XYLENE POWER LTD.

FNR BLANKET

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

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. In this matter we are concerned about the efficiency of that conversion.

A secondary function of the FNR blanket is to provide performance safety margin for the reactor cold shutdown safety systems by providing neutron absorption in the upper and lower FNR core zones.

Blanket vertical thickness = 1.35 m to 1.8 m.

In prolonged reactor operation the radial blanket outer fuel bundle ring may be composed of spent active fuel bundles. Hence the fuel bundles must be mounting plug compatible

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 calculated neutron random walk path length in the blanket should take this change in absorption cross section into account.

Neutrons that are not absorbed in the blanket must be totally absorbed by the gadolinium skirt and the 1.7 m wide liquid sodium guard band.

The combination of the core zone diameter, the blanket thickness, and the liquid sodium guard band thickness set the primary sodium pool dimensions, which determines 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 because they all require the same blanket thickness.

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. However, unproductive neutron absorption by the coolant, fuel tubes and fuel bundle steel significantly affect the performance of the breeding blanket. Since the heat dissipation in the blanket zone is much less than in the core zone it is feasible to make the blanket fuel rods larger in diameter than the core fuel rods which increases the average U-238 concentration in the blanket as compared to the core. It is also feasible to increase the diameter of the passive fuel tubes and positin them in a hexagonal array to increase the ratio of U-238 to other materials.

NEUTRON ABSORPTION:
For the simple case of only one type of absorbing atom:
Let N = number of remaining free neutrons;
Let X = distance along the free neutron path;
Na = concentration of absorbing atoms
Sigmaa = neutron absorption cross section of absorbing atoms
No - number of free neutrons at X = 0, t = 0.

Then:
dN / dX = - N(X) Na Sigmaa
N = No Exp[- Na Sigmaa (X-Xo)]

MATHEMATICAL IDENTITY:
For:
dN / dX = - C N
where C = constant

N = No Exp[- C (X - Xo)]

The average distance Xi travelled before absorption is given by:
Xi = [Integral from X = Xo to X = infinity of:
(-dN / dX)(X-Xo)dX] / No
= [Integral from X = Xo to X = infinity of:
N(X)(X-Xo)dX] / No
= [Integral from X = Xo to X = infinity of:
(No Exp[- C X])(X-Xo)dX] / No
= [Integral from X = Xo to X = infinity of:
(X - Xo) Exp[- C (X - Xo)]) d(X - Xo)]

Let U = C (X - Xo)
dU = C d(X - Xo)

Then:
Xi = Integral from U = 0 to U = infinity of:
U Exp(- U) dU / C

However:
Integral from U = 0 to U = infinity of:
U Exp(- U) dU
= 1

Hence:
Xi = 1 / C
 

NEUTRON ABSORPTION:
Consider a burst of N neutrons propagating through a large body. As the neutrons move some of the neutrons are absorbed by atoms of type "a" in the solid. Other neutrons are absorbed by species "b". The change dN in the number of neutrons N is given by:
dN = - N (Na Sigmaia + Nb Sigmab) dX
where:
Na = average concentration of type "a" atoms
Nb = average concentration of type "b" atoms
Sigmaa = neutron absorption cross section of type "a" atoms
Sigmab = neutron absorption cross section of type "b" atoms
dX = element of distance along the neutron propagation path

Now assume that there is concentration Nf of fissionable atoms present.
Sigmaf = cross section of fissionable atoms

Let No be the initial value of N. Let Xo be the intial value of X. Then this differential equation can be rewritten as:
dN = -N (Na Sigmaa + Nb Sigmab - G Nf Sigmaf) dN / N = -N (Na Sigmaa + Nb Sigmab - G Nf Sigmaf) which has the solution:
N = No Exp[- (Na Sigmaa + Nb Sigmab - G Nf Sigmaf)(X - Xo)]

CALCULATE BLANKET EFFICENCY
Blanket Efficiency
= [(Number of neutrons absorbed in blanket) / (Number of neutons emitted by core zone)]
X
[(Neutron absorption per unit neutron path length by U-238)
/ (Total neutron absorption per unit neutron path length)]

I the blanket is thick enough the first term is very close to unity. However, the second term is a bigger problem. The second term is:
[(Neutron absorption per unit neutron path length by U-238)
/ (Total neutron absorption per unit neutron path length)]
 
= (Sigmaaa Naa) / (Sigmaaa Naa + Sigmaab Nab + Sigmaac Nac + Sigmaad Nad + Sigmaae Nae + Sigmaaf Naf)
where:
Sigmaaa= absorption cross section of species a;
Sigmaab = absorption cross section of species b;
Sigmaac = absorption cross section of species c;
Sigmaad = absorption cross section of species d;
Sigmaae = absorption cross section of species e;
Sigmaaf = absorption cross section of species f;
and
Naa = average concentration of species a;
Nab = average concentration of species b;
Nac = average concentration of species c;
Nad = average concentration of species d;
Nae = average concentration of species e;

Naa = (Volume fraction of species a)(Rhoa)(atoms / mole)(1 / atomic weight a);
Nab = (Volume fraction of species b)(Rhob)(atoms / mole)(1 / atomic weight b);
Nac = (Volume fraction of species c)(Rhoc)(atoms / mole)(1 / atomic weight c);
Nad = (Volume fraction of species d)(Rhod)(atoms / mole)(1 / atomic weight d);
Nae = (Volume fraction of species e)(Rhoe)(atoms / mole)(1 / atomic weight e);
Naf = (Volume fraction of species f)(Rhof)(atoms / mole)(1 / atomic weight f);
where:
Rhoa = density of species a;
Rhob = density of species b;
Rhoc = density of species c;
Rhod = density of species d;
Rhoe = density of species e

