Question 9.6: A sodium-cooled fast reactor lattice is designed to have the......

Part A For a bare cylindrical reactor , from Eq. (8.3)

V=3.63P/\overline{{P}}_{\mathrm{max}}^{\prime \prime \prime}=3.63P/450=8.07\cdot10^{-3}P  ( volume in m³, power in MW)

H=(4V/\pi)^{1/3}=1.08V^{1/3}=1.08(8.07\cdot10^{-3})^{1/3}P^{1/3}

=>    H=0.217P^{1/3}

B^{2}=\left[\,\left(2\cdot2.405\right)^{2}+\pi^{2}\,\right]H^{-2}=33.0(0.217P^{1/3})^{-2}

=>     B^{2}=701P^{-2/3}

k_{\infty}=1+M^{2}B^{2}=1+0.18^{2}\cdot(33/H^{2})

=1+1.07{\ /}H^{2}=1+1.07\cdot(0.217{ p}^{1/3})^{-2}

=>     k_{\infty}=1+22.7P^{-2/3}

See table for numerical results

Part b: From Eq. (9.4)

{\frac{\Delta k}{k}}={\frac{\Delta k_{\infty}}{k_{\infty}}}-{\frac{M^{2}B^{2}}{1+M^{2}B^{2}}}\bigg({\frac{\Delta M^{2}}{M^{2}}}\bigg)

=\frac{\Delta k_{\infty}}{k_{\infty}}-\frac{22.7P^{-2/3}}{1+22.7P^{-2/3}}\left(\frac{\Delta M^{2}}{M^{2}}\right)

={\frac{\Delta k_{\infty}}{k_{\infty}}}-{\frac{1}{1+0.044P^{2/3}}}\biggl({\frac{\Delta M^{2}}{M^{2}}}\biggr)

=>     \frac{\Delta k}{k}=0.01-\frac{0.01}{1+0.044{P}^{2/3}}

See table for results

Part c In a fast reactor sodium voiding causes light weight coolant to be lost, decreasing the slowing down of neutrons, and shifting the spectrum up in energy. At the same time the coolant loss increases the migration length, because neutrons travel further in the less dense core. This increases the leakage, causing the reactivity to decrease, as indicated by Eq. (9.4). Smaller cores have larger leakage probabilities, amplifying the negative effect. Thus as the core power therefore size increases, the leakage becomes smaller. The net effect is for the voiding reactivity effect to become increasingly positive for larger cores.