# XYLENE POWER LTD.

## FNR CORE

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

This web page is concerned with calculating the average concentrations of various atomic species in the FNR core zone and in the adjacent FNR upper and lower regions.

As shown on the web page titled: FNR FUEL BUNDLE in the core zone the material volume fractions are given by:

SUMMARY OF INITIAL CORE ZONE CROSS SECTIONAL AREA FRACTIONS:
Sodium = 0.60101
Core Fuel = 0.180882
Steel = 0.21730

Hence:
volume fraction of steel = Fsteel = 0.21730
Volume fraction of core fuel = Ffuel = 0.180882
Volume fraction of sodium = Fsodium = 0.60101

Thus 1 m^3 of reactor core contains 0.21730 m^3 of steel, 0.180882 m^3 of core fuel and 0.60101 m^3 of liquid sodium. Using this information we need to calculate the average concentration of each atomic species in the reactor core.

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

For the core fuel the guessed 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 exact Pu concentration in the new fuel should be set 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 = 23
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

Ns = initial number of sodium atoms per m^3 in core zone
Ni = initial number of iron atoms per m^3 in core zone
Nc = number of chromium atoms / m^3 in core zone
Np = number of plutonium atoms / m^3 in core zone
Nu = number of uranium atoms per m^3 in core zone
Nz = number of zirconium atoms per m^3 in core zone
Nf = number of fission product atoms / m^3 in core zone
Vi = volume of iron in core zone
Vc = volume of chromium in core zone
Vt = total steel volume in core zone

FIX FROM HERE

FIND Ns:
Ns = (Fsodium) X 10^6 cm^3 / m^3 X Rhos (gm / cm^3) X Av / Aws (atoms / gm)
= 0.5852505 X 10^6 cm^3 / m^3 X 0.927 gm / cm^3 X (6.023 X 10^23 atoms / 23 gm)
= 0.14207 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)
or
(Fc / (Fc + Fi)) = 1 / [1 + (Fi / Fc)]
= 1 / [1 + (.88 / .12)(Rhoc / Rhoi)]
= 1 / [1 + 6.5 (7.19 / 7.874)]
= 1 / [6.935356871]
= 0.1441886868

Then:
Fi / [Fc + Fi] = 1 - 0.1441886868
= 0.8558113132

Thus:
Ni = Fsteel X {Fi / [Fc + Fi]} X 10^6 cm^3 / m^3 X Rhoi X Av / Awi
= 0.21555 X 0.8558113132 X 10^6 cm^3 / m^3 X 7.874 gm / cm^3 X 6.023 X 10^23 atoms / 55.845 gm
= 0.156657 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
= 0.21555 X 0.144188686 X 10^6 cm^3 / m^3 X 7.19 gm / cm^3 X 6.023 X 10^23 atoms / 51.9961 gm
= 0.0258851207 X 10^29 atoms chromium / m^3

FIND Nu, Np AND Nz:
The web page titled: FNR Fuel Rods shows that for new 0.35 m long core fuel rods:
Average mass Mu of U-238 in each core fuel rod is:
Mu = .7 (0.356390 kg / core rod)
= .249473 kg

Average mass of Pu in each core fuel rod is:
Mp = 0.2 (0.356390 kg)
= 0.071278 kg

Mass Mz of Zr in each core fuel rod is:
Mz = 0.1 (0.356390 kg)
= 0.0356390 kg

The number of fuel rods in a fixed fuel bundle plus a movable fuel bundle is:
416 + 280 = 696

From the web page titled: FNR Fuel Bundles the area occupied by each octagonal plus square fuel bundle pair with no swelling allowance is:
0.2222777 m^2

Thus for new fuel with 0.35 m core rod overlap:
For U average mass density in core zone
= (696 fuel rods) X (.249473 kg /fuel rod) / (0.35 m X 0.2222777 m^2)
= 2231.869872 kg / m^3

For Pu average mass density in core zone is:
(2 / 7) X 2231.869872 kg / m^3 = 637.6771062 kg / m^3

For Zr average mass density in core zone is:
(1 / 7) X 2231.869872 kg / m^3 = 318.8385531 kg / m^3

Calculate Nu, Np, Nz for new fuel.

Nu = 2231.869872 kg / m^3 X 6.023 X 10^23 atoms / mole X 1 mole / 0.238 kg
= 56.4813 X 10^26 U atoms / m^3

Np = 637.6771062 kg / m^3 X 6.023 X 10^23 atoms / mole X 1 mole / 0.239 kg
= 16.070 X 10^26 atoms / m^3

Nz = 318.8385531 kg / m^3 X 6.023 X 10^23 atoms / mole X 1 mole / 0.09122 kg
= 21.052 X 10^26 atoms / m^3

NEW FUEL DATA SUMMARY:
Ns = 0.14207 X 10^29 Na atoms / m^3
Ni = 0.156657 X 10^29 Fe atoms / m^3
Nc = 0.0258851207 X 10^29 Cr atoms / m^3
Nu = 56.4813 X 10^26 U atoms / m^3
Np = 16.070 X 10^26 Pu atoms / m^3
Nz = 21.052 X 10^26 Zr atoms / m^3

CALCULATE Nu AND Np AFTER 15% burnup:

After 15% fuel burnup:
Average Pu weight fraction drops from 20% to 12.7%
Average U weight fraction drops from 70% to 62.3%
Fission product weight fraction increases fro 0% to 15%
Zirconium weight fraction remains unchanged at 10%.

Thus after 15% fuel burnup the average concentrations of core fuel atoms are given by:

For Uranium:
Nu = 2231.869872 kg / m^3 X (62.3 / 70) X 10^3 gm / kg X 6.023 X 10^23 atoms / 238 gm
= 0.0502683676 X 10^29 U atoms / m^3

For plutonium:
Np = 637.6771062 kg / m^3 X (12.7 / 20) X 10^3 gm / kg X 6.023 X 10^23 atoms / 239 gm
= 0.0102044479 X 10^29 Pu atoms / m^3

For Fission Products:
Nf = 637.6771062 kg / m^3 X (3 / 4) X 10^3 gm / kg X 2 X 6.023 X 10^23 atoms / 239 gm
= 0.024104995 X 10^29 FP atoms / m^3

For Zirconium:
Nz = = 318.8385531 kg / m^3 X 10^3 gm / kg X 6.023 X 10^23 atoms / 91.22 gm
= 0.0209966455 X 10^29 Zr atoms / m^3

SUMMARY FOR FNR MIDDLE CORE 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

CALCULATE THE AVERAGE ATOMS / UNIT VOLUME IN UPPER AND LOWER CORE REGIONS:
Na =
Fe =
Cr =
U =
Zr =

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 May 26, 2022