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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 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

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

**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

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