Than Yan Ren - Department of Physics

Neutronics Computation with Angular Flux
Transport Equation
Than Yan Ren
supervised by
Prof. Lim Hock
Assoc. Prof. Chung Keng Yeow
An Honours Thesis submitted in partial fulfilment of the requirements for
the
Degree of Bachelor of Science
(Honours)
to the
Department of Physics
National University of Singapore
AY2015-2016
Abstract
Computer simulations are crucial in the analysis of nuclear reactors both
in the developmental phase and for accident scenarios due to the fact that they
provide a risk-free method for experimenting with the situation. Huge collections
of programs that can together simulate the entire meltdown process of a nuclear
power plant have been developed. The aim of this study is to investigate the
basics in building a simulation for core nuclear kinetics based on the solving the
time independent neutron transport equation. MATLAB codes were developed
from first principles based on methods from E. Lewis for various set ups in the
spherical geometry to investigate steady state behaviours. Special attention has
also been given to the choice of input parameters they have to fit approximations
taken in solving the transport equation. As initial exercises to test the code, the
criticality of pure uranium spheres were investigated. The results were found
to be in good agreement with known results like from diffusion approximation.
After we are satisfied with the workings of the code, we look at how it can be
applied to more complex situations. This will be illustrated with examples in
analysing the yield of a uranium bomb and the behaviour of the pebble bed
reactor in steady state conditions.
Acknowledgements
I would like express my sincere gratitude to Prof Lim Hock and Assoc Prof Chung Keng
Yeow for taking time out of their busy schedules for our regular meetings, and also for their
guidance and mentorship throughout the duration of this project.
I would also like to thank my family and my friends for their continued support and understanding throughout this demanding period.
Contents
1 Introduction
1.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Scope of project . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
1
2 Theory
2.1 Neutron chain fission . . . . . . . . . . . .
2.2 Cross section of interaction . . . . . . . . .
2.3 Neutron transport equation . . . . . . . .
2.4 Diffusion approximation . . . . . . . . . .
2.5 Three energy groups in moderated reactors
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4 Initial results
4.1 Critical mass of pure U-235 sphere . . . . . . . . . . . . . . . . . . . . . . .
4.2 Critical mass of enriched uranium sphere . . . . . . . . . . . . . . . . . . . .
4.3 Temperature profile of water-cooled uranium sphere . . . . . . . . . . . . . .
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5 Possible Applications
5.1 Estimating efficiency of
5.1.1 Theory . . . . .
5.1.2 Results . . . . .
5.2 Pebble bed reactor . .
5.2.1 Theory . . . . .
5.2.2 Results . . . . .
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3 Solving the time-independent neutron transport equation
3.1 Derivation of discretised equations . . . . . . . . . . . . . . .
3.2 Algorithm for solving equations . . . . . . . . . . . . . . . .
3.3 Meaning behind the iteration process . . . . . . . . . . . . .
3.4 Determining input cross section values and energy groups . .
3.4.1 Angular dependence of scattering cross sections . . .
3.4.2 Group-to-group scattering . . . . . . . . . . . . . . .
uranium
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bomb
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6 Conclusion
47
A Alternate derivations of discretised equations
50
B Obtaining temperature profile from given neutron flux
51
C Derivation of energy loss from scattering
53
D MATLAB codes
55
1
1.1
Introduction
Motivation
Nuclear power plants are power plants which obtain energy from fission of radioactive nuclei
(most commonly Uranium-235) in a nuclear reactor core. It provides a far more efficient
source of energy than the traditional fossil fuels and generates much less waste in comparison, about 1100GWh per ton of uranium[1] in a typical Pressurised Water Reactor (PWR)
plant to about only 2.7MWh per ton of coal[2], thus it is an attractive option that many
countries have begun to adopt. However the safety of such power plants has been a huge
concern due to the volatility of the radioactive materials, and indeed large scale accidents
have happened like the Chernobyl and Fukushima Daiichi disaster. On top of the immediate
loss of lives incurred, the radioactive material released could potentially spread hundreds
of miles away through being carried by wind or water and cause widespread health issues.
Thus many have been aiming to create safer reactor designs that are less prone to melting
down in the event of an accident. Crucial to the design process would be using computer
simulation is a safe way to analyse the dynamics of the hypothetical reactors.
Besides that, accident scenarios can also be analysed with computer simulation to determine the optimal response to minimise the damage done. Large scale and complex sets of
simulation codes like the Accident Source Term Evaluation Code (ASTEC) have been developed for such purposes and with the introduction of the upcoming Generation IV reactors,
even more simulations will have to be developed.
For anyone who wishes to work on or even develop such simulations, it will be very helpful
that he or she understands the basics working principles behind the simulations well in order
to identify and handle possible shortcomings in the code.
1.2
Scope of project
The aim of this project is then to conduct a study on these basic working principles of a
simulation. The type of simulation that will be investigated here is done through numerical
approximation of the neutron transport equation that describes the dynamics within fissionable materials. The other major type of simulation is done through Monte-Carlo and this
is also being looked at in a seperate project. We will aim to understand the physical prin-
1
ciples behind how and why certain approximations are taken. The first part of the project
will consist of writing up a basic algorithm for solving time independent problems based on
material from [3] with proper understanding of its workings. After writing up the code, it
will be applied to simple examples in order to verify that it is working as intended. Once
we are satisfied with that, we will then go on to discuss possible ways of adapting the code
into the analysis of more complex situations. Initial results will also be shown for a more
complete illustration of the method.
2
2.1
Theory
Neutron chain fission
Fissionable materials release energy primarily through the neutron-induced fission, which
causes a heavy nucleus like Uranium-235 to fission into two smaller daughter nuclei and
possibly other small fragments like protons and neutrons. The energy released comes from
the decrease in total binding energy of the fission products in comparison to the original
nucleus and thus it is this extra binding energy that is released from the fission event.
Figure 1: Plot of binding energy versus nucleon number [4]
As one can see from figure (1), after the element Iron, the binding energy tends decrease
as the nuclei gets heavier. Due to the fact that there are often a few neutrons present in
the fission products, these second generation neutrons can go on to trigger even more fission
2
events. If the conditions are favourable, an exponentially increasing chain of fissions events
can occur, leading to the release of large amounts of energy. When this happens, the materials is said to have reached criticality. This possibility is determined by whether the neutrons
produced from fission is able to compensate for those lost to leakage or other interactions.
The easiest way to reach criticality is simply to increase the amount of fissile material used,
since volume increases by r3 while the surface area only increases by r2 , the fission rate
that occurs within the volume of the material will increase faster than the leakage that only
happens on the surface. Thus for fissionable materials, the critical mass or size will be of
great interest.
It is the possibility of creating this chain reaction that allows significant amounts of energy
to be extracted from fissile materials. This indicates that properties of fissionable materials
are best studied by looking at the dynamics of the neutrons within the material.
The neutrons in general can be described with the quantity N (~r, Ω̂, E, t), which is the total
number of neutrons in a dV volume about position ~r travelling in the direction within a dΩ
cone about directional vector Ω̂ and having energy from E to E + dE at time t, as illustrated
in figure (2).
Figure 2: Illustration of quantity N (~r, Ω̂, E, t).
The next step is now to describe the interaction of these neutrons with fissionable materials
3
and other matter. This is done using the concept of the cross section of interaction.
2.2
Cross section of interaction
The cross section value describes the probability for the neutron to interact with the atomic
nuclei present, in the sense that these nuclei will appear to the neutrons to have a certain
cross sectional area as shown in figure (3). So should the neutron pass through that area,
Figure 3: Interpretation of the cross section of interaction. The neutron marked with a cross
does not interact. Note that this cross section may not necessarily coincide with the physical
size of the nucleus.
the neutron will interact with the nucleus.
Each individual type of interaction will have their own respective cross section and the combined cross section of separate interactions will just be the sum of the individual cross section
values. For example, when an interaction happens, the neutron can either be absorbed or
scattered, so we have
σt = σs + σa
(2.1)
where σt is the total cross section and σs and σa are the scattering and absorption cross
sections respectively. This is not the full picture yet however, as these cross sections do
change according to the incoming energy of the neutron and scattering cross sections can
also be split into the angle and energy that the incoming neutron is scattered into. So
4
equation (2.1) should instead read
Z
σt (~r, E) =
dE
0
Z
dΩ0 σs (~r, E 0 → E, Ω̂ · Ω̂0 ) + σa (~r, E).
(2.2)
Note that due to the symmetry of a collision event, σs will depend on just Ω̂ · Ω̂0 instead of
both Ω̂ and Ω̂0 separately. Equation (2.2) can also depend on time since materials can be
moving or expand due to heat and such, but in this case we will not be considering such
effects.
The most important interaction here is of course the fission, of which cross section is given
by σf (~r, E) and is considered a subset of the absorption reaction.
The reaction rate at each dV volume in space using this concept can then be given as
the total number of neutrons passing through the total cross section contributed by the
nucleons in the dV volume per unit time. This can be written more formally as
Z
Rt (~r, t) =
Z
dE
dΩ σt (~r, E)v(E)N (~r, E, Ω̂, t).
(2.3)
The rate of each specific type of interaction can also be written is very similar form, which
will be used in describing the dynamics of the system. Thus it is useful to define the quantity
ψ(~r, Ω̂, E, t) = v(E)N (~r, E, Ω̂, t)
which is also called the neutron flux.
