XYLENE POWER LTD.

FNR CORE

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

References:
Breed and burn Reactor theory;
and
ASTRID like Fast Reactor Cores
and
Regional and World Level Scenarios for Sodium Cooled Fast Neutron Reactors

This web page is concerned with calculating the average concentrations of various atomic species in the FNR core zone with (3 / 8) inch OD fuel tubes and various values of X = fuel tube center to center distance.(0.500 inch, 0.5625 inch0

Additional information is that for the steel the weight fractions are 12% Cr and 88% Fe.

For the core fuel the assumed initial weight fractions are 20% Pu-239, 70% U-238 and 10% Zr. As shown on the web page titled: FNR Fuel Rods with new fuel the FNR should become critical at a core zone thickness of 0.44 m_____. The Pu concentration in the new fuel should be sufficient to meet this core zone thickness target.

Define:
Av = Avogadro's Number = 6.023 X 10^23 atoms / mole
Aw = atomic weight
Aws = atomic weight of sodium = 22.9898
Awi = atomic weight of iron = 55.845
Awc = atomic weight of chromium = 51.9961
Awu = atomic weight of U-238 = 238
Awz = atomic weight of zirconium = 91.22
Awp = atomic weight of Pu-239 = 239
Rhos = mass density of liquid sodium = .927 gm / cm^3
Rhoi = mass density of iron = 7.874 gm / cm^3
Rhou = mass density of U-238 = 19.1 gm / cm^3
Rhop = mass density of Pu-239 = 19.8 gm / cm^3
Rhoc = density of chromium = 7.19 gm / cm^3
Rhoz = density of zirconium = 6.51 gm / cm^3
Rhof = density of fission products

Fi = volume fraction of iron in steel
Fc = volume fraction of chromium in steel
(Fi Rhoi) / [(Fi Rhoi) + (Fc Rhoc)] = 0.88
(Fc Rhoc) / [(Fi Rhoi) + (Fc Rhoc)] = 0.12

Fp = volume fraction of plutonium in fuel
Fu = volume fraction of uranium in fuel
Fz = volume fraction of zirconium in fuel
Ff = volume fraction of fission products

For new fuel:
(Fp Rhop) / [(Fp Rhop) + (Fu Rhou) + (Fz Rhoz)] = 0.20
(Fu Rhou) / [(Fp Rhop) + (Fu Rhou) + (Fz Rhoz)] = 0.70
(Fz Rhoz) / [(Fp Rhop) + (Fu Rhou) + (Fz Rhoz)] = 0.10
Ff = 0

For used fuel:
(Fp Rhop) / [(Fp Rhop) + (Fu Rhou) + (Fz Rhoz)] = 0.127
(Fu Rhou) / [(Fp Rhop) + (Fu Rhou) + (Fz Rhoz)] = 0.70 - .077 = .623
(Fz Rhoz) / [(Fp Rhop) + (Fu Rhou) + (Fz Rhoz)] = 0.10
Ff Rhof = 1.000 - 0.127 - 0.623 - 0.10 = 0.15

EVALUATION FOR X = 0.5:

As shown on the web page titled: FNR FUEL BUNDLE in the core zone the material volume fractions for X = (1 / 2) inch are given by:
FSodium = 0.60268
Ffuel = 0.149321
Fsteel = 0.24799
 

Thus 1 m^3 of reactor core height contains 0.24799 m^3 of steel, 0.149321 m^3 of core fuel and 0.60268 m^3 of liquid sodium. We can use this information we need to calculate the average concentration of each atomic species in the reactor core.

FIND Ns:
Ns = (Fsodium) X 10^6 cm^3 / m^3 X Rhos (gm / cm^3) X (Av / Aws) (atoms / gm)
= 0.60268 X 10^6 cm^3 / m^3 X 0.927 gm / cm^3 X (6.023 X 10^23 atoms / 22.9898 gm)
= 0.14637 X 10^29 atoms / m^3
 

FIND Ni AND Nc:
Recall that:
(Fi Rhoi) / [(Fi Rhoi) + (Fc Rhoc)] = 0.88
(Fc Rhoc) / [(Fi Rhoi) + (Fc Rhoc)] = 0.12
or
(Rhoi) / [(Rhoi) + ((Fc / Fi) Rhoc)] = 0.88
(Rhoc) / [((Fi / Fc) Rhoi) + (Rhoc)] = 0.12
or
(Rhoi/ 0.88) = [(Rhoi) + ((Fc / Fi) Rhoc)]
or
(1.0 / 0.88) = 1.000 + (Fc / Fi)(Rhoc / Rhoi)
or
(Fc / Fi) = (0.12 / 0.88)(Rhoi / Rhoc)
then
(Fc / (Fc + Fi)) = 1 / [1 + (Fi / Fc)]
= 1 / [1 + (.88 / .12)(Rhoc / Rhoi)]
= 1 / [1 + 7.33333 (7.19 / 7.874)]
= 1 / [7.69629]
= 0.1299327

