE
or in lethargy variable:
dq
= S(u) − Σa (u)ϕ(u)
du
22 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Solutions to the slowing-down equation
For scattering isotropic in the CM system:
Z E /α
Σ(E )ϕ(E ) =
E
dE ′
Σs (E ′ )
ϕ(E ′ ) + S0 δ(E − E0 )
(1 − α)E ′
4 cases to be considered
1 Slowing-down in hydrogen (A = 1) without absorption
(Σa = 0)
2
Slowing-down in hydrogen (A = 1) with absorption (Σa > 0)
3
Slowing-down (A > 1) without absorption (Σa = 0)
4
Slowing-down (A > 1) with absorption (Σa > 0)
The first three cases have analytical solutions.
23 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Case 1: A=1; Σa = 0 (1)
Z E0
Σs (E )ϕ(E ) =
dE ′
E
Σs (E ′ )
ϕ(E ′ ) + S0 δ(E − E0 )
E′
We will use:
Collision density: F (E ) ≡ Σ(E )ϕ(E ) = Σs (E )ϕ(E )
Separate Collided and Uncollided parts:
F (E ) = Fc (E ) + S0 δ(E − E0 )
Try to transform the integral equation into a differential
equation
d
dx
Z b(x)
′
′
Z b(x)
dx f (x, x ) =
a(x)
a(x)
dx ′
df (x, x ′ )
db
da
+f (x, b) −f (x, a)
dx
dx
dx
24 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Case 1: A=1; Σa = 0 (2)
We arrive at the solution:
S0
+ S0 δ(E − E0 )
E
S0
S0
ϕ(E ) =
+
δ(E − E0 )
Σs (E )E
Σs (E )
S0
S0
+
δ(u − u0 )
ϕ(u) =
Σs (u) Σs (u)
q(E ) = EFc (E ) = S0
F (E ) =
Remarks:
The flux ϕ(E ) is ≃ proportional to 1/E
The flux ϕ(u) is ≃ a constant
The slowing-down density q is just equal to the emission
density S0 (no absorption)
The solution is valid for any dependence Σs (E )!
25 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Case 2: A=1; Σa > 0 (1)
Z E0
Σ(E )ϕ(E ) =
E
dE ′
Σs (E ′ )
ϕ(E ′ ) + S0 δ(E − E0 )
E′
Using the same techniques as before we arrive at:
Z E0
Σs (E0 ) S0
dE ′ Σa (E ′ )
Fc (E ) =
exp −
Σ(E0 ) E
Σt (E ′ )E ′
E
q(E ) = EFc (E )
Remarks:
The solution is still valid for any dependences Σs (E ) and
Σa (E )!
Since hydrogen is a weak absorber, this model is called a
mixture of hydrogen plus an infinitely massive absorber
26 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Case 2: A=1; Σa > 0 (2)
From the previous expression we can identify:
Resonant escape probability p(E )
Z E0
q(E )
dE ′ Σa (E ′ )
p(E ) =
= exp −
S0
Σt (E ′ )E ′
E
This is valid (that is to say ”exact”) for slowing-down in hydrogen
plus an infinitely massive absorber.
27 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Case 3: A > 1; Σa = 0 (1)
Z E /α
Σs (E )ϕ(E ) =
E
dE ′
Σs (E ′ )
ϕ(E ′ ) + S0 δ(E − E0 )
(1 − α)E ′
When we try to transform this integral equation into a differential
equation we obtain:
dFc
1
=
[Fc (E /α) − Fc (E )]
dE
(1 − α)E
This is not an ordinary differential equation !!!
28 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Case 3: A > 1; Σa = 0 (2)
Z E /α
Σs (E )ϕ(E ) =
E
dE ′
Σs (E ′ )
ϕ(E ′ ) + S0 δ(E − E0 )
(1 − α)E ′
Neutrons that have made no collisions have a lethargy of
u = u0 = 0
Neutrons that have made exactly one collision have a lethargy
between 0 and ϵ: u ∈ [0, ϵ]
Neutrons that have made exactly two collisions have a
lethargy between 0 and 2ϵ: u ∈ [0, 2ϵ]
Neutrons that have made exactly three collisions have a
lethargy between 0 and 3ϵ: u ∈ [0, 3ϵ]
...
29 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Case 3: A > 1; Σa = 0 (3)
It turns out that:
The solution ϕ(u) has a
discontinuity at u = ϵ
The second derivative of the
solution d 2 ϕ(u)/du 2 has a
discontinuity at u = 3ϵ
102
[a.u.]
The derivative of the
solution dϕ(u)/du has a
discontinuity at u = 2ϵ
Neutron Spectrum
H1
H2
He4
C12
O16
101
10 4
10 3
E [MeV]
10 2
...
