Solving the neutron slowing down equation


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1 Solving the neutron slowing down equation Bertrand Mercier, Jinghan Peng To cite this version: Bertrand Mercier, Jinghan Peng. Solving the neutron slowing down equation <hal > HAL Id: hal Submitted on 11 Nov 2014 HAL is a multidisciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés. Distributed under a Creative Commons Attribution  NonCommercial 4.0 International License
2 Solving the neutron slowing down equation. Abstract : It is wellknown 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 selfshielded 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 (σ f =580 barns) for thermal neutrons (energy smaller than 1 ev) than for fast neutrons (σ f = 2 barns). This basic feature is used in most nuclear reactors to reduce the size of the core. A moderator (see e.g. 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 ev (see further references) and it is not easy to precisely evaluate the socalled 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. MonteCarlo analysis of the replacement of the by the 5. Exact solution vs MonteCarlo solution for the 6. Benefits of an exact solution to the neutron slowing down equation. 7. Reduction of the number of groups with selfshielding. 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 (E 0 / E) where E 0 = 10 MeV. If Σ s () denotes the scattering cross section and Σ a () the absorption cross section in the core of a nuclear reactor assumed to be homogeneous, then we have Σ() = Σ s () + Σ a (), so that in case of collision the neutron is absorbed with a probability Σ a ()/ Σ() or reappears at another lethargy > with a probability Σ s ()/ 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,
3 (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 we have 1 : = Σ Φ, (2.1) = Σ 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 ξ 0 h then we have (2.3) ξ + Σ Σ =ξσ. Σ Indeed, when, is given by (2.2), we have : so that = ξ Σ Φ /ξ /ξ = ξ Σ Φ /ξ ξ /ξ =ξ Σ Φ, = ξ ΣΦ (2.4) Φ = Σ + ξ Σ From (2.1), we have = Σ Φ = Σ Σ + ξ we can conclude that ξ + Σ Σ =ξσ Σ =ξ Σ Σ Σ which is the expected result. Σ Now if we assume the source =, that is to be a Dirac measure, then it is equivalent to solve Nov 10,
4 ξ + Σ Σ =0 0 = Σ Σ therefore, we have (2.5) =0 ξ Σ. Σ Indeed the function has a jump in =0 : this is physically obvious since all the neutrons start at lethargy =0 but only a fraction Σ go further. We have 0 Σ =1 but 0 = Σ ) Σ We note that the function 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 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 0 = 0 h We check that =1 = + =ξ, = = += +ξ = On the other hand, when, is given by (2.2), we have,= where We check that = ξ /ξ =1 0 0 h = ξ /ξ = ξ ξ /ξ + ξ /ξ, = = += +ξ A comparative plot of functions and is given below, in the case A = 2. = ξ /ξ =ξ We shall call the exact transition probability (1.1) and the approximate transition probability (2.2). Nov 10,
5 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 loi p loi q 4. MonteCarlo analysis of the replacement of the by the We shall analyze the effect of such an approximation with a MonteCarlo analysis in the case =2 that is in the case where the moderator is heavy water sigmac U238 vs lethargy 4792g NG Fig.2. capture cross section for (in barns) in the range The core is assumed to be homogeneous and made with a mixture of 3.7% enriched UO2 and heavy water D2O. Nov 10,
6 The dilution ratio (which is the ratio of the number of Deuterium atoms per to the number of U atoms per ) is assumed to be equal to. We shall make the MonteCarlo 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, Σ 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 (see Fig 2.) The capture cross sections for are given in Fig.3 (in arithmetic scale). 1.00E E E E E E E E E E E E+01 U8 capture U 3.7% 1.00E E E03 Fig.3. capture cross section for (in barns) in the range Solving the neutron slowing down equation (1.2) is quite easy. Each time a MonteCarlo neutron has a collision at some lethargy, the collision can be either an absorption (with probability Σ ) or an elastic collision (with probability Σ ) and in such a case, Σ Σ 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 MonteCarlo neutrons. The neutron current is obtained by counting the MonteCarlo neutrons which go further than lethargy. Function 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. Nov 10,
7 We used a pointsource in =9.6. In other words the source is a Dirac measure located in = glaw flaw 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 flaw glaw Fig.5 Comparison of the and the for =22 Nov 10,
8 The probability Σ for a neutron to be absorbed in the lethargy range , where the Σ 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 MonteCarlo solution for the To assess the accuracy of the MonteCarlo method we used, we have compared the MonteCarlo 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 ξ Σ Σ ). The results are given below flaw glaw exact Fig.6 Comparison of the with the exact solution (3.2) =22 The accuracy of our MonteCarlo calculations is then quite good. 6. Benefits of an exact solution to the neutron slowing down equation. Formula (2.4) shows that where =0, we have (6.1) Φ = or equivalently =ξ ΣΦ ξ Σ As we know, Σ 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 Nov 10,
