Energy Effects in Modeling Neutron Diffusion: Two-Group Models

Abstract

In this chapter we derive the multigroup 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 is determining the accurate multigroup cross sections for the design problem under consideration.

Problems

• Problem 4.1: List and describe the three approximations made in deriving the multigroup 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 of 370 cm and a diameter of 340 cm.

  Group constant	Group 1	Group 2
  vΣ gfission	0.001348	0.040783
  Σ gabsorption	0.001469	0.040249
  D g	0.859599	0.142481
  Σ gremoval	0.044587	0.040249
  χ g	1.0	0.0

• Problem 4.3: Using the group constants from the previous Problem 4.1, calculate ${k}_{\infty }$, 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
  vΣ gfission	0.002562	0.15180
  Σ gabsorption	0.001930	0.09509
  Σ gtransport	0.466100	2.95420
  Σ gdownstream	0.055140	–
  χ g	1.0	0.0

  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
  vΣ gfission	0.011040	0.005486
  Σ gabsorption	0.004598	0.004108
  Σ gdownstream	0.023420	–
  Σ gtransport	0.028020	0.242200
  χ g	0.55	0.45

  $$ \mathrm{Core}\ \mathrm{radius} = 111\ \mathrm{cm}\kern1em \mathrm{Core}\ \mathrm{height} = 91.44\ \mathrm{cm} $$

• Problem 4.6 Derive an equation for k _eff in case of the three-group problem described by the following equations:

  $$ \left[\begin{array}{ccc}\hfill {D}_1{B}^2+{\varSigma}_{\mathrm{r}}^1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill -{\varSigma}_{{\mathrm{s}}_0}^{1\to 2}\hfill & \hfill {D}_2{B}^2+{\varSigma}_{\mathrm{r}}^2\hfill & \hfill 0\hfill \\ {}\hfill -{\varSigma}_{{\mathrm{s}}_0}^{1\to 3}\hfill & \hfill -{\varSigma}_{{\mathrm{s}}_0}^{2\to 3}\hfill & \hfill {D}_3{B}^2+{\varSigma}_{\mathrm{r}}^3\hfill \end{array}\right]\left\{\begin{array}{c}\hfill {\phi}_1\hfill \\ {}\hfill {\phi}_2\hfill \\ {}\hfill {\phi}_3\hfill \end{array}\right\}=\left[\begin{array}{ccc}\hfill v{\varSigma}_{\mathrm{f}}^1\hfill & \hfill v{\varSigma}_{\mathrm{f}}^2\hfill & \hfill \mathrm{v}{\varSigma}_{\mathrm{f}}^3\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\left\{\begin{array}{c}\hfill {\phi}_1\hfill \\ {}\hfill {\phi}_2\hfill \\ {}\hfill {\phi}_3\hfill \end{array}\right\} $$

• Problem 4.7 Compute an infinite-medium spectrum for the following six-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
  Σ r	0.08175	0.05189	0.00813	0.03373	0.51380	0.37846
  ${\varSigma}^{\mathrm{s}g\to g+1}$	0.07066	0.041489	0.002214	0.015413	0.00192	–
  vΣ 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 multigroup 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 $\left({E}_{g-1}/{E}_g\right)=100$?

• Problem 4.11 Write out the detailed of the multigroup diffusion equations, $\boldsymbol{M}\phi ={k}^{-1}\boldsymbol{F}\phi$, 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

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

  by restricting our attention to $E<{E}_0$ and using the source as boundary condition as $E\to {E}_0$

• 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

  $$ \begin{array}{ccc}\hfill F(E)=\frac{S_0}{E\xi}\hfill & \hfill \mathrm{and}\hfill & \hfill \overline{\xi}=\frac{{\displaystyle \sum_{i=1}^N{\varSigma}_{\mathrm{s}i}{\xi}_i}}{\varSigma_{\mathrm{s}}}\hfill \end{array} $$

• Problem 4.15 Show that for $A>1$ and a monoenergetic source S ₀ at E ₀, we have the following relation:

  $$ \begin{array}{cc}\hfill F(E)=\frac{S_0{E}_0^{\left(\alpha /\left(1-\alpha \right)\right)}}{E\overline{\xi}}\frac{1}{E^{\left(\alpha /\left(1-\alpha \right)\right)}},\hfill & \hfill \kern1.2em \alpha {E}_0<E<{E}_0\hfill \end{array} $$

• Problem 4.16 Show that for $A>1$, compute and plot the collision density function F ₃(E) for neutrons that have had three collisions. Discuss the continuity of F ₃(E) and its derivatives at $E-{E}_0$, αE ₀, α ² E ₀, and α ³ E ₀:

• Problem 4.17 By means of the one-group treatment, estimate the critical radius of a spherical core of uranium-235 surrounded by a thick reflector of uranium-238. The cross sections may be taken as the average values for neutrons in the energy range of 0.4 to 1.35 MeV; then

  Uranium-235:	${\sigma}_{\mathrm{f}}=1.27\;\left(\mathrm{barn}\right)$	${\sigma}_{\mathrm{a}}=1.40\;\left(\mathrm{barn}\right)$	${\sigma}_{\mathrm{tr}}=5.7\;\left(\mathrm{barn}\right)$	$v=2.52$
  Uranium-238:	${\sigma}_{\mathrm{f}}=1.27\;\left(\mathrm{barn}\right)$	${\sigma}_{\mathrm{a}}=1.40\;\left(\mathrm{barn}\right)$	${\sigma}_{\mathrm{tr}}=5.7\;\left(\mathrm{barn}\right)$	$v=2.52$

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,

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

Thus,

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

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

For the reflector,

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

Thus,

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

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

In uranium-235 core, ${k}_{\infty }$ is equal to η, i.e., to vσ _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

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

or

$$ {B}_{\mathrm{c}}=0.27\;{\mathrm{c}\mathrm{m}}^{-1} $$

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:

$$ {B}_{\mathrm{c}}{R}_{\mathrm{c}} \coth \left({B}_{\mathrm{c}}{R}_{\mathrm{c}}\right)=-\frac{D_{\mathrm{r}}}{D}\left(1+\frac{R_{\mathrm{c}}}{L_{\mathrm{r}\mathrm{efl}.}}\right)+1 $$

However, since D and D _r are approximately equal, it reduces to

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

Hence, $\tan \left(0.27{R}_{\mathrm{c}}\right)=-(0.27)(14)=-3.8$ and R _c is 6.8 cm. Note that the experimental value is close to 6.0 cm.