2.3
Neutron transport equation
We would expect an equation describing neutron dynamics to have the form
∂
N (~r, Ω̂, E, t) =Factors removing or introducing neutrons into
∂t
particular phase space considered
5
(2.4)
These factors should consist, in some form, of the interaction terms as described in the
previous section. The actual neutron transport equation is in fact in this form and reads
∂
~ r, Ω̂, E, t) − σt (~r, E)ψ(~r, Ω̂, E, t) +
N (~r, Ω̂, E, t) = − Ω̂ · ∇ψ(~
∂t
Z
Z
0
qex (~r, Ω̂, E, t) + dE
dΩ0 σs (~r, E 0 → E, Ω̂ · Ω̂0 )ψ(~r, Ω̂0 , E 0 , t) +
Z
χ(E) dE 0 νσf (~r, E 0 )φ(~r, E 0 , t)
where
(2.5)
Z
φ(~r, E, t) =
dΩ ψ(~r, Ω̂, E, t).
(2.6)
The first term on the right describes the loss of neutrons from the net number of neutrons
moving out of the volume dV around ~r while the third term accounts for external sources.
Neutrons that interact and considered to be removed first, which is accounted for by the
second term, and then reinserted as their appropriate internal source terms according to
type of interaction. So the fourth term describes the neutrons that have been scattered from
E 0 and Ω̂0 into E and Ω̂. And finally the last term describes the neutrons produced from
fission events with energy E travelling in direction Ω̂; ν here is a constant describing the
average number of neutrons released per fission event (which depends on the fissile material)
and χ(E) is the energy distribution of the fission neutrons. One reasonable approximation
of χ(E) is as follows[5],
√
χ(E) = 0.453 exp(−1.036E) sinh( 2.29E)
(2.7)
in units of MeV. This approximation is found to be in very good fit with the experimental
data[6] as shown in the following figure.
6
Figure 4: MATLAB plot of equation (2.7) against actual experimental data
Note that for the full neutron dynamics will also consist of delayed neutrons, which are
neutrons not emitted at the instant of the fission event but rather at a later time by the
daughter nuclei produced by the fissioning as they could be in an excited state. However,
this will not matter here as we will only be looking at time independent form of the equation,
∂N
= 0. This is already very useful since only materials that are exactly critical
which is
∂t
can have time independent solutions. So in these steady state situations, it does not actually
matter when the neutrons are released. So finally, the equation that we will be handling is
given as
h
i
~ + σt (~r, E) ψ(~r, Ω̂, E) =
Ω̂ · ∇
Z
Z
0
qex (~r, Ω̂, E) + dE
dΩ0 σs (~r, E 0 → E, Ω̂0 · Ω̂)ψ(~r, Ω̂0 , E 0 ) +
Z
χ(E) dE 0 νσf (~r, E 0 )φ(~r, E 0 )
which is an integro-differential equation to be solved for ψ(~r, Ω̂, E, t).
7
(2.8)
2.4
Diffusion approximation
This section will describe a simple way to solve the time independent version of the neutron transport equation using the diffusion approximation. To illustrate this, first integrate
equation (2.8) through all energies and then all angles, obtaining
~ · J(~
~ r) = qex (~r) + σs (~r)φ(~r)
σt (~r)φ(~r) + ∇
(2.9)
+ νσf (~r)φ(~r)
noting that assumptions, which will be covered in detail in the next chapter, have been made
in order for the scattering and fission terms to be integrated into this form. Supposing that
there are sufficient neutrons moving in all directions, Fick’s first law of diffusion gives the
relation
~ r)
~ r) = − 1 ∇φ(~
(2.10)
J(~
3σt (~r)
thus resulting in the equation
ν
∇2 φ(~r) = 3σt (~r)(σa (~r) − σf (~r))φ(~r) + qex (~r).
k
(2.11)
One simple result from this approximation comes from considering the situation where the
fissile material is homogeneous, source-free and spherically symmetric, this greatly simplifies
the transport equation into
∂2
ν
rφ(r) = 3σt (σa − σf )rφ(r).
2
∂r
k
(2.12)
This is a simple ordinary differential equation which can be solved with the appropriate
boundary condition φ(R + 3σ2 t ) = 0 to obtain
C
φ(r) = sin
r
r
ν
3σt ( σf − σa ) r
k
(2.13)
as the only family of physically valid solutions under the condition that
3σt (νσf − σa ) =
8
π 2
.
R + 3σ2 t
(2.14)
The above condition also sets a restriction on the radius to be
s
π2
2
−
R=
3σt (νσf − σa ) 3σt
(2.15)
which we can identify as the critical radius, thus we can see that this gives a very quick way
to estimate the critical radius of any fissile material that roughly fits the assumptions taken.
This will be a very useful method to use as a quick check for the results obtained in this
project. Also note that for materials where νσf − σa < 0, criticality cannot be achieved as
square root term will become imaginary.
2.5
Three energy groups in moderated reactors
In analysing situations within the typical moderated reactor, it is common to just take three
energy groups labelled fast (F), intermediate (I) and thermal groups (T)[5]. This method is
chosen due to the properties of the fissile materials and the neutrons. Firstly, the fast range
covers all energies above 0.1MeV as this is the range where practically all fission neutrons
are released into. The intermediate range, which covers 1eV ≤ E ≤ 0.1MeV, is where the
scattering and absorption cross sections are dominated by closely packed resonance peaks
due to the atomic spectra of the nuclei. It is typically seen as an intermediate group that
fission neutrons go through to eventually reach the thermal range, where the fission cross
section increases dramatically (from 1 to 2) in the case of uranium-235, one of the commonly
used fuels in reactors. Finally the thermal range covers energies below 1eV as this is where
the neutron energies are comparable to the thermal energies of the nuclei. This also means
upscatter into the I and F ranges is impossible as they are beyond the reach of the thermal
energies.
The energy dependent fission cross section shown in figure (5) clearly illustrates the three
distinct regions and the tightly-packed resonance peaks can also be seen in the intermediate
region. The energy distributions of the neutrons within this context can also be roughly
described as
ψF (E) = χ(E)
(2.16)
for the fission group,
ψI (E) ∝
9
1
E
(2.17)
Figure 5: Plot of σf (E) of U-235 obtained from International Atomic Energy Agency[7]
for the intermediate group, and finally
ψT (E) =
−E E
exp
(kB T )2
kB T
(2.18)
for the thermal group, which unsurprisingly is the Maxwell-Botlzmann distribution for particles in thermal equilibrium.
3
Solving the time-independent neutron transport equation
3.1
Derivation of discretised equations
This section will describe the method adapted from [3] to discretise the neutron transport
equation such that it will be suitable for computational schemes to handle.
10
The equation to solve here is
Z
h
i
ν
~
Ω̂ · ∇ + σ(~r, E) ψ(~r, Ω̂, E) = χ(E) dE 0 σf (~r, E 0 )φ(~r, E 0 )
k
Z
Z
+ dE 0 dΩ0 σs (~r, E 0 → E, Ω̂ · Ω̂0 )ψ(~r, Ω̂0 , E 0 ).
(3.1)
This is then split in three energy groups fast, intermediate and thermal, which results in the
following,
h
i
X
XZ
ν
~ + σg (~r) ψg (~r, Ω̂) = χg
σf g0 (~r)φg0 (~r) +
dΩ0 σg0 g (~r, Ω̂ · Ω̂0 )ψg0 (~r, Ω̂0 ) (3.2)
Ω̂ · ∇
k g0
g0
where g = T, I, F is the index for the energy groups. This is done by assuming the energy
seperability of the angular flux
ψ(~r, Ω̂, E) ≈ fg (E)ψg (~r, Ω̂)
Eg < E 6 Eg−1
(3.3)
R
where the function fg (E) is normalised with g dE fg (E) = 1. The cross sections can then
be seen to be weighted and averaged over this distribution for example
Z
σg (~r) =
dE σ(~r, E)f (E).
(3.4)
g
The energy seperability need not be assumed however, in which case to arrive at the same
result, we would require
R
dE σ(~r, E)ψ(~r, Ω̂, E) ?
g
= σg (~r).
(3.5)
R
dE
ψ(~
r
,
Ω̂,
E)
g
We can see that mathematically, σg should have
speaking σg (~r) should be correct since even if the
should be randomly oriented thus on average all
this is not convincing, an alternative derivation
same result.
angular dependence. However, physically
nuclei are not spherically symmetric, they
incoming angles should be equal. Even if
shown in Appendix A will still yield the
Note that using the spherical symmetry of the problem, any neutron streaming through
the domain of the system can be broken down as shown in figure(6).
Thus it can seen that the vectors ~r and Ω̂ can be replaced with scalars r and µ = cos Θ. So
11
Figure 6: Particle streaming through a spherical domain with u being a coordinate along Ω̂
with impact parameter b
equation (3.2) can be re-written as
i
h ∂
XZ
1 − µ2 ∂
νX
+ σg (r) ψg (r, µ) = χg
σf g0 (r)φg0 (r) +
µ +
dΩ σg0 g (r, Ω̂ · Ω̂0 )ψg0 (r, µ0 )
∂r
r ∂µ
k g0
g0
(3.6)
with the quantity Ω̂ · Ω̂0 being related to µ and µ0 as follows,
Ω̂ · Ω̂0 = µµ0 +
p
p
1 − µ2 1 − µ02 cos(ω − ω 0 )
(3.7)
and ω is the angle shown in figure (7).
Then note that the scattering cross section is a function defined on −1 6 Ω̂ · Ω̂ 6 1, meaning
it can be expanded using the Legendre polynomials as follows,
σg0 g (r, Ω̂ · Ω̂0 ) =
∞
X
(2l + 1)σlg0 g (r)Pl (Ω̂ · Ω̂0 ).
