\section*{Problem 5}

\subsection*{Part A}

\subsubsection*{Solution}

The xenon-135 transient for the given power history is solved using the coupled differential equations for I-135 and Xe-135:

\[\frac{dI}{dt} = -\lambda_I I + \rho \gamma_I P_0\]

\[\frac{dX}{dt} = -\lambda_{Xe} X - \rho R^{Max} X + \lambda_I I + \rho \gamma_{Xe} P_0\]

where $\rho$ is the normalized power (1.0 = 100\% power).

\textbf{Power History:}
\begin{itemize}
    \item 0-5 hours: 100\% power
    \item 5-15 hours: Shutdown
    \item 15-50 hours: 100\% power
    \item 50-80 hours: 40\% power
    \item 80-100 hours: Shutdown
    \item 100-150 hours: 100\% power
\end{itemize}

\textbf{Key features of the xenon transient:}

\begin{enumerate}
    \item \textbf{Initial equilibrium (0-5 hours):} At 100\% power, xenon reactivity = -2900 pcm

    \item \textbf{First shutdown (5-15 hours):}
    \begin{itemize}
        \item Xenon burnout stops immediately (no neutron flux)
        \item I-135 continues to decay into Xe-135
        \item Xenon concentration rises, reaching a peak around 8-9 hours after shutdown
        \item Most negative xenon reactivity occurs
    \end{itemize}

    \item \textbf{Return to full power (t = 15 hours):}
    \begin{itemize}
        \item Xenon burnout resumes at full rate
        \item System returns to equilibrium at 100\% power
        \item Xenon reactivity returns to -2900 pcm
    \end{itemize}

    \item \textbf{Power reduction to 40\% (t = 50 hours):}
    \begin{itemize}
        \item Reduced burnout rate (40\% of full power)
        \item Xenon concentration increases
        \item System approaches new equilibrium at 40\% power
        \item Equilibrium xenon significantly higher at lower power
    \end{itemize}

    \item \textbf{Second shutdown (80-100 hours):}
    \begin{itemize}
        \item Similar xenon peak behavior to first shutdown
        \item Starting from 40\% power equilibrium
        \item Peak less pronounced due to lower initial I-135 inventory
    \end{itemize}

    \item \textbf{Return to full power (t = 100 hours):}
    \begin{itemize}
        \item Final return to 100\% power operation
        \item System approaches equilibrium xenon level
        \item Xenon reactivity returns to -2900 pcm
    \end{itemize}
\end{enumerate}

The xenon transient is shown in the figure below (computed using scipy.integrate.odeint):

\begin{center}
\includegraphics[width=0.95\textwidth]{../python/problem5_xenon_transient.png}
\end{center}

\subsection*{Part B}

\subsubsection*{Python Code}
\begin{lstlisting}[language=Python,
    basicstyle=\ttfamily\small,
    keywordstyle=\color{blue},
    commentstyle=\color{gray},
    stringstyle=\color{red},
    showstringspaces=false,
    numbers=left,
    numberstyle=\tiny,
    frame=single,
    breaklines=true]
from scipy.integrate import odeint

# Define ODE system
def xenon_ode(y, t, power_func):
    I, X = y
    t_hours = t / 3600
    rho = power_func(t_hours)

    dI_dt = -lambda_I * I + rho * gamma_I * P0
    dX_dt = -lambda_Xe * X - rho * R_max * X + lambda_I * I + rho * gamma_Xe * P0

    return [dI_dt, dX_dt]

# Initial conditions at full power equilibrium
I0 = gamma_I * P0 / lambda_I
X0 = abs(Xe_eq_reactivity) / K

# Solve ODE over time period
t_hours = np.linspace(0, 150, 2000)
t_seconds = t_hours * 3600
solution = odeint(xenon_ode, [I0, X0], t_seconds, args=(get_power,))

# Find peak after first shutdown (5-15 hours)
X_transient = solution[:, 1]
Xe_reactivity = -K * X_transient
mask = (t_hours >= 5) & (t_hours <= 15)
peak_idx = np.argmin(Xe_reactivity[mask])
\end{lstlisting}

\subsubsection*{Solution}

After the first shutdown at t = 5 hours (shutdown period: 5-15 hours), xenon-135 concentration increases due to:
\begin{enumerate}
    \item Continued decay of I-135 inventory into Xe-135
    \item Elimination of xenon burnout (no neutron flux)
\end{enumerate}

The peak occurs when the production rate from I-135 decay equals the Xe-135 decay rate. This typically happens 8-12 hours after shutdown from full power operation.

\textbf{Given parameters:}
\begin{itemize}
    \item $\gamma_I = 0.057$ (I-135 fission yield)
    \item $\gamma_{Xe} = 0.003$ (Xe-135 fission yield)
    \item $\lambda_I = 2.87 \times 10^{-5}$ sec$^{-1}$ (I-135 decay, $t_{1/2}$ = 6.7 hr)
    \item $\lambda_{Xe} = 2.09 \times 10^{-5}$ sec$^{-1}$ (Xe-135 decay, $t_{1/2}$ = 9.2 hr)
    \item $R^{Max} = 7.34 \times 10^{-5}$ sec$^{-1}$ (full power burnout)
    \item $K = 4.56$ pcm$\cdot$sec$^{-1}$
    \item Initial Xe reactivity = -2900 pcm (at 100\% power)
\end{itemize}

\textbf{Initial equilibrium concentrations (100\% power):}

At equilibrium with $\rho = 1.0$:
\[I_{eq} = \frac{\gamma_I P_0}{\lambda_I} = 1985.12 \text{ [arb. units]}\]

\[X_{eq} = \frac{|\text{Xe reactivity}|}{K} = \frac{2900}{4.56} = 635.96 \text{ [arb. units]}\]

\textbf{Results from numerical integration:}

\[\boxed{\text{Time of peak: } t = 13.36 \text{ hours}}\]

\[\boxed{\text{Time after shutdown: } \Delta t = 8.36 \text{ hours}}\]

\[\boxed{\text{Peak xenon reactivity: } -5261 \text{ pcm}}\]

\textbf{Interpretation:}
\begin{itemize}
    \item The xenon reactivity becomes 2361 pcm more negative than equilibrium
    \item Peak occurs approximately 8.4 hours after shutdown
    \item This represents a significant reactivity penalty that must be overcome to restart
    \item If reactivity worth is insufficient, the reactor cannot be restarted until xenon decays
    \item At t = 15 hours, when power returns to 100\%, xenon has already started to decay from peak
\end{itemize}

\textbf{Physical insight:}

The time to peak can be estimated analytically. After shutdown, I-135 decays with time constant $1/\lambda_I \approx 10$ hours, while Xe-135 decays with $1/\lambda_{Xe} \approx 13$ hours. The peak occurs when:

\[\frac{dX}{dt} = \lambda_I I(t) - \lambda_{Xe} X(t) = 0\]

This typically occurs at $t \approx 8-12$ hours after shutdown for thermal reactors, consistent with our computed value of 8.36 hours. The reactor restarts at t = 15 hours, which is about 1.6 hours after the xenon peak, when xenon has already begun to decay.