Question 4.4: Suppose the nonleakage probability for a sodium cooled fast ......

From Eq. (3.2) k_{\infty}=k/P_{NL}=1.0/0.9=1.111

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_{P u O_{2}}+V_{U O_{2}}

Let x=V_{PuO_2}\,\,/\,V_{f}\,\,{\mathrm{and}}\,1-x={V}_{UO_2}\,\,/\,V_{f}

Using volume weighting we may write

k_{\infty} = \frac{\underline{\nu}\Sigma_{f}}{\Sigma_{a}}=\frac{(V_{f}/V){\underline{{{\nu}}}}\Sigma_{f}^{f}}{(V_{f}/V)\Sigma_{a}^{f}+(V_{N a}/V)\Sigma_{a}^{N a}+(V_{F e}/V)\Sigma_{a}^{F a}} = \frac{0.30 \underline{\nu}\Sigma_{f}^{f}}{0.30\Sigma_{a}^{f}+0.50\Sigma_{a}^{Na}+0.20\Sigma_{a}^{F e}}

Using volume weighting for the fissile and fertile material in the fuel:

\nu{\Sigma}_{f}^{f}=x\nu^{49}\Sigma_{f}^{P uO_{2}}+(1-x)\nu^{28}\Sigma_{f}^{UO_{2}}

and

\Sigma_{a}^{f}=x\Sigma_{a}^{PuO_2}+(1-x)\Sigma_{a}^{UO_{2}}

Combining equations we then have for the multiplication

k_{\infty} = \frac{0.30[x\cdot{\nu}^{49}\Sigma_{f}^{P u O_{2}}+(1-x){\nu}^{28}\Sigma_{f}^{U O_{2}}]}{0.30[x\cdot\Sigma_{a}^{PuO_{2}}+(1-x)\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

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

\frac{0.07076x+0.001509(1-x)}{0.02922x+0.004938(1-x)+0.003796\cdot10^{-2}+0.002452\cdot10^{-1}}=1.1111

Simplify:

{\frac{0.06925x+0.001509}{0.02428x+0.005221}}=1.111

Solve for x:

x = 0.102

Thus the fuel should be 10.2% PuO_2 and 89.8% UO_2 by volume.