(3.8)
l=0
Here the Pl (Ω̂ · Ω̂0 ) are the Legendre polynomials and
Z
σlg0 g (r) =
Z
dΩ Pl (µr )σg0 g (r, µr ) =
dµr
Pl (µr )σg0 g (r, µr )
2
(3.9)
where µr = Ω̂ · Ω̂0 . The term Pl (Ω̂ · Ω̂0 ) from equation (3.8) can be further simplified using
12
Figure 7: Coordinate ω in relation to Ω̂ and r̂
the Legendre addition theorem to be
l
X
1
Y m∗ (Ω̂)Ylm (Ω̂0 )
Pl (Ω̂ · Ω̂ ) =
2l + 1 m=−l l
0
where
s
Ylm (θ, ϕ) =
(2l + 1)
(3.10)
(l − m)! m
P (cos(θ))eimϕ
(l + m)! l
(3.11)
are the spherical harmonics. The scattering term now becomes
XZ
0
0
0
dΩ σg0 g (r, Ω̂ · Ω̂ )ψg0 (r, µ ) =
g0
∞
XX
g0
σlg0 g (r)
l=0
l
X
Yl
m∗
Z
(Ω̂)
dΩ0 Ylm (Ω̂0 )ψ(r, µ0 ).
m=−l
(3.12)
In this problem, the angular flux ψ is not dependent in the ω coordinate referenced in figure
(7). Thus it can be shown that for m 6= 0,
Z
0
m
0
0
dΩ Yl (Ω̂ )ψ(r, µ ) =
Z
dΩ0 Ylm (Θ0 , ω 0 )ψ(r, µ0 ) = 0
13
(3.13)
0
since eimω is integrated over its entire period. This leads to
∞
XX
g0
σlg0 g (r)
l=0
l
X
Yl
m∗
Z
(Ω̂)
dΩ0 Ylm (Ω̂0 )ψ(r, µ0 ) =
∞
XX
g0
m=−l
(2l + 1)Pl (µ)σlg0 g (r)φlg0 (r)
l=0
(3.14)
where
Z
φlg (r) =
since
Yl0 (Ω̂) =
dµ0
Pl (µ0 )ψg (r, µ0 )
2
(3.15)
√
2l + 1 Pl (µ).
(3.16)
The next step is to discretise equation (3.6) in terms of angle, or more specifically the
µ coordinate. The method[8] of using a Gaussian quadrature here will be optimal since
R
integrals over µ will have to be approximated. Note that since the integral dµf (µ) is
unweighted and defined over the interval [−1, 1], it can be approximated using the GaussP
P
Legendre quadrature as n wn f (µn ) for some weightage wn > 0 normalised to n wn = 2.
The quadrature rule allows the approximated integral to be exact for polynomials degree
(2N − 1) or less since there are 2N parameters {wn } and {µn }. According to the quadrature
rule, the associated orthogonal polynomials are the Legendre polynomials, thus the {µn }
will satisfy
PN (µn ) = 0.
(3.17)
For the remaining N weightages, it is sufficient to demand that polynomials up to (N − 1)
be integrated correctly. A convenient choice of polynomials to check this condition can be
the Legendre polynomials themselves. This leads to the relation
1
2
Z
dµ Pl (µ) =
1X
wn Pl (µn ) = δlo
2 n
for 0 6 l 6 N − 1
(3.18)
which can be used to determine the weightage set {wn }. It is also important to note that
due to the symmetry of the situation, the {wn } and {µn } have to satisfy
µN +1−n = −µn
for 1 6 n 6
N
2
(3.19)
wN +1−n = wn
for 1 6 n 6
N
2
(3.20)
and
14
which fortunately is indeed satisfied by the Gauss-Legendre quadrature.
Before taking approximation of the differential term, it is important to re-write the differential operators in conservation form such that approximations will still retain particle
conservation. The conservation form is one that when integrated over all streaming angles,
satisfies the following neutron balance equation
Z
X
XZ
ν
~ J~g (~r)+σg (~r)φg (~r) = χg
σf g0 (~r)φg0 (~r)+
dΩ dΩ0 σg0 g (~r, Ω̂· Ω̂0 )ψg0 (~r, Ω̂0 ) (3.21)
∇·
k g0
g0
which itself is equation (3.2) integrated over all angles. This means that the differential
~ · Ω̂ψ. For this case, the conservation
operator has to be derived from its divergence form ∇
form is
∂
1 − µ2 ∂
µ ∂
∂ 1 − µ2
µ +
= 2 r2 +
.
(3.22)
∂r
r ∂µ
r ∂r
∂µ r
A simple way to approximate the angular differential term is then
(1 − µ2n+1/2 )ψg (r, µn+1/2 ) − (1 − µ2n−1/2 )ψg (r, µn−1/2 )
∂ 1 − µ2
ψg (r, µn ) '
∂µ r
(µn+1/2 − µn+1/2 )r
(3.23)
where half-integer indexes are introduced to aid the computation with the relation
ψg (r, µn ) =
i
1h
ψg (r, µn+1/2 ) + ψg (r, µn−1/2 )
2
(3.24)
with µ1/2 = −1 and µN +1/2 = 1. However, when one considers the trivial solution in infinite
medium limit ψ = constant, which results in
~ g (r, µ) = 0 =⇒
Ω̂ · ∇ψ
µ2n−1/2 − µ2n+1/2
µn
=0
2r
+
r2
(µn+1/2 − µn+1/2 )r
(3.25)
1
which implies condition µn = (µn+1/2 + µn−1/2 ). However, this happens to be incompatible
2
with the {µn } determined from the Gauss-Legendre quadrature in equation (3.17). To rectify
this, angular differencing coefficients αn±1/2 [8] is used to replace the µn±1/2 terms and the
weightages wn can be introduced into equation (3.23) such that
2αn+1/2 ψg (r, µn+1/2 ) − 2αn−1/2 ψg (r, µn−1/2 )
∂ 1 − µ2
ψg (r, µn ) ≈
.
∂µ r
rwn
15
(3.26)
Comparing this with the case in equation (3.23), we expect that
2αn+1/2 ' 1 − µ2n+1/2
(3.27)
which will be a consistency check for the replacement used. Then looking at the case of
~ = 0 is now
infinite medium again, the condition from demanding Ω̂ · ∇ψ
αn+1/2 − αn−1/2
µn
= 0 =⇒ αn+1/2 = αn−1/2 − µn wn .
2r
+
2
r2
rwn
(3.28)
For the resulting equation to satisfy the neutron balance equation, a further condition can
be obtained by just looking at the left hand side of equation (3.21)
Z
Z
∂
1
2
~ · J~g (~r) + σg (~r)φg (~r) =
r
+
σ
(~
r
)
dΩ
(r̂
·
Ω̂)ψ
(~
r
)
dΩ
ψ
(~
r
)
∇
g
g
g
r2 ∂r
Z
Z
1
1 ∂ 1
= 2 r2
dµ µ ψg (r, µ) + σg (r)
dµ ψg (r, µ)
r ∂r
2
2
1 X
X
1 ∂ 2 1
≈ 2 r
wn µn ψg (r, µn ) + σg (r)
wn ψg (r, µn ) .
r ∂r
2 n
2 n
(3.29)
Another way to obtain the above is to discretise the left hand side of equation (3.6) and
integrate over all streaming angles as shown,
h ∂
i
1 − µ2 ∂
dΩ µ +
+ σg (r) ψg (r, µ)
∂r
r ∂µ
Z
hµ ∂
i
∂ 1 − µ2
1
2
dµ 2 r +
+ σg (r) ψg (r, µ)
=
2
r ∂r
∂µ r
i
1 X h µn ∂ 2
∂ 1 − µ2n
≈
wn
r +
+ σg (r) ψg (r, µn )
(3.30)
2 n
r2 ∂r
∂µ r
X α
1 ∂ 1 X
n+1/2 ψg (r, µn+1/2 ) − αn−1/2 ψg (r, µn−1/2 )
≈ 2 r2
wn µn ψg (r, µn ) +
r ∂r
2 n
r
n
1 X
+ σg (r)
wn ψg (r, µn ) .
2 n
Z
16
Comparing the above two equations give the condition
X
αn+1/2 ψg (r, µn+1/2 ) − αn−1/2 ψg (r, µn−1/2 ) = 0
(3.31)
n
=⇒ α1/2 ψg (r, µ1/2 ) = αN +1/2 ψg (r, µN +1/2 )
and simply setting α1/2 = αN +1/2 = 0 will ensure the above is satisfied for any solution
ψg (r, µn ) thus also allowing the µn+1/2 s to be arbitrarily determined. This claim can be
made due to the symmetry of the {µn } and {wn } from equations (3.19) and (3.20) which
gives
αN +1/2 = α1/2 −
X
n
µn wn = α1/2 −
N/2
X
µn w n −
n=1
N
X
µn wn = α1/2 .
(3.32)
n=N/2+1
The resulting k-eigenvalue equation is as follows,
2αn+1/2 ψg,n+1/2 (r) − 2αn−1/2 ψg,n−1/2 (r)
µn ∂ 2
r
ψ
(r)
+
+ σg (r)ψgn (r, n) =
gn
r2 ∂r
rwn
N −1
X
X
ν X
1XX
χg
σf g0 (r)
wn0 ψg0 n0 (r) +
(2l + 1)Pl (µn )σlg0 g (r)
wn0 Pl (µn0 )ψg0 n0 (r)
2k g0
2
0
0
0
l=0
g
n
n
(3.33)
where ψgn (r) represents the computational solution for ψg (r, µn ). Substituting equation
(3.24) into (3.33) to remove the n + 1/2 indexes results in
4αn+1/2 ψgn (r) − 2(αn+1/2 + αn−1/2 )ψg,n−1/2 (r)
µn ∂ 2
r ψgn (r) +
+ σg (r)ψgn (r, n) =
2
r ∂r
rwn
N −1
X
X
ν X
1XX
χg
σf g0 (r)
wn0 ψgn0 (r) +
(2l + 1)Pl (µn )σlg0 g (r)
wn0 Pl (µn0 )ψg0 n0 (r)
2k g0
2
0
0
0
l=0
n
g
n
(3.34)
Since the k-eigenvalue equation does not contain any integrals over space, the spatial discretisation can be done simply by integrating equation (3.34) over an incremental spherical
shell
Z ri+1/2
Vi = 4π
dr r2 .
