Question 3.10: A power reactor is cooled by heavy water, ( D2O) but a leak ......

All the needed data is in Table 3.1, provided we take \Sigma_{s i}=(\xi_{i}\Sigma_{s i})\,/\,\xi_{i} yielding

\Sigma_{s}^{D_{2}O}=0.353,     \Sigma_{s}^{H_{2}O}=1.38

and \Sigma_{a i}=\xi_{i}\Sigma_{s i} / (\xi_{i}\Sigma_{s i}/ \Sigma_{a i})

\Sigma_{a}^{D_2O}=8.57\cdot 10^{-6}  \Sigma_{a}^{H_2O}=0.022

For a.: From Eq. (2.61) the averaged slowing down decrement is

{\overline{{\xi}}}={\frac{1}{\sum_{s}}}\sum_{i}{\xi_{i}}\Sigma_{s i}

For 1% contamination, the number densities and thus the macroscopic cross sections of heavy water and water are replaced by 0.99 and 0.01 of their nominal values. Thus

\overline{{{\xi}}}=\frac{\xi_{D_{2}O}0.99\Sigma_{s}^{D_{2}O}+\xi_{H_{2}O}0.01\Sigma_{s}^{H_{2}O}}{0.99\Sigma_{s}^{D_{2}O}+0.01\Sigma_{s}^{H_{2}O}}

\overline{{{\xi}}}=\frac{0.51\cdot0.99\cdot0.353+0.93\cdot0.01\cdot1.38}{0.99\cdot0.353+0.01\cdot1.38}=\frac{0.191}{0.363}=0.53

For b. We again use Eq. (2.61);

\overline{\sigma}_{a\tau}(T)=\left(T_{o}/T\right)^{1/2} \overline{\sigma}_{a\tau}(T_{o})

For c.

\overline{\xi}\Sigma_{_{s}}/\Sigma_{_{\alpha}}=\sum_{i}\xi_{i}\Sigma_{_{si}}/(0.99\Sigma_{_\alpha}^{{ D}_{2}O}+ 0.01\Sigma_{_\alpha}^{{ H}_{2}O}) = {\frac{0.19}{0.99\cdot8.57\cdot10^{-6}+0.01\cdot0.022}}=833

Thus while the contamination has only small effects on the slowing down decrement and power, it decreases the slowing down ratio substantially as a result of the much larger absorption cross section of water.