Solving the neutron slowing down equation

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Solving the neutron slowing down equationBertrand Mercier, Jinghan Peng

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Bertrand Mercier - Arthur Peng

Nov 10, 2014 1

Solving the neutron slowing down equation.

Abstract :

It is well-known that the neutron slowing down equation has an exact solution in the case where the

moderator is only made of hydrogen (A=1). In this paper we consider the case where the moderator

is made with heavy water (A=2) and show that it is possible to approximate the transition probability

by a law with the same expectation, but leads to an analytic solution of the same type as for A=1.

We also show that such an analytic solution gives a lot of information about what happens in the

core of a nuclear reactor, and gives a way to compute self-shielded cross sections.

Introduction :

Fission neutrons are fast neutrons (energy in the range 1 to 10 MeV).

However, Uranium 235 fission cross section is much higher (σf =580 barns) for thermal neutrons

(energy smaller than 1 eV) than for fast neutrons (σf = 2 barns).

This basic feature is used in most nuclear reactors to reduce the size of the core.

A moderator (see e.g. http://en.wikipedia.org/wiki/Neutron_moderator) is then needed to achieve

neutron slowing down.

Neutron slowing down in a reactor is a complex phenomenon due to the fact that nuclear fuel is

usually made of a combination of 238

U and 235

U, with proportions depending of uranium enrichment.

In fact, the 238

U absorption cross section is very noisy in the range 1-500 eV (see further references)

and it is not easy to precisely evaluate the so-called resonance escape probability factor p, i.e. the

probability for a neutron not to be captured in this energy range.

In the case of a homogeneous core, Uranium fuel and moderator are supposed to be intimately

mixed.

The outline of this paper is as follows

1. first form of the neutron slowing down equation

2. Second form of the neutron slowing down equation

3. Approximation of the exact transition probability �� − �� in the case A > 1

4. Monte-Carlo analysis of the replacement of the � − �� by the − ��

5. Exact solution vs Monte-Carlo solution for the − ��

6. Benefits of an exact solution to the neutron slowing down equation.

7. Reduction of the number of groups with self-shielding.

8. Computation with 652 groups data.

9. Non linear averaging on unit lethargy groups.

2. First form of the neutron slowing down equation.

Rather than using the neutron energy as a parameter, we shall use the lethargy defined as

= Log (E0 / E) where E0 = 10 MeV.

If Σs() denotes the scattering cross section and Σa() the absorption cross section in the core of a

nuclear reactor assumed to be homogeneous, then we have Σ() = Σs() + Σa(), so that in case of

collision the neutron is absorbed with a probability Σa()/ Σ() or reappears at another lethargy ’ >

with a probability Σs()/ Σ().

The probability density for the new lethargy ’ is p(,’) and is such that

� ��, �� = 1��

If the moderator is made with an atom whose atomic mass is equal to A, then it is well known that


Nov 10, 2014 2

(1.1) ��, � = ��

�� ≤ ≤ � + �0�� ! "

where # = $%��%&�'(

and � = −ln�#� (see e.g. +1,).

In case there is a homogeneous source S() in the reactor, then, in the steady state, the balance of

neutrons in the lethargy band [, + �] is the following :

S()� + � Σ-��Φ��, �� =�. Σ() Φ()�

which is equivalent to

(1.2) S() + � Σ-��Φ��, �� =�. Σ() Φ()

where is arbitrary in the range �0,∞�. Equation (1.2) is known as the first form of the neutron

slowing down equation.

2. Second form of the neutron slowing down equation.

As in reference +1,, let us introduce the neutron current

0�� = � �� Σ-��Φ��, ��

�.

we have +1,: (2.1)

11� 0�� = 2�� −Σ3��Φ��

which is known as the second form of the neutron slowing down equation.

Note that it is not very convenient to use since we have first to compute Φ.

However we shall prove that in the case where

(2.2) ��, � = �ξ��4�

��5/ξ�� ≤ 0!7ℎ�9��"

then we have

(2.3) ξ1:1�+

Σ;��Σ�� 0 = ξ

Σ<��Σ�� 2�� .

Indeed, when ��, � is given by (2.2), we have :

0�� = �ξ� Σ-��Φ��/ξ�� /ξ��

��.

