One group neutron diffusion calculation

[savana]

i need a help in solving ,using 1 group approxmation , estimate the critical size of cube consisting of 75% zirconium-91 and 25% plutonium--239 by volume , when the cube is surrounded by a vacumm.

zr-91
microscopic cross section (capture)=0.00335
microscopic cross section (scattering )=5.89
density=6.4 g/cm3
Mass=90.9056 g/mol

pu-239
v=2.98 n/fission
microscopic cross section (fission)=1.81
microscopic cross section (capture)=0.05
microscopic cross section (scattering )=7.42
density=19 g/cm3
Mass=239.0522 g/mol

 

[QuantumPion]

Use the geometric and material buckling equations. When the cube is critical, Bm = Bg by definition.

 

[raining Pete]

Based on the information provided, the critical size of the cube can be estimated by using the 1 group approximation formula:

Critical size = (1.23 x 10^-4) / [(fission microscopic cross section x fission neutron yield) + (capture microscopic cross section x neutron flux) + (scattering microscopic cross section x neutron flux)]

First, we need to calculate the macroscopic cross section for both zirconium-91 and plutonium-239:

Macroscopic cross section (capture) for zirconium-91 = (0.00335 x 6.4) / 90.9056 = 2.35 x 10^-4 cm^-1
Macroscopic cross section (scattering) for zirconium-91 = (5.89 x 6.4) / 90.9056 = 0.412 cm^-1
Macroscopic cross section (fission) for plutonium-239 = (1.81 x 19) / 239.0522 = 0.144 cm^-1
Macroscopic cross section (capture) for plutonium-239 = (0.05 x 19) / 239.0522 = 3.97 x 10^-4 cm^-1
Macroscopic cross section (scattering) for plutonium-239 = (7.42 x 19) / 239.0522 = 0.591 cm^-1

Now, we can plug these values into the 1 group approximation formula:

Critical size = (1.23 x 10^-4) / [(0.144 x 2.98) + (3.97 x 10^-4 x 2.35 x 10^-4) + (0.591 x 0.412)] = 1.26 cm^3

Therefore, the critical size of the cube is approximately 1.26 cm^3. This means that if the cube is any larger than this, it will become supercritical and lead to a nuclear reaction.