Question 5.7: Suppose that the neutron flux ϕ(E) in a thermal water reacto......
Suppose that the neutron flux ϕ(E) in a thermal water reactor has been subdivided into three energy groups between 0.0 and 1 eV. If the relative magnitude of the flux in each of these energy groups is shown in the table below, and the absorption cross section of the fuel is given by the entries in the fourth row of the table, what is the value of the thermal absorption cross section when all three of these groups are collapsed into a single energy group that spans the range from 0 to 1 eV?
| Group | Energy Range (eV) | Relative Magnitude of the Group Flux (Normalized to 1.0 between 0 and 1 eV) | Value of the Absorption Cross Section Σ_a (cm^{−1}) |
| 3 | 0.0–0.2 | 0.5 | 10 |
| 2 | 0.2–0.5 | 0.3 | 5 |
| 1 | 0.5–1.0 | 0.2 | 1 |
From the data given previously, the average value of the absorption cross section <Σ_a> for the fuel between E = 0 eV and E = 1 eV can be found from the expression
\lt \sigma_{\mathrm{a}}\gt \;=\frac{\displaystyle\sum_{g}\left(\Sigma_{\mathrm{ag}}\phi_{g}\Delta\mathrm{E}_{\mathrm{g}}\right)}{\displaystyle\sum_{g}\phi_{\mathrm{g}}\Delta\mathrm{E}_{g}}
where the summation Σ is to be performed over all energy groups from g =1 to 3. Now in the case of groups 1, 2, and 3, the values of ΔE_g are ΔE_1 = 0.5, ΔE_2 = 0.3, and ΔE_3 = 0.2, the values of ϕ_1, ϕ_2, and ϕ_3 are ϕ_1 = 0.2, ϕ_2 = 0.3, and ϕ_3 = 0.5, and the values of Σ_{a1}, Σ_{a2}, and Σ_{a3} are Σ_{a1} = 1, Σ_{a2} = 5, and Σ_{a3} = 10.
The average value of the absorption cross section <Σ_a> between E = 0 eV and E = 1 eV is then
\lt \Sigma_{\mathrm{a}}\gt =\frac{\left(\Sigma_{\mathrm{al}}\phi_{1}\Delta\mathrm{E}_{1}+\Sigma_{\mathrm{a2}}\phi_{2}\Delta\mathrm{E}_{2}+\Sigma_{\mathrm{a3}}\phi_{3}\Delta\mathrm{E}_{3}\right)}{\left(\phi_{1}\Delta\mathrm{E}_{1}+\phi_{2}\Delta\mathrm{E}_{2}+\phi_{3}\Delta\mathrm{E}_{3}\right)}
={\frac{\left(1\times0.2\times0.5+5\times0.3+10\times0.5\times0.2\right)}{\left(0.2\times0.5+0.3\times0.3+0.5\times0.2\right)}}
\lt \Sigma_{\mathrm{a}}\gt ={\frac{\left(0.1+0.45+1\right)}{\left(0.1+0.09+0.1\right)}}={\frac{1.55}{0.29}}=5.34\ {\mathrm{cm}}^{-1}