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**BLANKET OVERVIEW:**

The primary function of the FNR blanket is to absorb almost all the surplus neutrons from the nuclear chain reation in the FNR core zone and to use these neutrons to convert U-238 into Pu-239 and Pu-240.

A secondary function of the FNR blanket is to provide performance safety margin for the reactor passive cold shutdown safety systems by sufficient neutron absorption to compensate for changes in active fuel bundle reactivity.

Blanket thickness = 1.6 m. In prolonged reactor operation the blanket outer fuel bundle ring may be composed of spent active fuel bundles.

Neutrons diffuse through the blanket 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 least 99% of the neutrons that diffuse out of the core zone must be absorbed in the blanket. Hence the neutron random walk path length in the blanket must be long enouch to ensure 99% absorption. Otherwise the fuel breeding efficiency will be poor. As a neutron proceeds along its random walk path it loses kinetic energy, so the neutron absorption cross section changes. Hence the neutron random walk path length in the blanket must take this change in absorption cross section into account.

Neutrons that are not absorbed in the blanket must be totally absorbed in the liquid sodium guard band.

The combination of the core zone diameter, the blanket thickness, the liquid sodium guard band thickness set the primary sodium pool dimensions, which set the reactor cost. Thus from an overall system cost perspective the required blanket thickness is an important reactor design constraint. The required blanket thickness is relatively independent of reactor power. Hence from a cost perspective it is not practical to make very small fuel efficient breeding FNRs.

The average concentration of U-238 atoms in the blanket is a function of the blanket fuel design. This concentration determines the rate of absorption of neutrons along a neutron random walk path. Since the heat dissipation in the blanket zone is much less than in the core zone it is feasible to make the perimeter blanket fuel tubes larger in diameter which potentially increases the average U-238 concentration in the perimeter portion of the blanket. Whether or not there is any real economy in having different fuel tube sizes remains to be determined. Different fuel tube sizes complicate automation of the fuel manufacturing and fuel reprocessing.

**REQUIRED DATA:**

From Kaye & Laby the cross sections for high energy neutron scattering in a FNR blanket are:

SigmasNa = 3.9 b

SigmasFe = 4.5 b

SigmasCr = 4.7 b

SigmasU = 10.3 b

SigmasZr =

From Kaye & Laby the cross sections for high energy neutron absorption in a FNR blanket are:

SigmaaNa = 1.9 X 10^-3 b

SigmaaFe = 10 X 10^-3 b

SigmaaCr = 19 X 10^-3 b

SigmaaU = 340 X 10^-3 b

+ 21 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 blanket are:

Na =

Fe =

Cr =

U =

Zr =

**NEUTRON RANGE Lsb TO SCATTERING IN BLANKET:**

(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 La TO PARTIAL ABSORPTION IN BLANKET:

Fraction of initial neutrons unabsorbed is:

Exp(- La {[(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 blanket must be at least (6 La) to get the remaining fraction of neutrons under (1 / 400)

Hence solve for La

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

6 La / Lsb =

Hence the distance along one axis that a neutron travels before exiting the blanket = required blanket thickness is:

[(6 La / Lsb)^0.5] X [Lsb / (3^0.5)] =

**NEUTRON RANGE TO SCATTERING IN LIQUID SODIUM:**

The cross section for high energy neutron scattering in sodium is 2.62 b. Hence the distance Ls between successive high energy scatters in pure liquid sodium is given by:

Ls = 1 / [(2.62 X 10^-28 m^2 / atom) X (6.023 X 10^23 atoms / 23 gm) X (.927 gm / 10^-6 m^3)]

= 23 m / [ 2.62 X 6.023 X .927 X 10]

= .1572 m

In each scattering event total momentum is conserved.

