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XYLENE POWER LTD.

FNR CORE

By Charles Rhodes, P.Eng., Ph.D.

CORE OVERVIEW:
The function of the FNR core is to sustain a nuclear chain reaction via fissioning of Pu-239, Pu-240 and other transuranium actinides while emitting surplus neutrons to the FNR blanket. A fundamental question from a practical reactor engineering perspective is: "How thick must the core be?"

An important technical issue that must be addressed to answer the aforementioned questions is: "What is the ratio of neutron random walk path length to core thickness?"

Neutrons diffuse through the core by scattering. At each scatter a neutron loses a small fraction of its kinetic energy. Between successive scatters the number of neutrons reduces due to neutron absorption. Our first concern is that at about (1 / 3) of the neutrons that are released in the core zone should be absorbed in the core zone. Hence the neutron random walk path length in the core zone must be long enouch to ensure 33% absorption. Otherwise the nuclear reaction will not be sustained.

Neutrons that are not absorbed in the core should be almost totally absorbed in the blanket.

The required blanket thickness is relatively independent of reactor power.

The average concentrations of Pu-239 and U-238 atoms in the core is a function of the core fuel design. These concentrations determine the rate of absorption of neutrons along a neutron random walk path.

REQUIRED DATA:
From Kaye & Laby the cross sections for high energy neutron scattering in a FNR core are:
SigmasNa = 3.7 b
SigmasFe = 3.8 b
SigmasCr = 4.2 b
SigmasU = 9.4 b
SigmasZr =

From Kaye & Laby the cross sections for high energy neutron absorption in a FNR blanket are:
SigmaaNa = 1.4 X 10^-3 b
SigmaaFe = 8.6 X 10^-3 b
SigmaaCr = 14 X 10^-3 b
SigmaaU = 250 X 10^-3 b
+ 41 X 10^-3 b (fissioning)
SigmaaZr =

The atomic weights are:
Na = 23
Fe =
Cr =
U = 238
Zr =

The densities are:
Na =
Fe =
Cr =
U =
Zr =

The volume fractions in the core are:
Na =
Fe =
Cr =
U =
Zr =
 

NEUTRON RANGE Lsc TO SCATTERING IN CORE:
(1 / Lsb)
= [(2.62 X 10^-28 m^2 / atom) X (6.023 X 10^23 atoms / 23 gm) X (.927 gm / 10^-6 m^3) X (Na volume fraction)]
+ X (Fe volume fraction)
+ X (Cr volume fraction)
+ X (U volume fraction)
+ X (Zr Volume fraction)
 
=

NEUTRON RANGE Lac TO PARTIAL ABSORPTION IN CORE:
Fraction of initial neutrons unabsorbed is:
Exp(- Lac {[(SigmaaNa (6.023 X 10^23 atoms / 23 gm x (gm / m^3) X (Na volume fraction)]
+ [(SigmaaFe (6.023 X 10^23 atoms / ___ gm x (__gm / m^3) X (Fe volume fraction)]
+ [(SigmaaCr (6.023 X 10^23 atoms / ___ gm x (__gm / m^3) X (Fe volume fraction)]
+ [(SigmaaU (6.023 X 10^23 atoms / ___ gm x (__gm / m^3) X (U volume fraction)]
+ [(SigmaaZr (6.023 X 10^23 atoms / ___ gm x (__gm / m^3) X Zr volume fraction)]}
= (1 / 2.718)  

The neutron path length in the core must be sufficient to allow (2 / 3) of the neutrons to escape from the core into the blanket

Hence solve for Lac

Hence required minimum number of neutron scatters while in the blanket is:

Hence the distance along one axis that a neutron travels before exiting the core is:

 

This web page last updated December 14, 2018

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