= �ξ� Σ-��Φ��/ξ��. ξ��/ξ

= ξ� Σ-��Φ��, ��. = ξ�Σ��Φ�� − 2��

so that

(2.4) Φ�� = =��Σ��+

�ξ :��Σ��

From (2.1), we have

1:1� = 2�� −Σ3��Φ�� = 2�� −Σ3�� $=��Σ��+

�ξ:��Σ��'

we can conclude that

ξ1:1� +

Σ;��Σ�� 0 = ξ

Σ��Σ;��Σ�� 2�� = ξ

Σ<��Σ�� 2��

which is the expected result. ∎

Now if we assume the source 2�� = ?��, that is to be a Dirac measure, then it is equivalent to

solve


Nov 10, 2014 3

ξ1:1� +

Σ;��Σ�� 0 = 0

0�0&� = Σ<�.�Σ�.�

therefore, we have

(2.5) 0�� = 0�0&� �@� $− �ξ� Σ;�A�

Σ�A��. �B'.

Indeed the function → 0�� has a jump in = 0 : this is physically obvious since all the neutrons

start at lethargy = 0 but only a fraction Σ<�.�Σ�.� go further. We have 0�0�� = 1 but 0�0&� = Σ<�.�

Σ�.� )

We note that the function → 0�� is decreasing, which is physically obvious.

Formula (2.4) is well known in the case of hydrogen as a moderator.

It seems that it was not known for ξ < 1.

3. Approximation of the exact transition probability in the case A > 1.

In section 1, we have seen that when A > 1, the exact transition probability is not given by (2.2) but

by (1.1).

However, if we choose ξ = 1 − �E�� we claim that in both cases

(3.1) � ��, �� =� + ξ ,

in other words, they are different probability laws but they have the same expectation.

When � is the exact transition probability (2.1) we have

��, � = �� − �� where

�� = �� F

��G��0 ≤ � ≤ �0!7ℎ�9��H

We check that

� ��. = 1

� ��. = �

�� G��E. = �

�� 4+−��G,.E +� ��G��E. 5 =ξ

� ��, �� =�� − �� =�

�� +�� = ��. + ξ ∎

On the other hand, when ��, � is given by (2.2), we have

��, � = � − �� where

�� = �ξ��G/ξ�� ≥ 00!7ℎ�9�� "

We check that

� ��. = 1

� ��. = �

ξ� ��G/ξ��. =�

ξ$J−ξ��G/ξK.

� +� ξ��G/ξ�@�. ' = J−ξ��G/ξK.

� =ξ

� ��, �� =�� − �� =�

�� +�� = ��. + ξ ∎

A comparative plot of functions � and is given below, in the case A = 2.

We shall call � − �� the exact transition probability (1.1) and − �� the approximate transition

probability (2.2).


Nov 10, 2014 4

The advantage of changing from the � − �� to the − �� is that we shall benefit from the

existence of the exact solution (2.5) for the neutron current .

Fig 1. Comparison of functions � and

4. Monte-Carlo analysis of the replacement of the L − MNO by the P − MNO

We shall analyze the effect of such an approximation with a Monte-Carlo analysis in the case Q = 2

that is in the case where the moderator is heavy water.

Fig.2 . capture cross section for S(TU (in barns) in the range 9.6 ≤ ≤ 15.3

The core is assumed to be homogeneous and made with a mixture of 3.7% enriched UO2 and heavy

water D2O.

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 1 2 3 4 5 6

loi p

loi q

0.001

0.01

0.1

1

10

100

1000

10000

9.6 10.6 11.6 12.6 13.6 14.6

sigmac U238 vs lethargy 4792g

NG


Nov 10, 2014 5

The dilution ratio (which is the ratio of the number of Deuterium atoms per [\T to the number of U

atoms per [\T ) is assumed to be equal to �. We shall make the Monte-Carlo Analysis in 2 cases :

� = 3.75 and � = 22.

Note that � = 3.75 is an acceptable dilution factor for light water which is an excellent moderator ;

However for heavy water it leads to a quite low value for the resonance escape probability factor �,

as we shall see. This is the reason why we also used the � = 22 value.

Uranium cross sections used in this paper.

As we know, for Uranium, Σ3�� is a quite irregular function, with high values corresponding to the

Uranium resonance peaks and low values between. Then it is necessary to use a sufficiently high

number of values. In what follows, we have used 4792 values in the range 9.6 ≤ ≤ 15.3 (see Fig

2.)

The capture cross sections for S(T^ are given in Fig.3 (in arithmetic scale).

Fig.3 . capture cross section for S(T^ (in barns) in the range 9.6 ≤ ≤ 15.3

Solving the neutron slowing down equation (1.2) is quite easy.

