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XYLENE POWER LTD.

FNR CORE

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

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

As shown on the web page titled: FNR FUEL BUNDLE in the middle core region the material volume fractions are given by:
Volume fraction of steel = Fsteel = 0.21555
Volume fraction of fuel = Ffuel = 0.1991995
Volume fraction of sodium = Fsodium = 0.5852505

Thus 1 m^3 of reactor core contains 0.21555 m^3 of steel, 0.1991995 m^3 of core fuel and 0.58525 m^3 of liquid sodium. Using this information we need to calculate the average concentration of each atomic species in the reactor core.

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

FIND Ns:
Ns = 1 m^3 X Fsodium cm^3) X 10^6 cm^3 / m^3 X Rhos (gm / cm^3) X Av / Aws (atoms / gm)
= 1 m^3 X 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
 

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 = 1 m^3 X Fsteel X {Fi / [Fc + Fi]} X 10^6 cm^3 / m^3 X Rhoi X Av / Awi
= 1 m^3 X 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

Thus:
Nc = 1 m^3 X Fsteel X {Fc / [Fc + Fi]} X 10^6 cm^3 / m^3 X Rhoc X Av / Awc
= 1 m^3 X X 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
 

FIND Nu, Np AND Nz:
The web page titled: FNR Fuel Rods shows that for new 0.45 m long core fuel rods:
Average mass Mu of U-238 in each core fuel rod is:
Mu = .7 (0.45821597 kg)
= .3207510141 kg

Average mass of Pu in each core fuel rod is:
Mp = 0.2 (0.45821597 kg)
= 0.091643194 kg

Mass Mz of Zr in each core fuel rod is:
Mz = 0.1 (0.45821597 kg)
= 0.045821597 kg

The number of fuel rods in a octagonal fuel bundle plus a square 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 is:
0.2222777 m^2

Thus for new fuel:
For U average mass density in core zone
= (696 fuel rods) X (.3207510141 kg /fuel rod) / (0.45 m X 0.2222777 m^2)
= 6958.26287 X .3207510141 kg / m^3
= 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

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 =
 

This web page last updated March 20, 2020

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