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**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

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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