Xenon Concentration
The Xenon buildup in fission fuel is an important parameter to consider
during burnup of fuel. Xenon is a radioactive gas and causes many
problems within the fuel cycle. Besides these issues, calculation of
Xenon buildup is also an illustration of how to use the Bateman
equations, which has many other uses in nuclear engineering.
To start, we know that Xenon can be produced directly in the fission
process, at the rate of \(\gamma_{Xe}\), and can also come from the
decay of Iodine that was produced by fission (at the rate of
\(\gamma_{I}\)). Thus, we must define two equations
\[\frac{dN_{I}}{dt}=\underbrace{\gamma_{I}F\left(t\right)}_{\text{production}} -\underbrace{\overline{\lambda}_{I}N_{I}\left(t\right)}_{\text{loss}}\]
\[ \begin{align}\begin{aligned}\frac{dN_{Xe}}{dt}=\underbrace{\gamma_{Xe}F\left(t\right)+\lambda_{I}N_{I}\left(t\right)}_{\text{production}} -\underbrace{\overline{\lambda}_{Xe}N_{Xe}\left(t\right)}_{\text{loss}}\\with :math:`\sigma_{.}` the cross section for neutron absorption,\end{aligned}\end{align} \]
\(\phi\) the neutron flux, \(F\left( t \right)\) the fission
rate, and
\[\overline{\lambda}_{I}=\lambda_{I}+\sigma_{I}\phi\left(t\right)\]
\[ \begin{align}\begin{aligned}\overline{\lambda}_{Xe}=\lambda_{Xe}+\sigma_{Xe}\phi\left(t\right)\\with :math:`\lambda_{x}` the decay constant.\end{aligned}\end{align} \]
Conceptually, both of these equations are simple: they both read that
the change in the concentration of the atom is the production of that
atom (whether directly from fission or from decay of Iodine in the case
of xenon) less the loss of that atom. The loss of that atom is the sum
of the natural decay and the neutron absorption probability (which will
transmute the atom).
To solve these equations, the integrating factor is used, and these
steps are shown below
\[\frac{d}{dt}\left[N_{I}\left(t\right)e^{\overline{\lambda}_{I}t}\right]=\gamma_{I}Fe^{\overline{\lambda}_{I}t}\]
\[N_{I}\left(t\right)=N_{I}^{0}e^{\overline{\lambda}_{I}t}+\frac{\gamma_{I}F}{\lambda_{I}}\left(1-e^{\overline{\lambda}_{I}t}\right)\]
\[\frac{d}{dt}\left[N_{Xe}e^{\overline{\lambda}_{Xe}t}\right]=\left[\gamma_{Xe}F+\lambda_{I}N_{I}\left(t\right)\right]e^{\overline{\lambda}_{Xe}t}\]
\[\begin{split}\begin{multline*}
N_{Xe}\left(t\right)=N_{Xe}^{0}e^{\overline{\lambda}_{Xe}t}+\frac{\gamma_{Xe}F}{\lambda_{Xe}}\left(1-e^{-\overline{\lambda}_{Xe}t}\right)+\frac{\lambda_{I}N_{I}^{0}}{\overline{\lambda}_{Xe}-\overline{\lambda}_{I}}\left(e^{-\overline{\lambda}_{I}t}-e^{-\overline{\lambda}_{Xe}t}\right)\\
+\frac{\lambda_{I}\gamma_{I}F}{\overline{\lambda}_{I}}\left[\frac{1}{\overline{\lambda}_{Xe}}\left(1-e^{-\overline{\lambda}_{Xe}t}\right)-\frac{1}{\overline{\lambda}_{Xe}-\overline{\lambda}_{I}}\left(e^{-\overline{\lambda}_{I}t}-e^{-\overline{\lambda}_{Xe}t}\right)\right]
\end{multline*}\end{split}\]
It should be noted that this can be more easily solved in matrix form,
and modern computing makes it almost trivial to solve. The program
ORIGEN is often used for this purpose.