Each time a Monte-Carlo neutron has a collision at some lethargy , the collision can be either an

absorption (with probability Σ;��Σ�� ) or an elastic collision (with probability

Σ<��Σ�� ) and in such a case,

the lethargy of the next event is found by sampling the transition probability ��, � or equivalently

by sampling the lethargy increment � by using either the � − �� or the − ��.

To obtain the results represented on Fig.4 & 5, we have used 5000 Monte-Carlo neutrons.

The neutron current 0�� is obtained by counting the Monte-Carlo neutrons which go further than

lethargy . Function → 0�� is obviously decreasing as can be seen from (2.1) or (2.5).

This is what we observe on Fig.4 and 5 where we compare the � − �� and the − ��.

1.00E-03

1.00E-02

1.00E-01

1.00E+00

1.00E+01

1.00E+02

1.00E+03

1.00E+04

9.60E+00 1.06E+01 1.16E+01 1.26E+01 1.36E+01 1.46E+01 1.56E+01

U8 capture

U 3.7%


Nov 10, 2014 6

We used a point-source in = 9.6.

In other words the source 2 is a Dirac measure located in = 9.6.

Fig.4 Comparison of the � − �� and the − �� for � = 3.75

The resonance escape probability factor � is then equal to the value of the current obtained in

= 15.3, that is � = 0.175 for the � − �� or � = 0.166 for the − ��.

As anticipated, this is a quite low value, and the reason is that, in the lethargy range of interest, the

scattering cross section is Σ- =3.4 barn for the deuterium atom compared to 20 barn for the

hydrogen atom.

Fig.5 Comparison of the � − �� and the − �� for � = 22

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

9.6 10.6 11.6 12.6 13.6 14.6 15.6

g-law

f-law

0.4

0.5

0.6

0.7

0.8

0.9

1

9.6 10.6 11.6 12.6 13.6 14.6 15.6

f-law

g-law


Nov 10, 2014 7

The probability Σ;��Σ�� for a neutron to be absorbed in the lethargy range 9.6 ≤ ≤ 15.3, where the

S(TU traps are located, is significantly larger for heavy water than for light water.

For � = 22, we obtain � = 0.551 for the � − �� or � = 0.547 for the − ��.

From these results, we conclude that the − �� is a quite good approximation to the � − ��.

5. Exact solution vs Monte-Carlo solution for the P − MNO

To assess the accuracy of the Monte-Carlo method we used, we have compared the Monte-Carlo

results to the exact solution as given in (3.5). (Note that the exact solution contains a Dirac Measure

in = 9.6 which is not represented here, see +1,: the amplitude of this Dirac measure is precisely

equal to 1 − ξΣ<�.�Σ�.� ).

The results are given below.

Fig.6 Comparison of the − �� with the exact solution (3.2) �� = 22�

The accuracy of our Monte-Carlo calculations is then quite good.

6. Benefits of an exact solution to the neutron slowing down equation.

Formula (2.4) shows that where 2�� = 0, we have

(6.1) Φ�� = �ξ :��Σ�� or equivalently 0�� = ξΣ��Φ��

As we know, Σ3�� is a quite irregular function, with high values corresponding to the Uranium

resonance peaks and low values between.

The above formula shows that Σ() Φ() is not only decreasing but also much more regular than

0

20

40

60

80

100

120

9 10 11 12 13 14 15 16 17

f-law

g-law

exact


Nov 10, 2014 8

Σ3�� or Σ3��Φ��. It also shows that, from .to , neutrons are disappearing from the absorption phenomena.

By writing Σ3 = Σà + Σ`b + Σc we see that neutrons can be

• captured by hydrogen

• captured by Uranium

• absorbed to produce one fission

Neutrons who survived the slowing down process will be thermalized at a lethargy ~20, which

means that they will be captured by hydrogen with probability ΣefΣ;

by Uranium with probability ΣegΣ;

and lead to a fission with probability ΣhΣ;

.

Up to now we have been interested in the resonance escape probability factor � only.

We could also start from . = 1.6094 (that is 2 Mev) since this is the average energy for fission

neutrons.

In view of (2.1) we have

(6.2) 0�� = 0�.� − � Σ3��Φ��i�j

which proves that all the current decrease is due to absorption but one part may be due to capture

and another one to fission.

