Question 17.1: Suppose that a fuel assembly in a thermal water reactor cont......

Since the fuel contains 50 lb of Uranium-235, it must contain (96%/4%) ≈ 24 times more U-238 by weight than U-235. Therefore, the amount of U-238 present in the fuel at start-up is 24 × 50 lb = 1200 lb. Since U-235 and U-238 both obey the same rate equation:

\mathrm{U}{\big(}\mathrm{t}{\big)}=\mathrm{U}_{\mathrm{o}}\mathrm{e}^{-\mathrm{σau\phi t}}

the time-dependent U-235 concentration is

\mathrm{U}^{235}\left(\mathrm{t}\right)=\mathrm{U}_{\mathrm{o}}^{235}\mathrm{e}^{-\mathrm{σau}235\mathrm{\phi t}}

and time-dependent U-238 concentration is

\mathrm{U}^{238}\left(\mathrm{t}\right)=\mathrm{U}_{\mathrm{o}}^{238}\mathrm{e}^{-\mathrm{σau}238\mathrm{\phi t}}

where U^{235}_{o}  and  U^{238}_{o} are the initial concentrations, in atoms/cm³, at start-up. In the case of U-235, σ_{au}235 ≈ 700 barns, and in the case of U-238, σ_{au}238 ≈ 2.7 barns. After 400 days (or 3.45 × 10^7 seconds) of full-power operation with an average flux of 2 × 10^{13} neutrons/cm²/s, the concentration of U-235 is

\mathrm{U}^{235}\left({\bf t}\right)={\bf U}_{\mathrm{o}}^{235}\mathrm{e}^{-\mathrm{σau235}\phi{\bf t}}={\bf U}_{\mathrm{o}}^{235}\mathrm{e}^{-0.483}=0.617{\bf U}_{\mathrm{o}}^{235}

and the U-238 concentration is U^{238}(t) = U^{238}_{o}e^{σau238\phi t} = e^{-0.00186} = 0.998  U^{238}_{o} . Since U^{235}_{o}= 50  lb, and U^{238}_{o} = 1200 lb , the remaining amounts of U-235 and U-238 after 400 days are U^{235} = 50 lb × 0.617 = 30.85 lb of U-235 and U^{238} = 1200 lb × 0.998 = 1197.6 lb of U-238. So roughly 40% of the U-235 will be burned up, and over 99% of the U-238 will still remain. These results are quite typical of the behavior of a thermal water reactor under normal operating conditions.