Species Assignments:
a = U-238
b = Mo-92
c = Mo-94
d = Na-23
e - Fe-55.845
f = Cr-51.996

DENSITIES:
U-238 > Rhoa =
Mo-92 > Rhob =
Mo-94 > Rhoc =
Na-23 > Rhod =
Fe-55.845 > Rhoe =
Cr-51.996 > Rhof =

NEUTRON FAST ABSORPTION CROSS SECTIONS:
Sigmaaa =
Sigmaab =
Sigmaac =
Sigmaad =
Sigmaae =
Sigmaaf =

VOLUME FRACTIONS:
(Volume fraction of species a) = Vfa =
(Volume fraction of species b) = Vfb =
(Volume fraction of species c) = Vfc =
(Volume fraction of species d) = Vfd =
(Volume fraction of species e) = Vfe =
(Volume fraction of species f) = Vff =

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-238 = 10.3 b

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-238 = 340 X 10^-3 b
+ 21 X 10^-3 b (fissioning)
SigmaaNatural Mo = 21 to 50 X 10^-3 b (clearly an isotope separation is required)

Use of isotopically adjusted Mo as a FNR Fuel Structural Material

Attached are the graphs of the absorptions for total Mo and for the Mo isotopes, as well as for Fe-56 and for total Zr. The field of the graphs is transparent. (you may want to shift some of the graphs to a second page).

If you take your mouse and click on one of edge of any image, you can drag that image over any other. I have made the ordinate scales all the same. If you line up the ordinates (move the one that is on top), you can see the relative levels of the two graphs. Shift the top image back and forth so that you can see which is which.

While Mo has Mo-95 as about 70 times more absorbing than Zr in the low energy range, it is still 10 times higher in the high energy range.

However the average Mo is about the same as Fe-56 in the low energy range but is much more absorbing at high neutron energies. That’s not good for fast-spectrum operation.

Mo-97 is better in the low energy range but is identical to Mo-95, i.e. as bad, at high neutron energies.

The other Mo isotopes are better at high neutron energy, but still about 5 times higher than Zr.

The other concern is that Mo has a two-fold higher inelastic scattering cross section (Excel file) than iron / chromium. That means it is better at moderating neutrons than HT9, which is not what you want.

Your statement about elastic scatter being less due to the higher molecular weight is correct, but somewhat irrelevant. The major energy down-shift for high energy neutrons for anything but the very low MW atoms (e.g. H, D, C) is due to inelastic scatter. Fuel atoms are particularly good at that (see U-238 in Excel file), about 6 times worse than Na, atom for atom.

Mo Isotopes as compared to Fe-56 and Zr.

Fe-Cr-Ni-Mo inelastic cross sections

Iron is about 92% Fe-56, so it would not be much different than elemental iron.

Mo-92 is a bit better than the average Mo, but it still is worse than iron at high neutron energies.

All the data is from the Brookhaven Data bank at
https://www.nndc.bnl.gov/sigma/tree/index.html

If you go into
“Neutron Reactions”
And then
“Download Neutrons”

Click on the element, and it gives you the all the isotopes for which they have data. For some of them they have combined the isotopes into the natural compositions, but not for iron. One can do that oneself, but it is a painful exercise.

The NIST cross sections for thermal neutron absorption are:
SigmaaNa = 0.53 b
SigmaaFe = 2.56 b
SigmaaCr = 2.05 b
SigmaaU-238 = 2.68 b
SigmaaMo-92 = 19 X 10^-3 b
SigmaaMo-94 = 15 X 10^-3 b
SigmaaBe = 7.6 X 10^-3 b
SigmaaBi = 30 X 10^-3 b
SigmaaPb = 171 X 10^-3 b

BREEDING CONDITION:
The probability of surplus neutron capture by the surrounding U-238 needs to be at least 95%.

Let Lb = thickness of the blanket surrounding the reactor core.

The blanket is composed of fuel rods that do not contain fissionable material. Then the required minimum blanket thickness Lb is given by:
Lb = 3 /(Sigmaab Nub)
where Nub = average uranium atomic concentration in the blanket.

Nub =
[(3 X 476) blanket rods X (0.3112 kg / blanket rod) X (238 kg uranium / 260 kg blanket rod) X (6.023 X 10^23 atoms uranium / 0.238 kg uranium) / 0.2400 m^3
= 4.9201 X 10^27 atoms / m^3

Data from the UK Nuclear Data Library indicates that in the blanket the transport cross section is:
Sigmaab = 10.3 b.

Hence the minimum required blanket thickness Lb is given by:
Lb = 3 /(Sigmaab Nub)
= 3 / {10.3 X 10^-28 m^2 X 4.9201 X 10^27 atoms / m^3}
= 0.592 m

The contemplated blanket is 1.2 m thick to ensure completeness of neutron absorption by U-238.
 

OPERATIONAL NOTE:
The Pu-239 concentration in the blanket may be less than 2% by weight whereas in the core the initial Pu-239 concentration must be about 20% by weight. The Pu-239 concentration in the blanket is gradually bred up to 2%. Then blanket rod reprocessing is used to remove almost 90% of the uranium and zirconium from the blanket fuel rod material to leave behind 20% Pu-239 material which is suitable as fuel for the reactor core.
 

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)] =
 

This web page last updated March 13, 2021