Computing the exact decrease of the current 0�� with formula (6.1) gives us Φ() and then by

decomposition of Σ3 = Σà + Σ`b + Σc it gives us separately

• � Σà��Φ��i�j capture by hydrogen

• � Σ`b��Φ��i�j capture by Uranium

• � Σc��Φ��i�j absorption leading to fission

The first term is negligible in the lethargy range . = 1.6094 to � = 15.3, but this is not the case

for the fission part which leads to the � factor in Fermi's four factors formula.

So by solving the neutron slowing down equation with (6.1) we can not only obtain � but also �.

Now we could be tempted to use (2.5) for > 15.3 to evaluate also �and η however we should stop

at = 19,8 which corresponds to the average thermalization energy.

We shall show some example of such solutions for A = 1 (moderator = light water) in section 9.

Before that, we shall show how to reduce the number of groups by using "self-shielding".

7. Reduction of the number of groups with self-shielding.

As we have noticed before, 0�� = Σ() Φ() is a decreasing and regular function even though, both

functions Σ() and Φ() taken separately are irregular. As is shown above that Σ() has high peaks,

whereas (see Fig.7) Φ() has low peaks precisely at the same lethargy locations : this phenomenon is

known as self-shielding.

The usual way to reduce the number of groups is to replace on the large lethargy interval +�, m, the

curve → Σ3�� by its weighted average Σ∗∗ which is such that

(7.1) Σ∗∗ � Φ��o3 =� Σ3��Φ��o

3 �

We propose to use alternatively the following nonlinear averaging by computing

(7.2) ��Σ∗� = �o�3� ��Σ3��o

3 �

where � is the homographic function


Nov 10, 2014 9

��Σ3� = Σ;Σp&Σ;

for suitably choosen Σ1. In other words, Σ∗ = ��\!q��Σ3��.

Fig. 7 Exact flux Φ (x100) solution given by (3) with 652 groups for 3.7% enriched UO2.

We shall see that if Σ- is constant, the appropriate choice is to take Σ1 = Σ-.

We claim that in such a case, we obtain approximately the same result with (7.1) or (7.2). In other

words that Σ∗ ≅ Σ∗∗. For this, we note that 0�� being regular, 0∗ = \!q�0� = �

o�3� 0��o3 is a good approximation of

0 on +�, m,. We let

? = \�@ ∈ +�, m,|0�� − 0∗|

From the particular shape of �, we have successively

Σ∗

Σp&Σ∗ = ��Σ∗� = \!q $�4Σ3��5' = \!q $Σ;Σ'

Σ∗ u1 − \!q $Σ;Σ'v = Σ1 \!q $Σ;Σ '

Σ∗\!q $Σ�Σ;Σ' = Σ1 \!q $Σ;Σ '

Σ∗\!q $ΣpΣ' = Σ1 \!q $Σ;Σ '

Σ∗\!q $�Σ' = \!q $Σ;

Σ'

Σ∗\!q $:∗Σ' = \!q $:∗Σ;

Σ'

Σ∗\!q $:∗�:&:Σ

' = \!q $�:∗�:&:�Σ;Σ

'

(7.3) Σ∗\!q $:∗�:Σ' + Σ∗\!q $:

Σ' = \!q $�:∗�:�Σ;

Σ' + \!q $:Σ;

Σ'

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 5 10 15 20 25

flux phi vs lethargy

phi


Nov 10, 2014 10

We note that

�

Σ�� ≤�Σp

and Σ;��Σ�w� ≤ 1

so that (7.3) gives

Σ∗\!q $:Σ' − \!q $:Σ;

Σ' = \!q $�:∗�:�Σ;

Σ' + Σ∗\!q $:∗�:

Σ'

xΣ∗\!q $:Σ' − \!q $:Σ;

Σ'x ≤ $1 + Σ∗

Σp' ?

|Σ∗\!q�Φ� − \!q�ΦΣ3�| ≤ $1 + Σ∗Σp' ?

\!q�Φ�|Σ∗ −Σ∗∗| ≤ $1 + Σ∗Σp' ?

Therefore |Σ∗ −Σ∗∗| is small as long as \!q�Φ� is not too small, which is precisely the case for a

macrogroup.∎

Remark 7.1

Note that Σ∗like Σ∗∗ appears as a weighted average of Σ3. In one case the weight function is �

Σ�w�, in

the other one the weight function is Φ�� = :�w�Σ�w� . It should be no surprise that we get close results

since 0�u� is smooth.