(3.35)
ri−1/2
17
The discretised angular flux is then taken as
ψgni
4π
=
Vi
Z
ri+1/2
dr r2 ψgn (r)
(3.36)
ri−1/2
which is simply the average over the corresponding shell. Another approximation
Z
ri+1/2
Z
ri+1/2
dr rψgn (r) ≈ 4π
4π
ri−1/2
ri−1/2
1
dr rψgni = (Ai+1/2 − Ai+1/2 )ψgni
2
(3.37)
where Ai = 4πri2 is well justified if ri+1/2 − ri−1/2 is small and finally the integration of the
spatial derivative term will be
Z
ri+1/2
4π
dr
ri−1/2
µn
r2 2
r
∂ 2
r ψgn (r) = 4π
∂r
Z
ri+1/2
∂
dr µn r2 ψgn (r) + 2µn rψgn (r)
∂r
ri−1/2
Z ri+1/2
ri+1/2
2
dr 2µn rψgn (r) − 2µn rψgn (r)
+ 4π
= 4πµn r ψgn (r)
ri−1/2
ri−1/2
= µn (Ai+1/2 ψgn,i+1/2 − Ai−1/2 ψgn,i−1/2 ).
(3.38)
Then as with the angular case,
1
ψgni = (ψgn,i+1/2 + ψgn,i−1/2 )
2
(3.39)
will relate the integer indexes to the half integer indexes. Note that similar to the angular
case r1/2 = 0 and rI+1/2 = R is taken, where R is the radius of the U-235 ball. Taking the
integral using the relations defined above and then dividing by Vi results in the equation
nµ
n
Vi
(Ai+1/2 − Ai+1/2 ) 2αn+1/2 ψgni − (αn+1/2 + αn−1/2 )ψg,n−1/2,i
Vi wn
N −1
n ν X
oj
X
X
1XX
σf g0 i
wn0 ψg0 n0 i +
(2l + 1)Pl (µn )σlg0 gi
wn0 Pl (µn0 )ψg0 n0 i
= χg
2k g0
2 g0 l=0
n0
n0
(Ai+1/2 ψgn,i+1/2 − Ai−1/2 ψgn,i−1/2 ) +
+σgi ψgni
oj+1
(3.40)
where j is the iteration index. This method is known as iteration over source since only
the source terms are from the previous iteration. Now that the set of discretised equations
are defined, the boundary conditions for the problem can be introduced. First is due to the
18
symmetry at the center of the ball, which is most frequently approximated by
ψg,N +1−n,1/2 = ψgn,1/2
for 1 6 n 6
N
.
2
(3.41)
Also since the U-235 ball is taken to be in vacuum, another boundary condition is
ψgn,I+1/2 = 0
for 1 6 n 6
N
.
2
(3.42)
Now from equations (3.24), (3.39) and (3.40) and the boundary conditions, there are a total
of 3N I + N equations within a group g while there are a total of 3N I + I + N unknowns
since all ψg,n+1/2,i+1/2 are not solved for. To account for the lack of I equations, one method
is to extend the boundary condition (3.42),
ψg,1/2,I+1/2 = 0
(3.43)
since equation (3.40) solves from µn−1/2 to µn . Then another equation needs to be introduced
to link ψg,1/2,i+1/2 to ψg,1/2,i . This is done by attempting to extend equation (3.40) to µ1/2 .
2µ
∂ 1 − µ2
ψ =
ψ, thus no approximation of the angular
Conveniently, at µ1/2 = −1,
∂µ r
r
derivative is required. Taking integral then gives
Z
ri+1/2
4π
dr r2
ri−1/2
2µ1/2
ψg,1/2 (r) = −(Ai+1/2 ψg,1/2,i+1/2 − Ai−1/2 ψg,1/2,i−1/2 )
r
(3.44)
thus resulting in
n −2
Vi
(Ai+1/2 ψg,1/2,i+1/2 − Ai−1/2 ψg,1/2,i−1/2 ) + σgi ψg,1/2,i
oj+1
j
= Sgni
(3.45)
j
where Sgni
is simply the right hand side of equation (3.40). Further substituting in equation
(3.39) gives
n2
Vi
oj+1
j
2Ai−1/2 ψg,1/2,i − (Ai+1/2 + Ai−1/2 )ψg,1/2,i+1/2 + σgi ψg,1/2,i
= Sgni
.
(3.46)
While (3.46) provides I equations as required, they are not useful unless the rest of ψg,1/2,i+1/2
are known, therefore giving an additional I equations to be solved. Fortunately, since
19
ψg,1/2,I+1/2 is known, a simple extension of equation (3.39),
1
ψg,1/2,i = (ψg,1/2,i+1/2 + ψg,1/2,i−1/2 )
2
(3.47)
will ensure that all ψg,1/2,i and ψg,1/2,i+1/2 can be solved. Overall, this process created 2I
equations and also an additional I unknowns, not counting equation (3.43), thus exactly
balancing the total equations and unknowns to be solved for.
3.2
Algorithm for solving equations
N
The main idea in solving for all the ψgni is to start from ψg,1/2,I+1/2 and ψgn,I+1/2 for n 6
2
and then solve for increasing µ and decreasing r in half integer steps. This requires the
substitution of equation (3.39) into (3.40) which gives
n −µ
n
(2Ai−1/2 ψgni − (Ai+1/2 + Ai−1/2 )ψgn,i+1/2 )
Vi
oj+1
(Ai+1/2 − Ai+1/2 ) j
+
2αn+1/2 ψgni − (αn+1/2 + αn−1/2 )ψg,n−1/2,i + σgi ψgni
= Sgni
Vi wn
(3.48)
j+1
gives
and then solving for ψgni
j+1
ψgni
1
=
−2µn Ai−1/2 +
2(Ai+1/2 −Ai−1/2 )
αn+1/2
wn
+ Vi σgi
j+1
j
− µn (Ai+1/2 + Ai−1/2 )ψgn,i+1/2 + Vi Sgni .
1
j+1
(Ai+1/2 − Ai−1/2 )(αn+1/2 + αn−1/2 )ψg,n−1/2,i
wn
(3.49)
The same is done for its µ1/2 case yielding
j+1
ψg,1/2,i
=
1
j+1
j
2(Ai+1/2 + Ai−1/2 )ψg,1/2,i+1/2
+ Vi Sgni
.
4Ai−1/2 + Vi σgi
(3.50)
Figure (8) shows the direction in which the solutions can be obtained in the discretised
grid, the bold arrows indicate equations (3.49) and (3.50) while the dotted arrows indicate
equations (3.39) and (3.33), which are also known as the diamond difference relations. Note
that some of the points have two bold arrows leading to it, this is because equation (3.49)
20
Figure 8: Illustration of the direction in which each of the discretised points of µ and r are
N
solved for n 6
in the case of I = 3 and N = 4
2
N
requires two input points from the grid. Now for the case of n > , boundary condition
2
(3.41) can be used to obtain ψgn,1/2 , which can then be solved in increasing µ and increasing
r for the rest of the ψ values. So now the system will be solved from ri−1/2 to ri thus the
diamond difference relation (3.39) have to be subsituted again into equation (3.40) to obtain
j+1
ψgni
1
=
2µn Ai+1/2 +
2(Ai+1/2 −Ai−1/2 )
αn+1/2
wn
+ Vi σgi
j+1
j
+ µn (Ai+1/2 + Ai−1/2 )ψgn,i−1/2 + Vi Sgni .
1
j+1
(Ai+1/2 − Ai−1/2 )(αn+1/2 + αn−1/2 )ψg,n−1/2,i
wn
(3.51)
The way to solve the upper half of the grid is illustrated in figure (9). This process can be
done seperately for each of the energy groups to obtain the full set of solutions.
21
Figure 9: Illustration of the direction for solving the remaining points of the solution grid in
the same case as figure (8)
3.3
Meaning behind the iteration process
This entire iterative process can be seen as repeatedly applying a matrix M into a state
vector Ψ where


ψT,1,1
 . 
 .. 




ψT,N,I 

Ψ=
(3.52)
 .. 
.


 . 
 .. 


ψF,N,I
and M can be seen as the discretised form of the operator
~ + σt )−1 (F + S)
(Ω̂ · ∇
(3.53)
where the operators F and S are
Z
Fψ(~r, Ω̂, E, t) = χ(E)
and
Z
Sψ(~r, Ω̂, E, t) =
dE
0
Z
0
dE νσf (~r, E)
Z
dΩ0 ψ(~r, Ω̂0 , E 0 , t)
dΩ0 σs (~r, E 0 → E, Ω̂ · Ω̂0 )ψ(~r, Ω̂0 , E 0 , t).
22
(3.54)
(3.55)
Putting the iteration index in we get
Ψj+1 = MΨj .
(3.56)
To illustrate what happens after many iterations, we first break down the initial condition
Ψ0 into normalised eigenvectors ψk of the matrix M as
Ψ0 =
X
Ck ψk
(3.57)
k
which leads to
Ψj =
X
(λk )j Ck ψk .