9 Σ or Σ Φ. It also shows that, from to, neutrons are disappearing from the absorption phenomena. By writing Σ =Σ +Σ +Σ 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 Σ Σ by Uranium with probability Σ Σ and lead to a fission with probability Σ Σ. Up to now we have been interested in the resonance escape probability factor only. We could also start from = (that is 2 Mev) since this is the average energy for fission neutrons. In view of (2.1) we have (6.2) = Σ Φ 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 with formula (6.1) gives us Φ() and then by decomposition of Σ =Σ +Σ +Σ it gives us separately Σ Φ capture by hydrogen Σ Φ capture by Uranium Σ Φ absorption leading to fission The first term is negligible in the lethargy range = 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 "selfshielding". 7. Reduction of the number of groups with selfshielding. As we have noticed before, = 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 selfshielding. The usual way to reduce the number of groups is to replace on the large lethargy interval, the curve Σ by its weighted average Σ which is such that (7.1) Σ Φ = Σ Φ We propose to use alternatively the following nonlinear averaging by computing (7.2) Σ = Σ where is the homographic function Nov 10,
10 Σ Σ = Σ Σ for suitably choosen Σ. In other words, Σ = Σ. 1.4 flux phi vs lethargy phi 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 Σ =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 being regular, == on,. We let =, From the particular shape of, we have successively Σ Σ Σ =Σ = Σ = Σ Σ Σ 1 Σ Σ =Σ Σ Σ Σ ΣΣ Σ =Σ Σ Σ Σ Σ Σ =Σ Σ Σ Σ Σ = Σ Σ Σ Σ = Σ Σ Σ = Σ Σ Σ (7.3) Σ Σ +Σ = Σ Σ + Σ Σ is a good approximation of Nov 10, Σ
11 We note that Σ Σ so that (7.3) gives and Σ Σ 1 Σ Σ Σ = Σ Σ +Σ Σ Σ Σ Σ Σ 1 + Σ Σ Σ Φ ΦΣ 1 + Σ Σ Φ Σ Σ 1 + Σ Σ Therefore Σ Σ is small as long as 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 Σ. In one case the weight function is, in Σ the other one the weight function is Φ=. It should be no surprise that we get close results Σ since u is smooth. Σ phi 1/sigmat 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 Σ by Σ on an interval, and if we start from the same then it does not change! How to choose Σ (or )? We shall introduce microscopic cross sections. Nov 10,
12 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 Σ =α Σ =1 α where = / and = /, where =1 + denotes the microscopic cross section of the enriched uranium used in the fuel. We note that Σ Σ Σ = α = α α where = where = α α is the dilution factor. We choose =75 for =20 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 =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 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 (defined above) = 75. They obtained 380 groups in the range 9.6 < <15.3 which they complemented with 217 groups in the range < <9.6 and 55 groups in the range 15.3< u < Note that it is not appropriate to go beyond =19.97 because it corresponds to an energy =0.025 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 = Σ() Φ( ). We now comment the results we obtained with our 652 groups for =3.7% and =75. What we give in Fig.8 is a comparison of the neutron currents,, by using the following capture cross sections in the core (1) 1 + only, (2) (3) + where = α α = 3.75 in our case. Nov 10,
13 q1 q2 q Fig.9. Neutron current =Σ Φ in the range Note that, even though the enrichment =3.7% is small, the influence of is significant. On the contrary, the influence of is rather small. This is because the capture cross section of hydrogen is quite small in the range We note once again that the function 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 Σ =Σ +Σ +Σ 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 Σ Φ =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 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 (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% % (capture by H + absorption by U) and then split 34.19% = 3.20% % % (capture by U8 + capture by U5 + fission). Finally we obtained = 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 neutrons we loose 609 neutrons which are Nov 10,
14 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 < ; 3140 neutrons are captured which means = =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 Then, we get =76,3% which gives η=ν. = Finally, we obtain = = 0.615, but there are many ways of defining p or..η. 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 < < u Such a table immediately tells us that = x 2.45 = Remark : Relation with the effective integral The effective integral is introduced in Reuss With our notations, it is defined as :. = when we solve the slowing down equation, this is a quantity we evaluate. In our case where ξ=1, the relationship between and is =exp. Since we have evaluated it gives a way of computing. Here we would get =35.5. This value is relatively high since we use =1.6 and = Had we used =9.6 and = 15.3 as it is usually evaluated in books, we would have obtained When we solve the slowing down equation, we compute. However 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 easily. With our method, we have been able to compute Φ= Σ Σ 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,
15 group Σ Σ Σ 5 <u < < u < < u < < u < < u < < u < < u < < u < < u < < u < < u < < u < < u < < u < < u < Tab.2 Average absorption cross sections for =75 and 3.7% enriched UO2 These 15 values which cover the lethargy range 5<<20 correspond exactly to a subset of 600 values extracted from the 652 described above. We checked that if we replace the 600 values of σ by the 15 values if column 3 above (noted Σ although we should have indicated σ ) we obtain the same current 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). Nov 10,
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