Question 4.7: Suppose the fuel rods from problem [4.6] are to be used in a......

For part a: From Eq. (4.40) the resonance escape probability is

p=\exp[-(V_{f}/{V_{m}})\frac{N_{f}}{{\xi}\Sigma_{s}}I]\ \ \mathrm{Thus}    Thus

(V_{f}/V_{m})\frac{\xcancel{N_f}}{\xi\Sigma_{s}}\xcancel{I}\Biggl|_{D_{2}O} = (V_{f}/V_{m})\frac{\xcancel{N_f}}{\xi\Sigma_{s}}\xcancel{I}\Biggl|_{H_{2}O}    or using the data from Table 3.1

(V_{f}\,/V_{m})\frac{1}{0.18}\bigg|_{D_{2}{ O}}=(1/2)\frac{1}{1.28}        Thus \left.{\frac{V_{m}}{V_{f}}}\right|_{D_{2}O}={\frac{2\cdot1.28}{0.18}}=14.2

For part b: From Eq. (4.48) the thermal utilization is

f = \frac{1}{1+\varsigma{\frac{V_{m}N_{m}}{V_{f}N_{f}}}{\frac{\sigma_{a}^{m}}{\sigma_{a}^{f}}}}    With the thermal disadvantage factor held constant:

\left.\frac{V_{m}N_{m}}{V_{f}\, \bcancel{N_{f}}}\,\frac{\sigma_{a}^{m}}{\bcancel{\sigma_{\alpha}^{f}}}\right|_{D_{2}O} = \left.\frac{V_{m}N_{m}}{V_{f}\, \bcancel{N_{f}}}\,\frac{\sigma_{a}^{m}}{\bcancel{\sigma_{\alpha}^{f}}}\right|_{H_{2}O}    or \left.{\frac{V_{m}}{V_{f}}}\right|_{D_{2}O} = 2\frac{N_{H_{2}O}\sigma_{a}^{H_{2}O}}{N_{D_2O}\sigma_{a}^{D_{2}O}} = 2{\frac{1.0\cdot0.5896}{1.1\cdot0.0010}}=1072

where the number density and cross section data is from Table E-3.
These results explain, in part, why heavy water reactors have much larger moderator to fuel volume ratios than light water reactors