Then:
Fi / [Fc + Fi] = 1 - 0.1299327
= 0.8700673

Thus:
Ni = Fsteel X {Fi / [Fc + Fi]} X 10^6 cm^3 / m^3 X Rhoi X Av / Awi
= 0.24799 X 0.8700673 X 10^6 cm^3 / m^3 X 7.874 gm / cm^3 X 6.023 X 10^23 atoms / 55.845 gm
= 0.183236 X 10^29 atoms iron / m^3

Thus:
Nc = Fsteel X {Fc / [Fc + Fi]} X 10^6 cm^3 / m^3 X Rhoc X Av / Awc
= 0.24799 X 0.1299327 X 10^6 cm^3 / m^3 X 7.19 gm / cm^3 X (6.023 X 10^23 atoms / 51.9961 gm)
= 0.02683636 X 10^29 atoms chromium / m^3
 

FIND Nu, Np AND Nz:
The web page titled: FNR Fuel Rods shows that for new core fuel rods with 70% U, 20% Pu, 10% Zr:
Average density = 16.006 gm / cm^3.
In a core fuel rod the average density of U is:
RhoU =.7 (16.006 gm / cm^3)
= 11.2042 g /cm^3

Average density of Pu in each core fuel rod is:
RhoPu = 0.2 (16.006 g / cm^3)
= 3.2012 g / cm^3

Average density of Zr in each core fuel rod is:
RhoZr = 0.1 (16.006)
= 1.6002 gm / cm^3

Recall that Volume fraction of core fuel = Ffuel = 0.149321
Calculate Nu, Np, Nz for new fuel.

Nu = Ffuel X RhoU X Av / 238 gm
= 0.149321 X 11.2042 g / cm^3 X 6.023 X 10^23 atoms/ 238 gm X 10^6 cm^3 / m^3
= 0.04234 X 10^29 U atoms / m^3

Np = Ffuel X RhoPu X Av / 239 gm
= 0.149321 X 3.2012 g / cm^3 X 6.023 X 10^23 atoms / 239 g X 10^6 cm^3 / m^3
= .012046 X 10^29 Pu atoms / m^3

Nz = Ffuel X RhoZr X Av / 91.22 gm
= 0.149321 X 1.6002 gm / cm^3 X 6.023 X 10^23 atoms / 91.22 gm X 10^6 cm^3 / m^3
= 0.01578 X 10^29 Zr atoms / m^3
 

NEW FUEL DATA SUMMARY FOR X = 0.500 inch:
Ns = 0.14637 X 10^29 Na atoms / m^3
Ni = 0.183236 X 10^29 Fe atoms / m^3 Nc = 0.02683636 X 10^29 Cr atoms / m^3
Nu = 0.04234 X 10^29 U atoms / m^3
Np = .012046 X 10^29 Pu atoms / m^3
Nz = 0.01578 X 10^29 Zr atoms / m^3
 

SUMMARY FOR FNR CORE FOR X = 0.500 AT 15% FUEL BURNUP:
Ns = 1.4207 X 10^28 sodium atoms / m^3
Ni = 1.56657 X 10^28 iron atoms / m^3
Nc = 0.258851207 X 10^28 chromium atoms / m^3
Nu = 0.502683676 X 10^28 uranium atoms / m^3
Np = 0.102044479 X 10^28 plutonium atoms / m^3
Nf = 0.24104995 X 10^28 fission product atoms / m^3
Nz = 0.209966455 X 10^28 zirconium atoms / m^3
 

With respect to embrittlement or annealing, the only thing that comes to mind is a rule of thumb that says significant annealing only occurs at temperatures that are higher than 0.7 of the melting temperature in degrees Kelvin.

Supposedly the metal fuel becomes spongy at around 2% burn-up and then no longer applies undue pressure on the canister walls but instead squeezes upwards into the “plenum”, the empty space above the fuel rod. One might wonder whether such a sponginess would also allow the fuel to squeeze past any points of adherence.
 

This web page last updated October 31, 2023