Question 5.9: Using Equations 5.75 through 5.78, estimate the percentage e......
Using Equations 5.75 through 5.78, estimate the percentage error in the fission and absorption rates when the fast neutrons are ignored in a thermal water reactor fuel rod.
\mathrm{R}_{\mathrm{f}}=\Sigma_{\mathrm{f2}}\phi_{2}=0.9310\times10^{15}\,\mathrm{firssions/s/cm^{3}} (5.75)
\mathrm{R}_{\mathrm{a}}=\Sigma_{\mathrm{a2}}\phi_{2}=1.085\times10^{15}\,\mathrm{absorptions/s/cm^{3}} (5.76)
\mathrm{R}_{\mathrm{f}}=\Sigma_{\mathrm{f1}}\phi_{1}+\Sigma_{\mathrm{f2}}\phi_{2}=0.9370\times10^{15}\,\mathrm{firssions/s/cm^{3}} (5.77)
\mathrm{R}_{\mathrm{a}}=\Sigma_{\mathrm{al}}{\phi}_{1}+\Sigma_{\mathrm{a2}}{\phi}_{2}=1.1042\times10^{15}\,\mathrm{absorptions/s/cm^{3}} (5.78)
According to Equations 5.75 and 5.77, the difference in the fission rate is 0.9370 × 10^{15} fissions/s/cm³ – 0.9310 × 10^{15} fissions/s/cm³ = 0.006 × 10^{15} fissions/s/cm³ ≈ 0.6%. When we use Equations 5.76 and 5.78, we find that the difference in the absorption rate is 1.1042 × 10^{15} absorptions/s/cm³ – 1.085 × 10^{15} absorptions/s/cm³ = 0.0192 × 10^{15} absorptions/s/cm³ ≈ 1.7%. Hence, there is a greater error in the absorption rate than there is in the fission rate in this case.