 Solution

ID: past-exam-of-the-mphil-in-nuclear-energy-course-of-the-university-of-cambridge/2019/paper-ne1/3/e/solution

The power coefficient is approximately $- 2.28 \times 1 0^{- 11} W^{- 1}$ , or $- 2.28 \times 1 0^{- 5} M W^{- 1}$ . From part (c),

R_{1} = R_{2} = \frac{75}{236 \times 1 0 ^{6}} = 3.17797 \times 1 0^{- 7} K W^{- 1}, \frac{1}{2 W _{C} c _{C}} = 2.11864 \times 1 0^{- 7} K W^{- 1} .

α_{P} = = (- 2.0 \times 1 0^{- 5}) (3.17797 \times 1 0^{- 7}) + (- 1.5 \times 1 0^{- 5}) (3.17797 \times 1 0^{- 7}) + (- 5.5 \times 1 0^{- 5}) (2.11864 \times 1 0^{- 7}) - 2.27754 \times 1 0^{- 11} W^{- 1} .

The negative sign means that a slow increase in thermal power produces a decrease in reactor reactivity along this family. This is a restoring thermal feedback control effect despite the positive moderator coefficient. A negative quasistatic slope alone does not prove stability of every transient: the separate heat capacities, transport delays and neutron-population dynamics matter for a time-dependent stability analysis.

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