A steady source injects fission neutrons into an infinite

PHYSICS · Multi-part · Fermi University Physics · Nuclear Physics · Nuclear reactions · Fission, fusion and nuclear energy

The Nucleus Resolved, Ch Neutrons: the reagent with no barrier

Problem

A steady source injects fission neutrons into an infinite, non-absorbing moderator at a steady rate SS. In the 1/E1/E slowing-down region the flux per unit lethargy is Eϕ(E)=S/(ξΣs)E\phi(E) = S/(\xi\Sigma_s), where SS is the number of neutrons crossing each energy per unit volume per second and ξΣs\xi\Sigma_s is fixed by the moderator. ((a)) For a source strength S=5.0×105 cm−3 s−1S = 5.0\times10^{5}\,\mathrm{cm^{-3}\,s^{-1}}, compute the flux per unit lethargy in graphite, for which ξΣs=0.0620 cm−1\xi\Sigma_s = 0.0620\,\mathrm{cm^{-1}}, and in heavy water, for which ξΣs=0.177 cm−1\xi\Sigma_s = 0.177\,\mathrm{cm^{-1}}. ((b)) A 238U^{238}\mathrm{U} resonance at Er=6.67 eVE_r = 6.67\,\mathrm{eV} has natural width Γ=24.5 meV\Gamma = 24.5\,\mathrm{meV}. Using ΓD=2ErkBT/A\Gamma_D = 2\sqrt{E_rk_BT/A} with A=238A=238, compute the Doppler width at T=300 KT=300\,\mathrm{K} and at T=900 KT=900\,\mathrm{K}, and give the ratio of the two widths. ((c)) State, without further calculation, whether raising the fuel temperature from 300 K300\,\mathrm{K} to 900 K900\,\mathrm{K} increases or decreases the resonance escape probability pp for lumped UO2\mathrm{UO_2} fuel, and explain why the same statement would not hold for infinitely dilute 238U^{238}\mathrm{U}.

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