(3.58)
k
Then suppose that the eigenvector index k is ordered such that λn > λm if n < m. So if the
iteration index j is large enough, we have (λ1 )j C1 (λ2 )j C2 which gives
Ψj ≈ (λ1 )j C1 ψ1 .
(3.59)
Should this λ1 be simply equal to 1, it would mean that
~ + σt )ψ1 = (F + S)ψ1
(Ω̂ · ∇
(3.60)
thus ψ1 will be the solution to the time independent neutron transport equation that we
have been seeking and Ψj would naturally converge to it. So as we can see, as long as Ψj
is chosen such that it has a component along ψ1 , the algorithm will naturally output the
correct solution.
Obtaining λ1 6= 1 is also useful as that would mean
~ + σt )λ1 ψ1 = (F + S)ψ1
(Ω̂ · ∇
which translates to
(3.61)
∂
~ + σt )ψ1 .
ψ1 = (λ1 − 1)(Ω̂ · ∇
(3.62)
∂t
~ is a neutron conserving term, the
Given that σt and ψ1 are positive definite, and that Ω̂ · ∇
∂
overall sign of ψ1 should be decided by the (λ1 − 1) factor. Thus we can see that the value
∂t
of λ1 will also tell us if the system is subcritical or supercritical, allowing us to adjust the
23
parameters accordingly.
3.4
Determining input cross section values and energy groups
Determining the cross section values could prove to be a difficult task as there may not be
well-established theories on the energy and angular dependence of the cross sections. Instead,
these values are determined experimentally and a collection of such data is made publicly
available by the International Atomic Energy Agency (IAEA)[7]. Each set of data provides
roughly 10000 data points across 13 orders of magnitudes of energy in terms of eV. Assigning
an energy group to each data point could prove to be rather computational exhausting,
thus one will typically have to construct broader energy groups based on observations or
assumptions and then construct the group cross section based on the assumed in-group
energy spectrum if required based on equation (3.4). This work will use rough values for the
cross sections as the aim of this project is to study the possible methods in which calculations
regarding fissionable materials can be done. Thus being able to obtain a decent estimate
will good enough as further improvements to the results can always be come by taking finer
discretisations.
3.4.1
Angular dependence of scattering cross sections
For simplicity’s sake the angular dependence of σg0 g (r, µr ) will be determined using classical
hard sphere scattering throughout this work. It is safe to assume that the impact point of
the neutron is uniformly distributed over the scattering cross section as shown in figure (10).
Hard sphere scattering with the approximation m M gives the relation
Θ 1
s
dΘs
db = − Rn sin
2
2
(3.63)
where Θs = cos−1 (µr ) and Rn is the radius of the nucleus. The distribution of incoming neu2πb
db, giving the corresponding distribution to be just
trons on the impact parameter is
πRn2
1
dµr . So assuming that σg0 g (r) is known, σg0 g (r, µr ) can be obtained by distributing evenly
2
over the cosine of the scattering angle µr . The fact that this distribution is uniform also has
implications on the σlg0 g (r) as the identity from equation (3.18), together with the definition
of σlg0 g (r) in equation (3.9) would mean that σlg0 g (r) = δl0 σg0 g (r), thus greatly simplifying
the expression of Sgni . So in situations where there is only heavy nuclei (like uranium), it
is sufficient to use this simple expression for the cross section, thus making calculations faster.
24
Figure 10: Visualisation of the scattering cross section
For situations with light nuclei, a more accurate expression for σs (µ) dµ will need to be
found without considering the infinite mass limit. This can be done by looking at the following two identities[9]
σs (θ) = σs (θCM ) ×
m
M
and
−1
θCM = sin
q
2
2
m2
cos θ + 1 − M 2 sin θ
q
2
m2
1− M
2 sin θ
m
(3.64)
sin θ + θ
(3.65)
M
where the cross section in the CM frame is just the cross section from the infinite mass limit.
So the scattering cross section in lab frame turns out to be
σs (µ) dµ = σs (θ)
dθ
dµ
dµ
mp
= sin sin−1
1 − µ2 + cos−1 µ ×
M
×
1
p
dµ
− 1 − µ2
25
m
µ
M
q
2
m2
2
+ 1 − M 2 (1 − µ )
q
m2
2
1− M
2 (1 − µ )
(3.66)
Z
dµ
σs (µ) = σs . This can then be incorporated in the
2
discrete ordinate scheme by obtaining the scattering moments
up to a constant factor such that
Z
σlg0 g (r) =
3.4.2
dµ
Pl (µ)σg0 g (r, µ).
2
(3.67)
Group-to-group scattering
To even begin determining group-to-group scattering cross sections, we must look at the
energy dependence of the scattering, again using the hard-sphere model for the target nucleus. The details of this calculation will be left in Appendix C but the final result is that
(M − m)2
for
E 6 E 0 6 E, we have
(M + m)2
p(E → E 0 ) =
(M + m)2 2
πRn .
4mM E
(3.68)
1
dependence. So after normalising,
Thus we can see that the cross section will simply have
E
we get
σs (E)
dE 0
for αE 6 E 0 6 E
(3.69)
σs (E → E 0 ) dE 0 =
(1 − α)E
A − 1 2
where α =
for collision with nucleus of nucleon number A. Handling group-toA+1
group scattering can be tricky, as the respective cross sections also depends on the deflection
angle of the scattering event. This is more so when the energy groups are broad, as will be
the case when the three-group approximation is used. The reason is that the algorithm will
be insensitive to energy losses from within-group scattering. A more rigid approach would
be to use Monte-Carlo methods to simulate the accurate down-scatter probabilities given
that fission neutrons are generated with a specific distribution. For the case of this project,
an energy distribution of the neutrons will need to be assumed in order for there to be a
quick way of determining the approximate values for these cross sections. Also, to further
simplify matters, the down-scatter probability will not be coupled with the angular part of
the cross section. To put in simply, we have
Z
σg0 g = σsg0
dE
g0
φg0 (E) Eg − αE
v(E) (1 − α)E
where φg0 (E) is assumed.
26
for Eg 6 E 6
Eg
α
(3.70)
4
4.1
Initial results
Critical mass of pure U-235 sphere
In this example, only the fast energy group will be used, since that is where fission neutrons
are generated. Furthermore, neutrons will only lose up to 1.7% of their energy per collision,
σa
≈ 0.17 in this range, means neutrons are unlikely to
which together with the fact that
σt
lose much energy before being either absorbed or simply leaking out of the material. This
justifies using only the fast energy group since φ(E) will then not deviate very far from χ(E).
Using parameters from [5], the critical radius obtained for the pure U-235 sphere is 8.09 ±
0.01cm, compared to 8.70cm obtained from diffusion approximation. In terms of mass this
will be 42.4kg compared to 52.7kg from diffusion theory. In contrast, the critical mass by the
European Nuclear Society[10] (ENS) reads 46.7kg. Not much conclusion can be drawn from
the ENS value as it is not stated how their value was obtained. However, what can be said
is that the three values are in good agreement with the highest percentage difference being
21.6%. What is more convincing is that the solution for φ(r) in both cases agree very well
with each other as shown in figure (11). Note that neutron flux plots throughout this work
obtained in time-independent settings like this have arbitrary units, since the total number
of neutrons will necessarily be the same as in the initial input flux. Figure (12) shows the
convergence of the solution through the iteration process as ψ1 starts to dominate over all
other eigenvectors. This plot is important also as it shows that the algorithm did indeed
converge in the desired manner. The algorithm is also capable of giving us plots of the angular neutron flux like the ones shown in figure (14) and figure (13) for this example. This
is something that the simple diffusion approximation will not be capable of. In spherically
symmetric situations, this is most useful in comparing the outward and inward fluxes. In
this situation the boundary condition can be clearly seen in the 3D plot while the greater
outward flux compared to the inward flux is more clearly seen in the contour plot.
27
Figure 11: Plot of φ(r) solution obtained (lower curve) in comparison with solution from
diffusion approximation (upper curve).
Figure 12: Plot of ratio of total neutrons in current iteration in comparison to previous
generation, showing convergence of solution.
28
Figure 13: Contour plot of angular flux solution.
Figure 14: 3D plot of angular flux solution.
4.2
Critical mass of enriched uranium sphere
The critical mass of a homogeneous sphere of uranium enriched to arbitrary levels is obtained.
In this case, it is appropriate to model it with at least two distinct energy groups, due to
29
the existence of a threshold fission energy in U-238. It is typically agreed that this threshold
energy lies at around 1MeV and this can also be clearly seen in figure (15). As follows
Figure 15: Plot of σf (E) for U-238 in log scale, showing the threshold energy at around
1MeV[7].
from the case of pure U-235, only the fast energy group needs to be considered, but the
difference here is that it will have to be split into two at 1MeV due to the threshold property
of the cross section. The two isotopes have similar neutron yield per fission, 2.536 for U-235
and 2.819 for U238, and similar fission neutron spectrum as shown in figure (16). In fact,
given the two-energy groups taken here, 70.0% of neutrons generated for U-235 have energy
greater than 1MeV while for U-238 it is 69.5%. This means that we can simply use the same
parameters as the pure U-235 case for ν and χ(E). The results obtained can be summarised
in the figure (17). Note that as the enrichment approaches 0, the critical radius is diverging.
This is the expected result as pure U-238 is known to be unable to reach criticality. In fact,
natural uranium (0.7% enrichment) is also known to be unable to reach criticality and it has
been verified using this method up to a sphere of 10m in radius, equivalent to 80000 tonnes
of uranium!