Define:

Mn = neutron mass

Ms = scattering mass

Vn = neutron velocity

Vs = scattering mass velocity

Conservation of momentum gives:

Mn Vn = Ms Vs

or

(Vs / Vn) = (Mn / Ms)

The fractional loss of neutron kinetic energy in each scattering event is:

Ms Vs^2 / Mn Vn^2 = (Ms / Mn)(Mn / Ms)^2

= (Mn / Ms)

= 1 / (23)

Hence the neutron energy after a scattering event is (22 / 23) of its energy before the scattering event until the neutron kinetic energy reaches thermal energy.

(22 / 23)^2 = .9149338374

(22 / 23)^4 = .8371039268

(22 / 23)^8 = .7007429843

(22 / 23)^16 = .49104073

(22 / 23)^32 = .2411209985

(22 / 23)^64 = .0581393359

(22 / 23)^128 = .0033801824

Thus after about 128 scattering events a 3 MeV neutron has lost sufficient energy to drop to 10 keV.

Since scattering takes place in a 3 dimensional random walk the required thickness of liquid sodium required to provide this energy loss along a single axis is:

[(128)^0.5 X (.1572 m / 3^0.5)] = **1.03 m**

For neutrons with energies of less than 10 keV the average sodium atom absorption cross section is 0.310 barns = 0.31 X 10^-28 m^2 and the scattering cross section is 122.68 barns.
Hence the probable number of scatters before absorption is 122.68 / .31 = **395.74**.

The density of liquid sodium is 927 gm / lit. The atomic weight of sodium is 23. Avogadro's number is 6.023 X 10^23 atoms / mole. Hence the range Ls between scatters in liquid sodium for neutrons with midrange energies less than 10 keV is given by:

Ls = 1 / [927 gm / lit X 1000 lit /m^3 X 1 mole / 23 gm X 6.023 X 10^23 atoms / mole X 122.68 X 10^-28 m^2 / atom]

= 23 m / [927 X 1000 X 6.023 X 10^23 X 122.68 X 10^-28]

= 23 m / [9.27 X 6.023 X 122.68]

= **.003357 m**

Hence the expected distance along one axis that a neutron travels before absorption is:

(395.74)^0.5 X (.003357 m / 1.732) = **.0385 m**

However, after that distance there are still a lot of thermal neutrons. For thermal neutrons the scattering cross section is 3.090 b and the absorption cross section is 0.417 b.
Hence the average number of scatters before absorption is:

3.090 / 0.417 = 7.41 scatters.

The distance between scatters is:

Ls = 1 / [927 gm / lit X 1000 lit /m^3 X 1 mole / 23 gm X 6.023 X 10^23 atoms / mole X 3.090 X 10^-28 m^2 / atom]

= 23 m / [927 X 1000 X 6.023 X 10^23 X 3.090 X 10^-28]

= 23 m / [9.27 X 6.023 X 3.090]

= **0.1333 m**

Thus the expectation distance for thermal neutron travel along a single axis for a single scatter is:

0.1333 m / 1.732 = 0.07696 m

(N)^0.5 = 1.7 m / .07696 m

= 22.089

or

N = (22.089)^2 = 487.9

Assume that in 7.41 scatters the neutron flux is reduced by a factor of 2.71. Thus the neutron flux reduction factor is about:

exp (487.9 / 7.41) = exp(65.847)

which is sufficient.

**Thus a 2.8 m wide liquid sodium guard band around the reactor is sufficient for absorbing all emitted neutrons.**

**The issue with a breeder reactor is that as the breeding progresses neutrons start to be emitted by the blanket. To absorb all these neutrons we need 2.8 m of sodium between the outside edge of the blanket and the reactor wall and the heat exchange bundles. This issue substantially impacts the overall sodium pool dimensions.**

**Hence the corresponding liquid sodium pool dimensions are:
20 m diameter X 15.5 m deep**.

The neutrons exiting the reactor blanket vertically can avoid the sodium between the tubes by travelling through the fuel tube plenums. Hence there must be 3 m of liquid sodium both above and below the fuel tubes. The fuel tube end plugs may be subject to severe neutron irradiation.

This web page last updated December 16, 2018

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