Fig 8. Comparison of the weight functions leading to Σ∗ (in red) and Σ∗∗ (in blue).

Moreover, in view of formula (2.5) if we replace Σ3�� by Σ∗ on an interval �., �� and if we start

from the same 0�.� then it does not change �� !

How to choose ΣΣΣΣz (or {z) ?

We shall introduce microscopic cross sections.

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0 5 10 15 20 25

phi

1/sigmat


Nov 10, 2014 11

Let # be the partial volume fraction of fuel in the core and �1 − α� the partial volume fraction for the

moderator which is liquid water (note that we can neglect the O cross section compared to H).

We have

Σ3�� = α|b}b�� Σ1 = �1 − α�|a}a

where |b = 0.024510(~�7/[\T and |a = 0.048310(~�7/[\T, where

}b�� = �1 − ��}U�� + �}^�� denotes the microscopic cross section of the enriched uranium used in the fuel.

We note that

Σ;��

Σp&Σ;��= α�g�g��

��α��f�f&α�g�g��= �g��

�p&�g��

where }1 = �}a where � = ��α��fα�g

is the dilution factor.

We choose }1 = 75m for }a = 20m which gives # = 0.34 which corresponds to a moderation ratio

��α�α

=1.94 which is typical in a PWR.∎

8. Computation with 652 groups data.

From now on, we are in the case Q = 1 and then ξ = 1.

A table of 652 groups cross sections has been prepared by Olivia Feng, Alexis Jin and Arthur Peng

during their 2013-2014 bachelor project.

They started from the 4792 values plotted in fig 1 and 2 combining them with an enrichment

� = 3.7%. The lethargy range for these values extracted from the ENDF files is 9.6 < < 15.3.

They used the nonlinear averaging introduced here with }1(defined above) = 75m.

They obtained 380 groups in the range 9.6 < < 15.3 which they complemented with 217 groups in

the range 1.6094 < < 9.6 and 55 groups in the range 15.3 < u < 19.9723.

Note that it is not appropriate to go beyond = 19.97 because it corresponds to an energy

� = 0.025�B which is the thermalisation energy at room temperature.

With such a table of cross sections, we are able to solve exactly the neutron slowing down equation

(2.2) by using (3.5) and 0�� = Σ() Φ().

We now comment the results we obtained with our 652 groups for � = 3.7% and }1 = 75m.

What we give in Fig.8 is a comparison of the neutron currents 0�, 0(, 0Tby using the following

capture cross sections in the core

(1) �1 − ��}U�� + �}c�� only,

(2) }b�� (3) }b�� + �}a�� where � = ��α��f

α�g = 3.75 in our case.


Nov 10, 2014 12

Fig.9 . Neutron current 0�� = Σ��Φ�� in the range 1.6094 ≤ ≤ 19.8904.

Note that, even though the enrichment � = 3.7% is small, the influence of }^�� is significant.

On the contrary, the influence of }a�� is rather small. This is because the capture cross section of

hydrogen is quite small in the range 1.6 ≤ ≤ 15.3.

We note once again that the function → 0�� is decreasing and quite smooth.

We summarize in table 1 the results obtained by solving exactly the neutron slowing down equation

in this homogeneous core, where formula (7.2) and the decomposition Σ3 = Σà + Σ`b + Σc allows

us to compute in columns 3, 4, 5 and 6, the different cumulative captures and fission.

As an example, at = 15.3 we get � Σc��Φ��j = 0,0609 which means that 6.09% of the

neutrons which where created at the initial lethargy =1.609 have produced a fission in the

epithermal range and that 18.84% of them have produced a fission in the thermal range.

u Current captU8 captU5 captH fission

1.609 0.9987 0 0 0 0

15.305 0.6224 0.2786 0.0331 0.0023 0.0609

19.890 0.3577 0.3072 0.0672 0.0149 0.2493

Tab.1 : Capture and fission in the fast range and in the thermal range.

Now we see that 35.77% of them have reached the thermalisation energy � = 0.025�B (that is

= 19.890). From standard argument we can evaluate the thermal utilization factor � = 0.956 so

that we can split 35.77% = 1.58% + 34.19% (capture by H + absorption by U) and then split

34.19% = 3.20% + 4.57% + 26.42% (capture by U8 + capture by U5 + fission).

Finally we obtained 0.2493+0.2642=0.5135 fissions. With ν = 2.45 neutrons generated by fission it

gives �� = 1.26 which is comfortable.