These are called the
Plazcek’s transients
Figure: Plazcek’s transients
30 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Case 3: A > 1; Σa = 0 (4)
The asymptotic solution is
given by:
ϕ(u) −→
11 O
CHAMP NEUTRONIQUE
t f (u)
S0
ξΣs (u)
for a mixture of different
moderators i we have
S0
ϕ(u) −→
⟨ξ⟩ Σs (u)
P i
Σ (u)ξ i
⟨ξ⟩ (u) = Pi s i
i Σs (u)
Transitoire de Placzek
0,5-r---t---+--_j_
0,25 r---t--~L---1o;----------+1--------J________l--f ~2
3
Figure: Plazcek’s transients
31 / 32
Introduction
Scattering
Slowing down in infinite media
The slowing-down equations
Solutions to the slowing-down equation
Case 4: A > 1; Σa > 0 (1)
Z E /α
Σ(E )ϕ(E ) =
dE ′
E
Σs (E ′ )
ϕ(E ′ ) + S0 δ(E − E0 )
(1 − α)E ′
No ”analytical” solution . . .
32 / 32
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Resonant Absorption
Fausto Malvagi
IRSN - CEA - INSTN
2024
1 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
1
Introduction
The self shielding phenomena
2
The isolated resonance
3
Flux factorisation
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
4
Doppler broadening
2 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
The self shielding phenomena
Outline
1
Introduction
The self shielding phenomena
2
The isolated resonance
3
Flux factorisation
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
4
Doppler broadening
3 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
The self shielding phenomena
Self shielding phenomena (1)
Reaction rates
τα (E ) = Σα (E )ϕ(E )
Resonances are localized intervals in energy where the cross
sections vary significantly (several orders of magnitude)
Correspondingly, the neutron flux can also plunge significantly
This depression in the flux limits the increase in reaction rates
This phenomena is called self-shielding effect.
4 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
The self shielding phenomena
Self shielding phenomena (2)
Slowing down in H1 + U238
101
t
105
104
10 1
103
10 2
102
10 3
101
[b]
[a.u.]
100
Flux
Reaction Rates
10 4 6
10
10 5
E [MeV]
100
10 4
5 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
The self shielding phenomena
Self shielding phenomena (3)
Not taking into account self shielding would increase U238
(simulated) absorption by a factor 14 in a PWR
Z
Z
Z
dE Σa (E )
dE ϕ(E ) ≈ 14
dE Σa (E )ϕ(E )
RD
RD
RD
RD Resonance Domain
U238 absorption affects the resonance escape probability p
and the U238 conversion into Pu239
We recall that plutonium counts for 40 % of total fissions in
the lifetime of a UO2 pellet
6 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
The self shielding phenomena
Breit-Wigner single level resonances
U238 (T=300 K)
104
(n,n)
(n, )
103
[b]
102
101
100
10 1
5.0
5.5
6.0
6.5
σγ (E ) = σ0
σa (E ) = σ0
7.0 7.5
E [MeV]
8.0
8.5
9.0
1e 6
Γγ Γ
(E − E0 )2 + Γ2 /4
Γ2
Γ(E − E0 )
+ σ1
+p
2
2
(E − E0 ) + Γ /4
(E − E0 )2 + Γ2 /4
7 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
The self shielding phenomena
Metaphor of the kangaroos (1)
P.Reuss: the resonance is a trap along the path
(slowing-down) of the kangaroos (neutrons)
ξ
u
σ
γ
All neutrons that fall into the trap are absorbed
All the others are unaffected
8 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
The self shielding phenomena
Metaphor of the kangaroos (2)
The number of kangaroos which fall into the trap does not
depend on the depth (σ) of the trap, but only on the width(γ)
of the trap, compared to the size of the jumps(ξ) of the
kangaroos
The width of a resonance of 238U is of the order of the eV,
while the neutrons lose on average, in each collisions with
water, half of their kinetic energy. We thus have, for E > 100
eV:
γ
≪1
ξ
The resonances are narrow with respect to the average energy
loss per collision in the moderator
Very few neutrons are absorbed in a single resonance (at high
energy).
