Energy Effects in Modeling Neutron Diffusion: Two-Group Models

Abstract

In this chapter, we derive the multi-group diffusion equation (MGDE) and we illustrate how do we solve them in a way that allows us to calculate an accurate eigenvalue and accurate reaction rates. Since the cross sections vary wildly by multiple orders of magnitude over the energy range in a typical nuclear reactor, the major problem becomes one of determining accurate multi-group cross sections for the design problem under consideration.

Problems

Problem 4.1

List and describe the three approximations made in deriving the multi-group diffusion equations from the neutron transport equation.

Problem 4.2

Listed below are a set of typical group constants for a BWR. Calculate the multiplication factor for this reactor if it can be treated as a right circular cylinder of height 370 cm and a diameter of 340 cm.

Group Constant	Group 1	Group 2
$\mathrm{v}{\sum}_{\mathrm{fission}}^g$	0.001348	0.040783
${\sum}_{\mathrm{absorption}}^g$	0.001469	0.040249
D g	0.859599	0.142481
${\sum}_{\mathrm{removal}}^g$	0.044587	0.040249
χ g	1.0	0.0

Problem 4.3

Using the group constants from the previous Problem 4.1, calculate k_∞, the resonance escape probability, the fast fission factor, and the fast and thermal non-leakage probabilities for this reactor.

Problem 4.4

Calculate the critical radius for an unreflected right circular cylinder reactor characterized by the following group constants.

Group Constant	Group 1	Group 2
$\mathrm{v}{\sum}_{\mathrm{fission}}^g$	0.002562	0.15180
${\sum}_{\mathrm{absorption}}^g$	0.001930	0.09509
${\sum}_{\mathrm{transport}}^g$	0.466100	2.95420
${\sum}_{\mathrm{downstream}}^g$	0.055140	-------
χ g	1.0	0.0

1. Reactor height = 25 cm.

Problem 4.5

Calculate the multiplication factor for an unreflected fast reactor with the following cross-sectional data and core size:

Group constant	Group 1	Group 2
$\mathrm{v}{\sum}_{\mathrm{fission}}^g$	0.011040	0.005486
${\sum}_{\mathrm{absorption}}^g$	0.004598	0.004108
${\sum}_{\mathrm{downstream}}^g$	0.023420	--------
${\sum}_{\mathrm{transport}}^g$	0.028020	0.242200
χ g	0.55	0.45
Core Radius = 111 cm.	Core Height = 91.44 cm

Problem 4.6

Derive an equation for k_eff in case of the three-group problem described by the following equations.

$$ \left[\begin{array}{ccc}{D}_1{B}^2+{\sum}_r^1& 0& 0\\ {}-{\sum}_{s_0}^{1\to 2}& {D}_2{B}^2+{\sum}_r^2& 0\\ {}-{\sum}_{s_0}^{1\to 3}& -{\sum}_{s_0}^{2\to 3}& {D}_3{B}^2+{\sum}_r^3\end{array}\right]\left\{\begin{array}{l}{\phi}_1\\ {}{\phi}_2\\ {}{\phi}_3\end{array}\right\}=\left[\begin{array}{ccc}\mathrm{v}{\sum}_f^1& \mathrm{v}{\sum}_f^2& \mathrm{v}{\sum}_f^3\\ {}0& 0& 0\\ {}0& 0& 0\end{array}\right]\left\{\begin{array}{l}{\phi}_1\\ {}{\phi}_2\\ {}\phi \end{array}\right\} $$

Problem 4.7

Compute an infinite medium spectrum for the following 6 group fast reactor problem that is directly coupled.

	Group 1	Group 2	Group 3	Group 4	Group 5	Group 6
D	3.1348	2.5292	1.8095	0.7256	0.30601	0.1350
∑r	0.08175	0.05189	0.00813	0.03373	0.51380	0.37846
∑sg → g+1	0.07066	0.041489	0.002214	0.015413	0.00192	-----------
v∑f	0.037445	0.023684	0.010181	0.020241	0.42601	4.1648
χ	0.052579	0.52040	0.41085	0.016081	0.0008	0.0

Problem 4.8

Use the spectrum calculated as solution for Problem 4.7 to collapse the upper three groups to one group and the lower three groups to a second group for broad group cross sections. Calculate all five cross sections.

Problem 4.9

Estimate the minimum group spacing that will yield directly coupled multi-group for equations for C¹², D², Be⁹, and Na²².

Problem 4.10

What percentage of the neutrons slowing down in hydrogen will tend to skip groups if the group structure is chosen that (E_g − 1/E_g) = 100?

Problem 4.11

Write out the details of the multi-group diffusion equations, Mϕ = k⁻¹ Fϕ, for a four-group model in which:

• (a) There is direct coupling.