How to evaluate the resonance escape probability factor from our computation ?

To evaluate � (the fast fission factor) we shall start from the 6.09% figure which appears in table 1 for

fast fission. This means that if we start with 10000 neutrons we loose 609 neutrons which are

0

0.2

0.4

0.6

0.8

1

1.2

0 5 10 15 20 25

q1

q2

q3


Nov 10, 2014 13

absorbed but we gain 1492 fission neutrons. The net balance is 883 additional neutrons which

corresponds to � = 1.09.

To evaluate p we count the captured neutrons in the lethargy range < 15.305 ; 3140 neutrons are

captured which means � = �(^��T�� = 0,666.

We could evaluate η by counting the number of fissions in the thermal domain (4526) and divide by

the number of absorptions which is 5930. Then, we get �c�- = 76,3% which gives η = ν. �c�- =1.896.

Finally, we obtain � = ��.η.E =0.615 , but there are many ways of defining p or �.

However, rather than trying to fit by all means within the 4 factors formula approach, is it not better

to summarize what happens in the core by a table like the following one ?

Capt U8 Capt U5 Capt H Fission

u < 15.3 0.2786 0.0331 0.0023 0.0609

15.3 < u 0.0606 0.0798 0.0284 0.4526

0.3392 0.1129 0.0307 0.5135

Such a table immediately tells us that �� =0.5135 x 2.45 =1.258.

Remark : Relation with the effective integral

The effective integral is introduced in Reuss §8.1.4. With our notations, it is defined as :

��cc = }1 � �;4��5�p&�;��

��i�j .

when we solve the slowing down equation, this is a quantity we evaluate.

In our case where ξ=1, the relationship between � and ��cc is � = exp�− ��hh�p � .

Since we have evaluated � it gives a way of computing ��cc.

Here we would get ��cc = 35.5m. This value is relatively high since we use . = 1.6 and �= 15.3.

Had we used . = 9.6 and �= 15.3 as it is usually evaluated in books, we would have obtained

28.8 m.

When we solve the slowing down equation, we compute ��cc. However ��cc leads only to p. The

way we solve the slowing down equation gives not only � but also �� and the 3 other factors of the 4

factors formula.

9. Non linear averaging on unit lethargy groups.

Of course tables give �

Σ�w�easily. With our method, we have been able to compute Φ�� = :�w�Σ�w�

exactly. The 652 values of both are plotted in Fig 7. Then, to illustrate the efficiency of our averaging

method we shall compare Σ∗, Σ∗∗ and the arithmetic average Σ�. We carry out the computations on

Table 2 where we have introduced unit lethargy groups.


Nov 10, 2014 14

group Σ∗∗ Σ∗ Σ�

5 <u <6 0.43655407 0.43668413 0.43816014

6 < u < 7 0.69983904 0.70005838 0.72358126

7 < u < 8 0.64539982 0.64552069 0.64948876

8 < u < 9 1.30459053 1.30369732 1.35222406

9 < u < 10 4.25194273 4.28431402 6.5199155

10 < u < 11 3.20163301 3.21261903 5.33694795

11 < u < 12 4.77618344 4.80592045 14.4293756

12 < u < 13 5.37284543 5.385717 40.1310686

13 < u < 14 6.63458438 6.51378881 62.6851997

14 < u < 15 10.4902327 10.2708093 117.004049

15 < u < 16 1.57314787 1.57365626 1.67438029

16 < u < 17 3.79242095 3.79142838 3.65192972

17 < u < 18 8.47902342 8.51880873 8.91876532

18 < u < 19 13.0166584 13.1141147 13.0177133

19 < u < 20 23.5539949 23.8300991 24.0063245

Tab.2 Average absorption cross sections for }1 = 75m and 3.7% enriched UO2

These 15 values which cover the lethargy range 5 < < 20correspond exactly to a subset of 600

values extracted from the 652 described above. We checked that if we replace the 600 values of σ3

by the 15 values if column 3 above (noted Σ∗ although we should have indicated σ∗) we obtain the

same current 0 at the common points : this is consistent with remark 7.1 The same is approximatively

true for column 2 (indicated Σ∗∗). However the arithmetic averages obtained in column 4 are

completely inadequate.

Biblographical references

+1, Paul Reuss, Neutron Physics, EDP sciences, 2008, 696 pages

+2, Serge Marguet, La Physique des Réacteurs Nucléaires, Collection EDF R&D, Editions Lavoisier (2013).

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