Question 4.3: A sodium-cooled fast reactor is fueled with PuO2 , mixed wit......
A sodium-cooled fast reactor is fueled with PuO_2 , mixed with depleted UO_2. the structural material is iron. Averaged over the spectrum of fast neutrons, the microscopic cross sections and densities are the following:
The fuel is 15 % PuO_2 and 85 % UO_2 by volume. The volumetric composition of the core is 30% fuel, 50% coolant and 20% structural material. Calculate k_∞ assuming that the values of ν for plutonium and uranium in the fast spectrum are 2.98 and 2.47, respectively, and that the cross sections of oxygen can be neglected. What fraction of the mass of the core does the fuel account for?
| σ_f b | σ_{a} b | σ_t b | ρ g/cm³ | |
| PuO_2 | 1.95 | 2.40 | 8.6 | 11.0 |
| UO_2 | 0.05 | 0.404 | 8.2 | 11.0 |
| Na | – | 0.0018 | 3.7 | 0.97 |
| Fe | – | 0.0087 | 3.6 | 7.87 |
In this core we know that V=V_{f}+V_{N a}+V_{F e} and are given: V_{f}\,/\,V=0.30\,,\,V_{N a}\,/\,V=0.50\,, {V}_{F_{e}}/{ V}=0.20\,.
We also know that V_{f}=V_{PuO_{2}}+V_{UO_{2}} and are given V_{PuO_{2}}/V_{f}=0.15\mathrm{~and~}V_{UO_{2}}/V_{f}=0.85
Using volume weighting we may write
k_{∞}={\frac{\underline{\nu}\Sigma_{f}}{\Sigma_{a}}}={\frac{(V_{f}/V){\underline{\nu}\Sigma_{f}^{f}}}{(V_{f}/V)\Sigma_{a}^{f}+(V_{Na}/V)\Sigma_{a}^{Na}+(V_{F e}/V)\Sigma_{a}^{F e}}}= \frac{0.30{\underline{\nu}}\Sigma_{f}^{f}}{0.30\Sigma_{a}^{f}+0.50\Sigma_{a}^{N a}+0.20\Sigma_{a}^{F e}}
Using volume weighting for the fissile and fertile material in the fuel:
\nu{\Sigma}_{f}^{f}=(V_{P u O_{2}}/V_{f})\nu^{49}\Sigma_{f}^{P u O_{2}}+(V_{UO_{2}}/V_{f})\nu^{28}\Sigma_{f}^{UO_{2}}=0.15\nu^{49}\Sigma_{f}^{P u O_{2}}+0.85\nu^{28}\Sigma_{f}^{UO_{2}}
and
\Sigma_{a}^{f}=(V_{PuO_{2}}/V_{f})\Sigma_{a}^{PuO_{2}}+(V_{UO_{2}}/V_{f})\Sigma_{a}^{UO_{2}}=0.15\Sigma_{a}^{PuO_{2}}+0.85\Sigma_{a}^{UO_{2}}
Combining equations we then have for the multiplication
k_{\infty}=\frac{0.045\nu^{49}\Sigma_{f}^{P u O_{2}}+0.255\nu^{28}\Sigma_{f}^{UO_{2}}}{0.045\Sigma_{a}^{P uO_{2}}+0.255\Sigma_{a}^{U O_{2}}+0.50\Sigma_{a}^{N a}+0.20\Sigma_{a}^{F e}}
Using \,\Sigma={ N}\sigma=\frac{\rho N_{o}}{ A}\,\sigma and ignoring the cross sections of oxygen, we have
k_{\infty}=\frac{0.045{\frac{\rho_{P u O_{2}}}{A_{P uO_{2}}}}N_{o}V^{49}\sigma_{f}^{49}+0.255{\frac{\rho_{U O_{2}}}{A_{U O_{2}}}}N_{o}V^{28}\sigma_{f}^{28}}{0.045{\frac{\rho_{P u O_{2}}}{\displaystyle{{A}_{P u O_{2}}}}}\,N_{o}\sigma_{a}^{49}+0.255{\frac{\rho_{U O_{2}}}{\displaystyle{{A}_{U O_{2}}}}}\,N_{o}\sigma_{a}^{28}+0.50{\frac{\rho_{N a}}{\displaystyle{{A}_{N a}}}}\,N_{o}\sigma_{a}^{N a}+0.20{\frac{\rho_{F e}}{\displaystyle{{A}_{F e}}}}\,N_{o}\sigma_{a}^{F e}}
Note that A_{P u O_{2}}=239+2^{*}16=271\,,\ A_{U O_{2}}=238+2^{*}16=270\,,\ A_{N a}=23\,,\ A_{F e}=55.85.
Thus canceling N_o from numerator and denominator, and using the densities and cross sections given we have
k_{\infty}=\frac{0.045{\frac{11}{271}}2.98\cdot1.95+0.255{\frac{11}{270}}2.47\cdot0.05}{0.045{\frac{11}{271}}2.40+0.255{\frac{11}{270}}0.404+0.50{\frac{0.97}{23}}0.0018+0.20{\frac{7.87}{55.85}}0.0087}
k_{\infty}={\frac{0.011+0.001283}{0.004384+0.004197+0.003796\cdot10^{-2}+0.002452\cdot10^{-1}}}
k_{\infty}=1.342