30
Figure 16: Plot of fission neutron spectrum for U-238 in comparison with U-235. The upper
curve is for U-238[7].
Figure 17: Plot of critical radius versus enrichment of uranium
4.3
Temperature profile of water-cooled uranium sphere
In this section, we look at the case of a tampered fissile material. The configuration chosen
here is the pure U-235 immmersed in flowing, pressurised water to draw some similarities
31
with the water-cooled fuel in Pressurised Water Reactors (PWR) that are common nowadays.
Pure uranium is used here even though uranium dioxide is what is usually used as fuel so as
to make comparisons with the previously obtained results. Since this is a moderated system,
using the threee energy groups mentioned at the end of Chapter 2 will be appropriate. Figure
Figure 18: Plot critical radius of uranium vs thickness of water tamper.
(18) shows the expected trend of decreasing critical radius as more tamper is used. The next
two figures show the neutron flux and the temperature profile taken for the case of 5cm of
tamper. Note the presence of a slight bump in the flux near the interface, this is likely due
to the better downscattering effect of the tamper layer, which feeds thermal neutrons back
into the core, giving rise more fission in the outer layer of the core. Indeed figure (21) shows
the incoming thermal neutrons generated in the moderation layer. This effect is interesting
to note as we could also expect it to happen for other geometries given its mechanism. This
shows that one may not even need to use the appropriate geometry in initial studies of
certain situations as the effects may already be present in the spherical geometry.
32
Figure 19: Plot of scalar flux versus radius in water moderated U-235 sphere.
Figure 20: Close up of bump in neutron flux near core-moderator interface.
33
Figure 21: Plot of angular flux of thermal neutrons in water moderated uranium sphere.
Core radius here is 6.98cm.
Figure 22: Temperature profile of water moderate uranium sphere.
34
Figure (22) shows the temperature profile taken for only the uranium core. This is done
by solving the time independent heat equation with the energy generated from the fission
events as the source term. The flowing water surrounding the core is taken to be at 300◦ C[11],
typical in PWR reactors, which gives the boundary condition needed. The exact method for
this is included in Appendix B. The specific value of the temperature here is not important
since it can be adjusted by changing the total number of neutrons in the system. The
important part is that we can see that the temperature profile is very similar in shape to
the neutron flux, which is not surprising and can be useful in making quick evaluations of
situations. Being able to take the temperature profile like this can also be very useful for
full simulations of reactors.
5
Possible Applications
From its ability to reproduce the correct results in a few simple configurations, we should
now be confident that the algorithm is working as we intended it to. With that, we can start
to use it to analyse interesting situations like for example the yield of nuclear weapons or
designs of new and upcoming Generation IV nuclear reactors. The purpose of this section
is then to illustrate how the algorithm developed in Chapter 3 can be applied in the study
of such situations. Initial results will also be shown as demonstration but further analysis
will not be made here as it is not within the scope of the project to study these individual
situations in detail.
5.1
5.1.1
Estimating efficiency of uranium bomb
Theory
The purpose of this section is to illustrate how the computational solution to the timeindependent neutron transport equation can also be used to tackle a time dependent problem
like the detonation of a uranium bomb. This is a faster alternative to solving the full time
dependent equation, which will involve
1
~ + σt + F + S ψ(ti + ∆t) = 1 ψ(t).
+ Ω̂ · ∇
v∆t
v∆t
35
(5.1)
We can see here that we will need to derive the discretised form of the inverse operator
1
−1
~ + σt + F + S
+ Ω̂ · ∇
v∆t
(5.2)
which could involve significant amount of work.
In this case, the efficiency of a spherical uranium bomb with tamper is estimated. First
begin by looking at the overall picture of the detonation process. The bomb is driven by
energy from the fissioning of the uranium nuclei, which is to be sustained as a chain reaction mediated by neutrons. This bomb would start as a supercritical assembly of fissionable
material which would have positive neutron multiplicity, allowing the neutron flux and thus
the fission rate to build up with time. While this is happening, the energy released from
the fission events will also cause the bomb to expand. This effectively decreases the fission
cross sections by lowering the density of the material, causing the neutron multiplicity to
decrease. Eventually, the bomb will expand to the point where it is no longer critical and
the fission would then die down. Preliminary calculations have determined that the bomb
in general will not be able to fission all of its material before it becomes subcritical, thus
it is of interest to estimate the actual efficiency of uranium bombs in different configurations.
With the concept of neutron multiplicity, it is natural to assume the time dependence of
the neutron population in the material to be
N (t) = N0 eαt
(5.3)
for some effective multiplicity factor α. In terms of the neutron flux, the assumption would
be
ψ(~r, Ω̂, E, t) = ψ0 (~r, Ω̂, E)eαt .
(5.4)
Now it is important to note that in actual fact, this α factor would change with time due to
the expansion of the material causing the neutron multiplicity to decrease, thus using α(t)
would be more appropriate. However, if we are to only consider a very small time interval,
this α can be taken as a constant and the whole calculation would then consist of evaluating
the situation at each of these small time intervals.
36
Inserting the approximation back into the neutron transport equation would yield
α
1∂
~ − σt (~r, E) + F + S ψ(~r, Ω̂, E, t).
ψ(~r, Ω̂, E, t) = ψ(~r, Ω̂, E, t) = − Ω̂ · ∇
v ∂t
v
(5.5)
In order to proceed, we now bring in the computational algorithm, which is able to solve for
~
− Ω̂ · ∇ − σt (~r, E) + F + S ψ(~r, Ω̂, E, t) = 0.
(5.6)
Simply introduce an additional factor κ which can be seen as an adjustment to the value of
σt (~r, E) such that the equation to solve is
~ − σt (~r, E) − κ + F + S ψ(~r, Ω̂, E, t) = 0.
− Ω̂ · ∇
(5.7)
Then should a solution be found to this equation, it will satisfy
α
1∂
ψ(~r, Ω̂, E, t) = ψ(~r, Ω̂, E, t) = κψ(~r, Ω̂, E, t)
v ∂t
v
(5.8)
ψ(~r, Ω̂, E, t) = ψ0 (~r, Ω̂, E)eκvt
(5.9)
and the expression
will be valid for the small time interval ∆t as taken in the approximation. We then have at
each time step ti ,
ψ(~r, Ω̂, E, ti + ∆t) = ψ(~r, Ω̂, E, ti )exp(κi v∆t)
(5.10)
and the total number of fissions from ti to ti + ∆t is given by
Z
Z
∆t
0
Z
dΩ σf (~r, E)ψ(~r, Ω̂, E, ti )exp(κi vt0 )
0
1
=
σf (~r)φ(~r, ti ) exp(κi v∆t) − 1
κi v
f i(~r, ti ) =
dE
dt
(5.11)
and the total energy released up to this time will be
U (~r, ti ) =
i
X
ξf f i(~r, tj )
(5.12)
j=0
where ξf is simply the average energy released per fission event. Now to determine how
this energy that is released contributes to the expansion of the material, we look at the
37
following picture[12]. Consider the fact that per kilogram of weapons-grade uranium (> 80%
enrichment), the potential yield is about 3.28 × 1011 J while the energy required to bring the
same amount of uranium to boiling point is only of the order 106 J. Thus it is appropriate
to model the uranium sphere as a gas and to simplify calculation, we can assume that
all uranium gas particles are expanding outwards at the same speed u(t). This is an okay
assumption since for spherical geometry we expect most of the fission to happen in the center,
thus most of the pressure pushing the material outward will originate near the center. This
means that we can assume the entire sphere is being pushed outward by pressure from the
center point, thus all layers of the sphere would indeed expand at the same rate. The rate of
change of this expansion is then determined by the total pressure exerted by the fission energy
2R 3
d ~r U (~r, ti ).
released, which for the case of non-relativistic gas pressure, should be P (ti ) =
3
But for per particle energies greater than around 2keV, radiation pressure arising from
photons emitted by the deceleration of fission fragments should dominate, as should be the
case here with around 200MeV released per fission event. So the pressure should instead be
1
P (~r, ti ) =
3
Z
d3~r U (~r, ti ).
(5.13)
To show that radiation pressure dominates one simply need to express the respective pressures in the form
NU
kB T
(5.14)
Pcl =
V
and
8π 5 k 4 B
Prad =
T4
(5.15)
3
45c h3
3
and then find the respective per-particle energy E = kB T such that Prad > Pcl .
2
The work-energy theorem then gives the following relation of the total pressure to the outward expansion speed of the material
dV
dKE
du
4
dR
P (t)
=
= Mu
= πR2
dt
dt
dt
3
dt
38
Z
d3~r U (~r, t).
(5.16)
Noting that
dR
= u(t) and taking the time discretised form of the above equation yields
dt
u(ti + ∆t) = u(ti ) +
4πR2 P (t ) i
M
∆t
(5.17)
and the radius of the expanding sphere at each time step is then
R(ti + ∆t) = R(ti ) + u(ti + ∆t)∆t.
(5.18)
This updated radius will then be subsequently used to update the cross sections which will
R(t ) 3
i
and with this, calculations can begin for the growth
decrease by a factor of
R(ti + ∆t)
factor κv for the next time step.
At some point in time, system will expand to the point that it is no longer critical, with
κv < 0. The fission will then start dying down but not completely stop at this point, thus
it will be safer to continue running the process till the fission rate drops below a reasonable
threshold (say 0.1% of the peak fission rate). At the end the total number of fissions is then
compared to the amount of uranium available at the beginning to obtain the efficiency of
the bomb.
A final point to discuss will be the initial condition, that is the flux at t = 0, to be inserted
into the process. In fact, this calculation should be insensitive to the initial conditions.
