Question 17.9: In the previous example, how much additional reactivity has ......

In Example Problem 17.8, the initial reactivity of the core at a temperature of 300°C was ρ = (k−1)/k = 0.0, and the final reactivity of the core (without moving the control rods) would be ρ = (k−1)/k = −0.01/0.99 = −0.0101. Hence, the control rods must add an additional 0.0101 of positive reactivity to the core to keep it critical. This is equivalent to adding 1010 pcm, or since the delayed neutron fraction of U-235 is β = 0.0065, Δρ = 0.0101/β = $1.55.