• (b) The fission source exists only in the upper two groups.

• (c) Only the lowest group contains thermal neutrons.

Problem 4.12

Solve the collision rate equation for Hydrogen as:

$$ F(E)={\int}_E^{E_0}d{E}^{\prime}\frac{F\left({E}^{\prime}\right)}{E^{\prime }}+{S}_0\delta \left(E-{E}_0\right) $$

By restricting our attention to E < E₀ and using the source as boundary condition as E → E₀.

Problem 4.13

Determine the neutron flux ϕ(E) resulting from an arbitrary source in an infinite hydrogenous medium by solving the infinite medium slowing down equation with a general source term S(E).

Problem 4.14

Show that for a mixture of N nuclides the collision rate density F(E) can be written in asymptotic region as:

$$ F(E)=\frac{S_0}{E\overline{\xi}}\kern1em \mathrm{and}\kern1em \overline{\xi}=\frac{\sum \limits_{i=1}^N{\sum}_{si}{\xi}_i}{\Sigma_s} $$

Problem 4.15

Show that for A > 1 and a monoenergetic source S₀ at E₀ we have the following relation:

$$ F(E)=\frac{S_0{E}_0^{\left(\alpha /\left(1-\alpha \right)\right)}}{E\overline{\xi}}\frac{1}{E^{\left(1/\left(1-\alpha \right)\right)}},\kern1em \alpha {E}_0<E<{E}_0 $$

Problem 4.16

Show that for A > 1 compute and plot the collision density function F₃(E) for neutrons that have had three collisions. Discuss the continuity of F₃(E) and its derivatives at E − E₀, αE₀, α²E₀, and α³E₀.

Problem 4.17

By means of the one-group treatment, estimate the critical radius of a spherical core of Uranuim-235 surrounded by a thick reflector of Uranium-238. The cross sections, etc. may be taken as the average values for neutrons in the energy range of 0.4 to 1.35 MeV, then:

Uranium-235:	σf = 1.27 (barn)	σa = 1.40 (barn)	σtr = 5.7 (barn)	ν = 2.52
Uranium-238:	σf = 1.27 (barn)	σa = 1.40 (barn)	σtr = 5.7 (barn)	ν = 2.52

Solution

The solution to this problem is as follows:

The density of uranium is approximately 19 g/cm³, so that atoms N for both Uranium-235 and Uranium-238 are close to 0.048 × 10²⁴ nuclei per cm³.

For the core,

$$ {\sum}_a=0.048\times 1.40=0.067\ {\mathrm{cm}}^{-1}\kern1em \mathrm{and}\kern1em {\sum}_{tr}=0.048\times 5.7=0.27\ {\mathrm{cm}}^{-1} $$

Thus:

$$ {L}^2=\frac{1}{3{\sum}_{tr}{\sum}_a}=\frac{1}{(3)(0.27)(0.067)}=18\;{\mathrm{cm}}^2 $$

$$ L=4.24\;\mathrm{cm} $$

For the reflector,

$$ {\sum}_a=0.048\times 0.13=0.0062\ {\mathrm{cm}}^{-1}\kern1em \mathrm{and}\kern1em {\sum}_{tr}=0.048\times 5.8=0.28\ {\mathrm{cm}}^{-1} $$

Thus.

$$ {L^2}_{refl.}=\frac{1}{3{\sum}_{tr}{\sum}_a}=\frac{1}{(3)(0.28)(0.0062)}=190\;{\mathrm{cm}}^2 $$

$$ {L}_{refl.}=13.78\;\mathrm{cm} $$

In Uranium-235 core, k_∞ is equal to η, i.e., to νσ_f/σ_a, so that:

$$ {k}_{\infty }=(2.52)(1.27)/(1.40)=2.3 $$

Hence, by the one-group critical equation, we can write:

$$ {B}_c^2=\frac{k_{\infty }-1}{L^2}=\frac{2.3-1}{18}=\frac{1.3}{18}=0.072\;{\mathrm{cm}}^{-2} $$

or

$$ {B}_c=0.27\;{\mathrm{cm}}^{-1} $$

It is found that R_c/L_refl. is not large in this case, so that for one-group critical equation for a spherical core with an infinitely thick reflector written as below must be used:

$$ {B}_c{R}_c\;\coth \left({B}_c{R}_c\right)=-\frac{D_r}{D}\left(1+\frac{R_c}{L_{refl.}}\right)+1 $$

However, since D and D_r are approximately equal, it reduces to:

$$ \tan \left({B}_c{R}_c\right)=-{B}_c{L}_{refl}. $$

Hence tan(0.27R_c) = − (0.27)(14) = − 3.8 and R_c is 6.8 cm. Note that the experimental value is close to 6.0 cm.