Question 10.11: Taking into account neutron capture in plutonium-239 and –24......
Taking into account neutron capture in plutonium-239 and –240:
a. Write a rate equation for the concentration of plutonium-240
b. Solve the equation from part a. using Eq. (10.37) for the concentration of plutonium-239
c. Using the data in Table 3.2 show that your result behaves similarly to the plutonium-240 plot in Fig. 10.3.
(This problem may be solved in a simplified form wich assumes that the flux is time independent. We treat the more general case here)
Part a: Since plutonium-240 is a fissile material, we can ignore its fission cross section
{\frac{d}{d t}}N^{40}(t)=\sigma_{\gamma}^{49}\phi(t)N^{49}(t)-\sigma_{\gamma}^{40}\phi(t)N^{40}(t)
Part b. Multiply by an integrating factor \exp\left[\,\sigma_{\gamma}^{40} \int_0^t\phi(t^{\prime})d t^{\prime}\,\right] (See appendix A for details) Since
\frac{d}{d t}\left\{N^{40}(t)\exp\!\!\left[\sigma_{\gamma}^{40}\!\int_{0}^{t}\phi(t^{\prime})d t^{\prime}\right]\right\}=\left[\frac{d}{d t}N^{40}(t)+\sigma_{\gamma}^{40}\phi(t)N^{40}(t)\right]\!\!\exp\!\!\left[\sigma_{\gamma}^{40}\!\int_{0}^{t}\phi(t^{\prime})d t^{\prime}\right]
we have
\frac{d}{d t}\Bigg\{N^{40}(t)\exp\Bigg[\sigma_{\gamma}^{40}\int_{0}^{t}\phi(t^{\prime})d t^{\prime}\Bigg]\Bigg \}=\sigma_{\gamma}^{49}\phi(t)N^{49}(t)\exp\Bigg[\sigma_{\gamma}^{40}\int_0^t \phi(t^{\prime})d t^{\prime}\Bigg]
Integrating between 0 and t, with N^{40} (0) = 0
N^{40}(t)e^{\sigma_{\gamma}^{40}\Phi(t)}=\sigma_{\gamma}^{49}\int_{0}^{t}\ \phi(t^{\prime})N^{49}(t^{\prime})e^{\sigma_{\gamma}^{40}\Phi(t^{\prime})}d t^{\prime}
Where
\Phi(t)\equiv\int_{0}^{t}\ \phi(t^{\prime})d t^{\prime}
Inserting Eq. (10.37), and solving for N^{40} (t)
N^{40}(t)=\frac{\sigma_{\gamma}^{49}\sigma_{\gamma}^{28}}{\sigma_{a}^{49}}\,N^{28}(0)e^{-\sigma_{\gamma}^{40}\Phi(t)} \int_0^t \phi (t^{\prime})\biggr[1-e^{-\sigma_{a}^{49}\Phi(t^{\prime})}\biggr]e^{\sigma_{\gamma}^{40}\Phi(t^{\prime})}d t^{\prime}
We next note that
d\Phi(t)=\phi(t)d t
so that we may rewrite the equation as
N^{40}(t)={\frac{\sigma_{\gamma}^{49}\sigma_{\gamma}^{28}}{\sigma_{a}^{49}}}N^{28}(0)e^{-\sigma_{\gamma}^{40}\Phi(t)}\int_{0}^{\Phi(t)}\left[e^{\sigma_{\gamma}^{40}\Phi}-e^{(\sigma_{\gamma}^{40}-\sigma_{a}^{49})\Phi}\right]d\Phi
=\frac{\sigma_{\gamma}^{49}\sigma_{\gamma}^{28}}{\sigma_{a}^{49}}N^{28}(0)e^{-\sigma_{\gamma}^{40}\Phi(t)}\left[\frac{e^{\sigma_{\gamma}^{40}\Phi(t)}-1}{\sigma_{\gamma}^{40}}-\frac{e^{(\sigma_{\gamma}^{40}-\sigma_{a}^{49})\Phi(t)}-1}{\left(\sigma_{\gamma}^{40}-\sigma_{a}^{49}\right)}\right]
=\frac{\sigma_{\gamma}^{49}\sigma_{\gamma}^{28}}{\sigma_{a}^{49}}\,N^{28}(0)\left[\frac{1-e^{-\sigma_{\gamma}^{40}\Phi(t)}}{\sigma_{\gamma}^{40}}-\frac{e^{-\sigma_{a}^{49}\Phi(t)}-e^{-\sigma_{\gamma}^{40}\Phi(t)}}{\left(\sigma_{\gamma}^{40}-\sigma_{a}^{49}\right)}\right]
N^{40}(t)=\sigma_{\gamma}^{49}\sigma_{\gamma}^{28}N^{28}(0)\left[\frac{1}{\sigma_{a}^{49}\sigma_{\gamma}^{40}}-\frac{e^{-\sigma_{a}^{49}\Phi(t)}}{\sigma_{a}^{49}\left(\sigma_{\gamma}^{40}-\sigma_{a}^{49}\right)}+\frac{e^{-\sigma_{\gamma}^{49}\Phi(t)}}{\sigma_{\gamma}^{40}\left(\sigma_{\gamma}^{40}-\sigma_{a}^{49}\right)}\right]