Question 4.6: A pressurized water reactor has 3% enriched UO2 fuel pins th......
A pressurized water reactor has 3% enriched UO_2 fuel pins that are 1.0 cm in diameter and have a density of 11.0 gm/cm³. The moderator to fuel volume ratio is 2:1. Calculate η_T, p, f, and k_∞ at room temperature under the assumptions that ε = 1.24, the thermal disadvantage factor ς = 1.16, and the Dancoff correction increases the fuel diameter for the resonance integral calculation by 10%.
Let N be the number density of UO2. From Eq. (4.49), and the enrichment is {\tilde{e}}=0.03
\eta_{T}=\frac{\nu\Sigma_{f}^{f}}{\Sigma_{a}^{f}} = \frac{\tilde{e}N \nu^{25}{\sigma}_{f}^{25}}{\tilde{e}N{\sigma}_{a}^{25}+(1-\tilde{e})N{\sigma}_{a}^{28}+2N{\sigma}_{a}^{o}} = \frac{{\tilde{e}}\nu^{25}\sigma_{f}^{25}}{{\tilde{e}}\sigma_{a}^{25}+(1-{\tilde{e}})\sigma_{a}^{28}+2\sigma_{a}^{o}}
Using data from Table 3.2 and oxygen data from Table E-3
\eta_{T}=\frac{0.03\cdot2.43\cdot505}{0.03\cdot591+(1-0.03)2.42+2\cdot0.0002}=1.82
From Eq. (4.40)
p=\exp\left[-\left(\frac{V_{f}}{V_{m}}\right)\frac{N_{f}}{{{\xi}}\Sigma_{s}}{I}\right]
For the fuel A=238+2\cdot16=270\,.
N_{f}=\rho N_{o}/\,A=11.0\cdot0.6023\cdot10^{24}\,/\,270=0.0245\cdot10^{24}\ \mathrm{gm/cm}^{3}
From Table 4.3 for UO2
I=4.45+26.6{\sqrt{4/{{\rho D}}}}=4.45+26.6{\sqrt{4/(11\cdot1.1)}}=19.74\,b
where we have increased the diameter by 10%: Taking ξΣ_s = 1.28 for water from Table 3.1, we obtain
p=\exp\left[-\left({\frac{1}{2}}\right){\frac{0.0245\cdot10^{24}}{1.28}}19.74\cdot10^{-24}\right]=0.828
From Eq. (4.55)
f=\frac{1}{1+\varsigma(V_{m}/V_{f})(N_{m}/N_{f})(\overline{{{\sigma}}}_{a T}^{m}/\overline{{{\sigma}}}_{a T}^{f})}
N_{m}/{N}_{f}=\frac{\rho_{m}N_{o}/{A}_{m}}{\rho_{f}N_{o}/A_{f}}=\frac{\rho_{m}A_{f}}{\rho_{f}A_{m}}=\frac{1.0\cdot270}{11\cdot18}=1.364
Taking the absorption cross sections from Table E-3:
{f}=\frac{1}{1+1.16\cdot2\cdot1.364(0.5896/6.540)}=0.778
Finally,
k_{\infty}=\varepsilon p f\eta_{T}=1.24\cdot0.828\cdot0.778\cdot1.82=1.45