9 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Outline
1
Introduction
The self shielding phenomena
2
The isolated resonance
3
Flux factorisation
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
4
Doppler broadening
10 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Isolated black resonance (1)
Black resonance
All neutrons that fall into the resonance are absorbed
Z ui +γ
Z ui +γ
Z u′
′
′
′
du ρ(u ) =
du
1 − pi =
du ′′ Σs (u ′′ → u ′ )ϕ(u ′′ )
ui
u ′ −ϵ
ui
γi
u
u′ − ϵ
ui
u′
ui + γ i
Figure: An isolated resonance
11 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Isolated black resonance (2)
Before the resonance the flux takes its asymptotic value:
ϕ(u) ≈ (ξΣs )−1 ;
u < ui
We neglect the flux inside the resonance:
ϕ(u) ≈ 0;
u ∈ [ui , ui + γi ] ≡ (i)
For isotropic elastic scattering we have
′
P(u ′ → u) =
e −(u−u )
1−α
12 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Isolated black resonance (3)
Putting everything together:
Z
Z u
1 − pi ≈
du
du ′ Σs (u ′ )P(u ′ → u)ϕ(u ′ )
(i)
Z ui
Z
1 − pi ≈
u−ϵ
du
(i)
du
u−ϵ
′e
−(u−u ′ )
(1 − α)ξ
=
1 − e −γ − αγ
(1 − α)ξ
For a narrow resonance:
γ→0
1 − pi −−−→
γ
ξ
13 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Isolated grey resonance
Grey resonance
The neutrons that fall into the resonance have a probability Σa /Σ
of being absorbed
Z
Σa (u ′ )
1 − pi ≈
du ′
ρ(u ′ )
′)
Σ(u
(i)
Using the asymptotic value of the flux ϕ(u) ≈ 1/[ξΣs (u)] outside
but also inside the resonance (γi ≪ ϵ) we have:
ρ(u) ≈ Σs (u)ϕ(u) ≈ 1/ξ
Z
Σa (u ′ )
du ′
1 − pi ≈
ξΣ(u ′ )
(i)
14 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
A succession of grey resonances
Since the absorption of each resonance is small, we can use
the limited Taylor development:
" Z
#
Z
′
′
′ Σa (u )
′ Σa (u )
≈ exp −
pi ≈ 1 −
du
du
ξΣ(u ′ )
ξΣ(u ′ )
(i)
(i)
If we have a succession of well-isolated grey resonances we
can write, for the escape probability
Z u
Y
Σa (u ′ )
p(u) =
pi ≈ exp
du ′
ξΣ(u ′ )
0
(i)<u
15 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Outline
1
Introduction
The self shielding phenomena
2
The isolated resonance
3
Flux factorisation
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
4
Doppler broadening
16 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Flux factorization
Factorisation of the flux
ϕ(E ) = Φ(E )φ(E )
Asymptotic flux Φ
Φ(E ) ≈
1
ξΣs (E )E
It has the dimensions of a flux
It corresponds to the flux outside of the resonances
Self-shielding factor or Fine-structure flux φ
Dimensionless
Goes to one outside of the resonances
17 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Mixture
We consider a mixture made of an absorber A and a moderator M
The moderator has no absorption and no resonances:
M
ΣM (E ) ≈ ΣM
s (E ) = Σs,p
The absorber has also a potential scattering and a resonant
scattering cross sections:
A
A
ΣA (E ) = ΣA
a (E ) + Σs (E ) + Σs,p
18 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
The slowing-down operator
We can rewrite the slowing-down equation as:
Σ(E )ϕ(E ) = RM (ϕ) + RA (ϕ)
where
Slowing-down operator
Z E /αβ
Rβ (ϕ) =
E
dE ′
Σβs
ϕ(E ′ )
(1 − αβ )E ′
with β = {M, A}
19 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Slowing down by the moderator
When a neutron of energy E scatters on the moderator, the
average loss of energy 0.5(1 − αM )E is large wrt the width of
the resonance
It follows that for most of the interval [E , E /αM ] the flux
assumes its asymptotic value ϕ ∼ 1/E
we can thus write
Z E /αM
RM (ϕ) =
dE ′
E
ΣM
s
≈
(1 − αM )
ΣM
s
ϕ(E ′ )
(1 − αM )E ′
Z E /αM
E
dE ′
E′
1
E′
=
ΣM
s
E
20 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Slowing down by the absorber
When the neutron of energy E scatters on the absorber, we
have two limiting cases:
Narrow Resonance: the average loss of energy 0.5(1 − αA )E
is large wrt the width of the resonance and we
write as before
ΣA
RA (ϕ) ≈ s
E
Wide Resonance: the interval [E , E /αA ] is so small that we
can consider the integrand a constant and thus
Z E /αA
′
ΣA
s (E )
RA (ϕ) =
dE ′
ϕ(E ′ ) ≈ ΣA
s (E )ϕ(E )
′
(1
−
α
)E
M
E
21 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Narrow/Wide resonance self-shielding factors (1)
In these two cases the slowing down equation can be solved
analytically:
Narrow resonance
Σ(E )ϕNR =
A
ΣM
s + Σp
E
φNR (E ) =
A
ΣM
s + Σp
Σ(E )
Wide resonance
Σ(E )ϕWR = ΣA
s ϕWR +
φWR (E ) =
ΣM
s
E
ΣM
s
Σ(E ) − ΣA
s (E )
22 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Narrow/Wide resonance self-shielding factors (2)
If we keep in mind that the width of the resonances tend to
have the same order of magnitude as a function of energy, but
the energy loss per collision is proportional to the energy
itself, we have that all resonances are narrow at high energy,
but not at low energy.