Consider two different cases with N0 = 1 in comparison to N0 = 1000, the first case will
log(1000)
reach the same number of neutrons as the second case after about
≈ 8 generations
log ν
1000 σt
with the release of roughly
× 200MeV ≈ 1.5 × 10−8 J of energy, which will have
ν σt − σa
negligible contribution to the expansion. So it turns out that the first case will evolve exactly
into the second case, thus one should expect that both cases yield the same efficiency only
that the first case will take ∼ 8 generations longer in time to detonate.
For the case of the uranium sphere covered with a tamper material, the calculation will
have a small difference, since heat is not directly generated in the tamper, it is not reasonable to expect the tamper material to also undergo thermal expansion but instead simply
be pushed outwards by the expanding core. For ease of calculation, we assume that the
tamper is at least in a melted state. which will be smeared out on the surface of the core as
39
it expands. So given that the tamper itself does not expand, its thickness at each step can
be obtained after the new core radius is calculated as per above, which gives
Rtamp =
5.1.2
q
3
3
3
Rinit
− Rc,init
+ Rc3 − Rc .
(5.19)
Results
Using the aforementioned method, we aim to give an estimate of the yield of the ’Little
Boy’ weapon used during World War II while identifying any interesting dynamics of the
detonation process. Given the weapon design [13][14], it is reasonable to approximate in this
case as a sphere containing 64kg of uranium enriched to 80% encased with a 9cm layer of
Tungsten Carbide as tamper. Only one energy group will be used here, similar pure U-235
case. This is because there is only 20% U-238 in the core, thus it the bomb will be mainly
running on U-235 fission, making it unnecessary to split the energy group for U-238. Cross
sections from only the fission energy group will be used since the time-scale of the detonation
is too short for neutrons to be slowed significantly. The time step used here will have to be
less than the growth parameter κv for accuracy, which is of order 10−7 in this case. For this
case, 10−8 s is used as further decreasing it 10−9 s does not significantly change the results
but would render the calculation much slower.
Given these parameters, the yield of the bomb is calculated to be 2.06% which is equivalent to 23kt of TNT. This is actually 53% higher than the actual reported yield of 15kt[15]
but nevertheless as just an estimate, it is a decent result. The yield obtained here also means
that assumptions taken regarding the physical state of the uranium core and the dominant
form of pressure in the theory are justified, which is important.
The dynamics of the detonation is also interesting to look at. As seen in figure (23), most
of the fission events and thus the energy release comes from just the final 0.2µs of the detonation. The rest of the time was just used to build the fission rate to the point where it
becomes significant. Thus we can see here that it is very important for the tamper of the
bomb to hold on to the core for as long as possible since every bit of time towards the end
will contribute greatly to the yield. This effect can also be seen from figure (24) where the
expansion of the core is tracked, again emphasizing the significance of the last 0.2µs. With
this, it will be interesting to then investigate the actual effect varying the tamper mass on the
bomb yield. This will not be done in this project however, as it is not our aim to investigate
40
Figure 23: Plot of fission rate versus time
Figure 24: Plot uranium core radius versus time
41
the fission bomb in detail.
5.2
5.2.1
Pebble bed reactor
Theory
With the call for safer and more efficient reactor designs, the pebble reactor is one of the
newer Generation IV designs that is being developed. The reactor core consists of a gascooled bed laden with fuel pebbles as shown in the figure below.
Figure 25: Schematic for a pebble bed reactor developed by MIT[16].
As can be seen from figure (26), the pebbles are simply thrown into the core without any
42
Figure 26: Image of the core of an actual pebble bed reactor.
specific configuration in mind. The basic idea is that each pebble alone will not be enough
to go critical, but will be when large numbers of them are assembled together as shown.
This also allows for easy maintenance as individual pebbles can easily be swapped out during operation. Each pebble also in turn consists of more than 10000 microspheres, which
contain the uranium-dioxide fuel, embedded in graphite, which acts as moderation for the
fuel. Each microsphere has a layered structure as shown in figure (27), which is perfect for
analysis using our code that is developed for spherical geometries.
We will analyse the reactor by looking at just a single microsphere. This microsphere will be
surrounded by many other identical microspheres and the ensemble itself is also surrounded
by many other identical fuel pebbles. Thus we can see the situation as each single microsphere being put in a constant bath of neutron flux, which is what allows the fuel inside to
be able to go critical despite not being of critical mass. So the boundary condition in this
case will be set as
ψ(R, µ) = const
for µ < 0.
(5.20)
Since there is an external source in this situation, a time-independent solution to the neutron
transport equation is guaranteed. The computational algorithm can then be used to find
43
Figure 27: Cross sectional diagram of each fuel pebble and the microspheres embedded
within.
this time-independent solution from which the leakage can be obtained as
1
Z
ψout =
dµ ψ(R, µ)
(5.21)
0
and then subsequently compared to the incoming flux
Z
0
dµ ψ(R, µ)
ψin =
(5.22)
−1
which is specified at the beginning of the problem to determine if the dimensions of the
microsphere are sufficient to sustain the constant bath imposed.
Another benefit of the design is that due to how spread out the fuel pieces are, the flux
and thus the temperature throughout the reactor is expected to be very even. This is ideal
as with the lack of hot-spots, the design will be less prone to melting down in an emergency
situation.
44
5.2.2
Results
Since this is a reactor situation with moderator included to slow the neutrons down to thermal range, the usual three energy groups and their respective averaged cross sections will
be used. Ideally the fast group should also be split into two here to account for the U-238
given the low enrichment of the fuel, but since the main purpose of this is to illustrate the
method, using only three groups should be good enough. The microspheres are taken to be
TRISO particles [17]
Figure (28) shows the flux within each micropebble. As expected, the flux is higher at
the center as that is where the fission is concentrated but also evens out towards the outside
due to the presence of incoming neutrons. Also, the overall deviation between the flux is
Figure 28: Plot of scalar flux against radius for individual micropebble
also very small, within 5%, which should produce a more even temperature profile due to
the direct relationship between temperature and flux we saw in the water moderated sphere
example. This will indeed achieve the goal of the design for a smoother temperature profile.
The plot for the neutron flux is obtained by assuming an equal amount of incoming flux
in each of the energy groups. Each micropebble will in this case generate about 0.13% of
45
the incoming flux, with the outgoing flux having the proportions
φout,F
= 1.0141
φin,F
φout,I
= 0.9964
φin,I
φout,T
= 0.9934.
φin,T
(5.23)
The next step could be then to determine if the moderation surrounding each micropebble to slow the neutrons down to its original energy distribution. Also if the overall ratio
φF : φI : φT in the fuel pebble could be found, then the incoming flux is already fixed and
further analysis should be straightforward.
An interesting observation made while changing around the incoming flux ratios is that
φout,T
do not seem to be very sensitive to adjustments to the ratio of φin,T with respect to
φin,T
the other two energy groups. For example, setting φin,F : φin,I : φin,T = 1 : 1 : 10 gives
φout,F
= 1.1353
φin,F
φout,I
= 0.9964
φin,I
φout,T
= 0.9932
φin,T
(5.24)
φout,T
. What changes significantly, however, is the flux
φin,T
generated by the sphere, which now increases to 0.533%.
which is only about 0.02% change in
Increasing the incoming thermal flux yet again to φin,F : φin,I : φin,T = 1 : 1 : 100 gives
similar effect on outgoing flux ratios but only increased the generated flux to 0.65%. Using
φin,F : φin,I : φin,T = 1 : 1 : 1000 again gives similar results. This could be indication that
neutron moderation should be designed to effective up till a certain threshold, after which
the returns would be minimal.
46
6
Conclusion
We have developed a working algorithm for solving the time independent form of the neutron transport equation, which allows us to investigate steady state situations in fissionable
materials. The equation in principle, can be solved as accurately as we want it to since the
approximations taken are based on taking discretisations. So it is up to ones skill to be able
to determine how finely to discretise in order to obtain results of desired accuracy without
sacrificing too much computation time. Throughout this project, the discretisation used was
mostly rough, especially in the energy variable since the aim of the project is mainly illustration. After writing out the code on MATLAB, it is tested on several toy model situations
like the pure uranium sphere in Chapter 4. The results obtained were satisfactory and in
line with known estimates like the diffusion approximation. In Chapter 5, possible studies
that can be done with this code are suggested with some initial results presented to illustrate
how the study can proceed. These initial results may not have the desired accuracy for a
proper study in each case but the method has be presented and it is up to future work to
improve on them if desired.
47
References
[1] U.S. Energy Information Administration. Table 3. annual spent fuel discharges and burnup, 1968 - 2002. http://www.eia.gov/nuclear/spent_fuel/ussnftab3.htm, October 2004.
[2] Juliya Fisher. Energy density of coal.
JuliyaFisher.shtml, 2003.
http://hypertextbook.com/facts/2003/
[3] E. E. Lewis and Jr. W. F. Miller. Computational Methods of Neutron Transport. John
Wiley & Sons, Inc., 1984.
[4] R Nave. Nuclear binding energy. http://hyperphysics.phy-astr.gsu.edu/hbase/
nucene/nucbin.html.
[5] Elmer E. Lewis. Fundamentals of Nuclear Reactor Physics. Academic Press, 2008.
[6] N. Kornilov, F. J. Hambsch, I. Fabry, S. Oberstedt, T. Belgya, Z. Kis, L. Szentmiklosi,
and S. Simakov. The 235u(n, f) prompt fission neutron spectrum at 100k input neutron energy. Nuclear Science and Engineering: The Journal of The American Nuclear
Society, August 2010.
