Question 4.8: A reactor lattice consists of uranium rods in a heavy water ......

Part a: From Eq. (4.40): p=\exp\left[-\left({\frac{V_{f}}{V_{m}}}\right){\frac{N_{f}}{\xi\Sigma_{s}}}\,I\right]

The only quantity changing is the slowing down power, ξΣ_s. From Table 3.1 we see that its value is larger for water than for heavy water. Thus the resonance escape probability will increase.

Part b. From Eq. (4.55): f=\frac{1}{1+ς(V_{m}/V_{f})(N_{m}/N_{f})(\overline{{{\sigma}}}_{a T}^{m}/\overline{{{\sigma}}}_{a T}^{f})}

The major change will be in the moderator thermal absorption cross section {\overline{{\sigma}}}_{a t}^{m}\ . The thermal disadvantage factor will change less. From Table E-3, we seen that {\overline{{\sigma}}}_{a t}^{m}\ is much smaller for heavy water than for water. Thus the thermal utilization will decrease.

Part c. Because of its very small thermal abortion cross section, heavy water is considered to be the best moderator; reactors can be built using natural uranium if heavy water is the moderator, but not with ordinary water. Thus the net effect of replacing heavy water with water would be to decrease the value of k_∞.