From the previous expressions we see that φNR < φWR
We can thus conclude that a ”same” resonance at low energy
absorbs more than at high energy.
23 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Precision of the NR/WR approximations
First resonances of U238
E0
Exact
NR
WR
6.67
21.0
36.9
81.3
90.0
117.5
192.0
212.0
0.1963
0.0676
0.0582
0.0096
0.0011
0.0092
0.0071
0.0050
+21.0%
+10.4%
-18.6%
+5.9%
+0.9%
-1.5%
-28.6%
-11.6%
+1.8%
+4.5%
+5.0%
-15.5%
-9.6%
+3.6%
+73.0%
+53.1%
Table: Effective resonance integral (1 − pi )
24 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Livolant-Jeanpierre (1)
We go back to the slowing-down equation in an infinite
medium
h
i
ΣM (E ) + ΣA (E ) ϕ(E ) = RM (ϕ) + RA (ϕ)
We use the flux factorization ϕ(E ) = Φ(E )φ(E ) where Φ(E )
is the slowing-varying part and φ(E ) is the fast-varying part
We choose Φ to be the solution of the slowing-down equation
in the moderator:
ΣM (E )Φ(E ) = RM (ϕ)
25 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Livolant-Jeanpierre (2)
Since Φ varies slowly, we can approximate
R A (Φφ) ≈ ΦR A (φ)
We obtain the equation for the fine structure:
h
i
ΣM (E ) + ΣA (E ) φ(E ) = ΣM (E )φ(E ) + RA (φ)
We define the dilution cross section
Dilution cross section
σd = ΣM
s /NA
N →0
Limit: σd −−A−−→ ∞
26 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Livolant-Jeanpierre (3)
Dividing the previous equation by NA we obtain
Fine structure equation
(σd + σ A )φ(E ) = σd r A (φ)
where
A
Z E /αA
r (φ) =
(∗)
dE ′ σsA (E ′ → E )φ(E )
E
Equation (∗) is still an integral equation, but it contains
microscopic cross sections only
It can be solved numerically for each resonance of each
isotope and tabulated as a function of σd once per library
(and including anisotropy)
27 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Effective resonance integral
For each resonance (i) we can compute:
Resonance integral
I (i) =
dE ′ A ′
σ (E )
′ a
(i) E
Z
If we take into account self-shielding, the relevant quantity is:
Effective resonance integral
Z
(i)
Ieff =
dE ′ A ′
σ (E )φ(E ′ )
′ a
E
(i)
28 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
Self-shielding models
We have looked at an infinite system
We have seen the analytical Narrow/Wide Resonance models
We have seen the tabulated numerical Livolant-Jeanpierre
model
We have not seen the sub-group model
These models have (must have) also heterogeneous versions
These models are not convergent: we don’t know how to
reduce the error by half if we want to!
29 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Outline
1
Introduction
The self shielding phenomena
2
The isolated resonance
3
Flux factorisation
Self-shielding factor
NR/WR approximations
Livolant-Jeanpierre
4
Doppler broadening
30 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Doppler effect (1)
Cross sections are given in the evaluation files at 0 Kelvin
Close to the resonances we must take the thermal motion of
the background into consideration
We need to perform a convolution of the 0 K cross sections
with a Maxwellian (distribution of the velocity of the target)
at the appropriate temperature
U238
105
T=0
T=300
T=600
T=900
T=1200
[b]
104
103
102
101
6.2
6.4
6.6
6.8
E [eV]
7.0
7.2
31 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Doppler effect (2)
The effect of higher temperatures is to lower the resonance
peak and widen its width
The area under the curve stays roughly the same, but the
resonant absorption increases
The absorption effective integral increases as the square root
of the temperature
For fissile materials the capture and fission effects partially
compensate
For U238 the effect is a net increase in captures
32 / 33
Introduction
The isolated resonance
Flux factorisation
Doppler broadening
Doppler effect (3)
If a local perturbation increases the reactivity → the power
increases → the temperature increases → the absorption
increases → the reactivity decreases
The effect is about 2 − 3 [pcm/K]
The effect is practically immediate: U238 is right next to
U235!
This is the main stabilizing feedback in a power reactor
33 / 33