[7] International Atomic Energy Agency. Evaluated nuclear data files endf. https://
www-nds.iaea.org/exfor/endf.htm.
[8] B. G. Carlson and K. D. Lathrop. Transport theory - the method of discrete ordinates.
In H. Greenspan, C. N. Kelber, and D. Okrent, editors, Computing Methods in Reactor
Physics. Gordon and Breach, New York, 1968.
[9] Stephen T. Thorton and Jerry B. Marion. Classical Dynamics of Particles and Systems.
Brooks/Cole Cengage Learning, 2008.
[10] European Nuclear Society. Critical mass.
encyclopedia/criticalmass.htm.
https://www.euronuclear.org/info/
[11] Jacopo Buongiorno. Pwr description. Engineering of Nuclear Systems Lecture, 2010.
[12] B. Cameroon Reed. Arthur compton’s 1941 report on explosive fission of u-235: A look
at the physics. American Journal of Physics, 2007.
48
[13] Bruce Cameron Reed. The Physics of the Manhattan Project. Springer-Verlag Berlin
Heidelberg, 2011.
[14] John Coster-Mullen. Atom Bombs: The Top Secret Inside Story of Little Boy and Fat
Man. John Coster-Mullen, 2002.
[15] John Malik. The yields of the hiroshima and nagasaki nuclear explosions, 1985.
[16] Andrew C. Kadak. A future for nuclear energy: pebble bed reactors. International
Journal of Critical Infrastructures, 2005.
[17] R. N. Morris, D. A. Pett, D. A. Powe, and B. E. Boyack. Triso-coated particle fuel
phenomenon identification and ranking tables (pirts) for fission product transport due to
manufacturing, operations, and accidents. NUREG/CR-6844, Volume 1: Main Report,
2004.
49
A
Alternate derivations of discretised equations
This section is motivated by the fact that the assumption made in equation (3.3) may not
be valid and thus a different way of discretising the equations may be needed. The desired
form is still
Z
?
dE σ(~r, E)ψ(~r, Ω̂, E) = σg (~r)ψg (~r, Ω̂).
(A.1)
g
However, it turns out that simply taking the integral like this would result in
Z
dE σ(~r, E)ψ(~r, Ω̂, E) = σ̃g (~r, Ω̂)ψg (~r, Ω̂)
(A.2)
g
since
R
g
dE σ(~r, E)ψ(~r, Ω̂, E)
= σ̃g (~r, Ω̂).
R
dE
ψ(~
r
,
Ω̂,
E)
g
(A.3)
This process results in an unwanted angular dependence in the resulting energy-averaged
total cross section. One way to circumvent this is to expand the angular dependence of the
angular flux in terms of Legendre polynomials, giving
∞
X
(2l + 1)Pl (µ)φl (r, E)
ψ(~r, Ω̂, E) = ψ(r, µ, E) =
(A.4)
l=0
where
Z
φl (r, E) =
dµ
Pl (µ)ψ(r, µ, E).
2
(A.5)
The integral over energy group can then be taken, resulting in
Z
dE σ(r, E)ψ(r, µ, E) =
g
∞
X
(2l + 1)Pl (µ)σlg (r)φlg (r)
(A.6)
l=0
where
R
σlg (r) =
g
dE σ(r, E)φl (r, E)
φlg (r)
(A.7)
and φlg (r) is exactly as defined in equation (3.15). Now note that the scattering source is
R
also expanded similarly in terms of the Legendre polynomials, thus taking g dE on equation
50
(3.1) gives
∞
hµ ∂
XX
X
∂ 1 − µ2 i
2
0
0
0
r
+
ψ
(r,
µ)
=
(2l+1)P
(µ)
σ
(r)−σ
(r)δ
φ
(r)+χ
ν
σf g0 (r)φg0 (r).
g
l
lg g
lg
gg
lg
g
r2 ∂r
∂µ r
0
0
l=0
g
g
(A.8)
If one wishes to retain a form similar to the main derivation, the following term
σ̃g (r)ψg (r, µ) = σ̃g (r)
∞
X
(2l + 1)Pl (µ)φl (r, E)
(A.9)
l=0
can be added back into the equation, resulting in
∞
i
hµ ∂
XX
X
∂ 1 − µ2
2
0 g (r)φlg 0 (r)+χg ν
r
+σ̃
(r)
ψ
(r,
µ)
=
(2l+1)P
(µ)σ̃
+
σf g0 (r)φg0 (r)
g
g
l
lg
r2 ∂r
∂µ r
0
0
l=0
g
g
(A.10)
which has the exact same form as the main derivation with the exception that the cross
sections indicated with a tilde indicate different quantities with
R
σ̃g (r) = σ0g (r) =
g
dE σ(r, E)φ(r, E)
φg (r)
(A.11)
and
σ̃lg0 g (r) = σlg0 g (r) + σ̃g (r) − σlg (r) δgg0 .
B
(A.12)
Obtaining temperature profile from given neutron
flux
The heat equation for stationary state is
α∇2 T (~r) = −S(~r)
(B.1)
where α is the thermal diffusivity. Due to the rotational symmetry of the problem, the heat
equation can be simplified into
1 d 2d
S(r)
r
T (r) = −
.
2
r dr dr
α
51
(B.2)
Setting U (r) = T 0 (r) then gives
2
S(r)
U 0 (r) = − U (r) −
r
α
(B.3)
which is to be solved together with the boundary conditions
U (0) = 0
(B.4)
−αU (R) = C(T (R) − Tc ) + σ(T (R)4 − Tc4 ).
(B.5)
and
The second boundary condition given here comes from the fact that heat loss on the surface,
which is the left-hand side term, is given by the convective heat loss for a cooling fluid in
contact with the surface and the radiation heat loss, which give the first and second terms
of the right respectively. The exact form of this boundary condition is given by the physics
of the situation and can change accordingly.
1
The factor in equation (B.3) means an implicit scheme will have to be used when solving
r
numerically such that the singularity is avoided.The step size h is chosen to be equal to each
half-integer step (i.e. ri+1/2 − ri ). Using a fully-implicit scheme, the differential equations
are
2
S(r0 + h) U (r0 + h) −
U (r0 + h) = U (r0 ) − h
r0 + h
α
1
S(r0 + h) =⇒ U (r0 + h) =
U (r0 ) − h
α
1 + r02h
+h
(B.6)
T (r0 + h) = T (r0 ) + hU (r0 + h).
(B.7)
and
To determine the source term, first note the relation
3
E = N kB T
2
(B.8)
where N is the number of neutrons. Taking this equation on a per unit volume basis leads
to
3
Qf φF (r)σf = N (r)kB T
(B.9)
2
52
with Qf = 200 MeV being the energy released per fission. So the source term can now be
written as
Qf φF (r)σf
(B.10)
S(r) = 3
N
(r)k
B
2
where
N (r) =
X φg (r)
g
vg
.
(B.11)
To solve this set of equations, shooting method is used, taking an initial guess G0 of the
value of T (0) and then adjusting it in order to best fit the condition given in (B.5). This in
turn is done by defining from equation (B.5),
residue(G0 ) = αU (R) + C(T (R) − Tc ) + σ(T (R)4 − Tc4 )
(B.12)
noting that U (R) and T (R) are now considered functions of G0 . So finding the correct G0 is
equivalent to finding the root of residue(G0 ), which can be easily done by bisection or other
root-finding algorithms.
C
Derivation of energy loss from scattering
The can be analysed in the center of mass frame picture as seen in figure (29). Consider a
Figure 29: Analysis of collision in center of mass frame with impact parameter b.
neutron coming in with mass m and initial speed u colliding with a nucleus of mass M . The
53
center of frame would then move with speed
VCM =
mu
m+M
(C.1)
relative to the lab frame, which means that the initial speed of the nucleus in CM frame is
given by
U 0 = VCM
(C.2)
with primed quantities indicate they are taken in the CM frame. Since the collision considered here is elastic, the CM speeds of both the nucleus and neutron would have to be equal
before and after the collision, giving
u0 = v 0
and
U0 = V 0
(C.3)
where v and V are the final speeds of the neutron and nucleus respectively. Solving for the
final velocities then give
mu
(C.4)
V0 =
m+M
and
Mu
.
(C.5)
v 0 = u0 = u − VCM =
m+M
In the CM frame, since the total momentum will be 0, the neutron and nucleus will always
travel in opposite directions even after collision. Furthermore, according to figure (29), the
neutron would be deflected by an angle π − 2α. The final speed of the neutron in lab frame
is then obtained by vectorially adding V~CM to ~v 0 , which gives
q
2
− 2v 0 VCM cos(2α).
v = v 02 + VCM
(C.6)
1
1
The initial and final energies are E = mu2 and E 0 = mv 2 respectively, which gives
2
2
!
2
2
M + m − 2mM cos(2α)
E0 = E
.
(C.7)
(M + m)2
Now note that sin(α) =
b
2b2
and thus cos(2α) = 1 − 2 , which leads to
Rn
Rn
E
b2 2
E =
(M − m) + 4mM 2 .
(M + m)2
Rn
0
54
(C.8)
So for a fixed E,
σs (E → E 0 ) dE 0 = 2πb db = π db2 .
(C.9)
Here, it is important to first note that from equation (C.7) that σs (E → E 0 ) is only non-zero
(M − m)2
at
E 6 E 0 6 E and so finally,
(M + m)2
σs (E → E 0 ) =
(M + m)2 2
πRn
4mM E
(C.10)
within this region, which can be seen is independent of scattering angle.
D
MATLAB codes
This section will include some important parts of the MATLAB codes that were written and
used for the project. They will be attached to the back of the report.
55