M .task/backlog.data M .task/completed.data M .task/pending.data M .task/undo.data M Class_Work/nuce2101/exam2/latex/main.aux M Class_Work/nuce2101/exam2/latex/main.fdb_latexmk M Class_Work/nuce2101/exam2/latex/main.fls M Class_Work/nuce2101/exam2/latex/main.log
168 lines
6.0 KiB
TeX
168 lines
6.0 KiB
TeX
\section*{Problem 5}
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\subsection*{Part A}
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\subsubsection*{Solution}
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The xenon-135 transient for the given power history is solved using the coupled differential equations for I-135 and Xe-135:
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\[\frac{dI}{dt} = -\lambda_I I + \rho \gamma_I P_0\]
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\[\frac{dX}{dt} = -\lambda_{Xe} X - \rho R^{Max} X + \lambda_I I + \rho \gamma_{Xe} P_0\]
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where $\rho$ is the normalized power (1.0 = 100\% power).
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\textbf{Power History:}
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\begin{itemize}
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\item 0-5 hours: 100\% power
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\item 5-15 hours: Shutdown
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\item 15-50 hours: 100\% power
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\item 50-80 hours: 40\% power
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\item 80-100 hours: Shutdown
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\item 100-150 hours: 100\% power
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\end{itemize}
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\textbf{Key features of the xenon transient:}
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\begin{enumerate}
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\item \textbf{Initial equilibrium (0-5 hours):} At 100\% power, xenon reactivity = -2900 pcm
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\item \textbf{First shutdown (5-15 hours):}
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\begin{itemize}
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\item Xenon burnout stops immediately (no neutron flux)
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\item I-135 continues to decay into Xe-135
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\item Xenon concentration rises, reaching a peak around 8-9 hours after shutdown
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\item Most negative xenon reactivity occurs
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\end{itemize}
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\item \textbf{Return to full power (t = 15 hours):}
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\begin{itemize}
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\item Xenon burnout resumes at full rate
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\item System returns to equilibrium at 100\% power
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\item Xenon reactivity returns to -2900 pcm
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\end{itemize}
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\item \textbf{Power reduction to 40\% (t = 50 hours):}
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\begin{itemize}
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\item Reduced burnout rate (40\% of full power)
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\item Xenon concentration increases
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\item System approaches new equilibrium at 40\% power
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\item Equilibrium xenon significantly higher at lower power
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\end{itemize}
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\item \textbf{Second shutdown (80-100 hours):}
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\begin{itemize}
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\item Similar xenon peak behavior to first shutdown
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\item Starting from 40\% power equilibrium
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\item Peak less pronounced due to lower initial I-135 inventory
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\end{itemize}
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\item \textbf{Return to full power (t = 100 hours):}
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\begin{itemize}
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\item Final return to 100\% power operation
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\item System approaches equilibrium xenon level
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\item Xenon reactivity returns to -2900 pcm
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\end{itemize}
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\end{enumerate}
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The xenon transient is shown in the figure below (computed using scipy.integrate.odeint):
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\begin{center}
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\includegraphics[width=0.95\textwidth]{../python/problem5_xenon_transient.png}
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\end{center}
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\subsection*{Part B}
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\subsubsection*{Python Code}
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\begin{lstlisting}[language=Python,
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basicstyle=\ttfamily\small,
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keywordstyle=\color{blue},
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commentstyle=\color{gray},
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stringstyle=\color{red},
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showstringspaces=false,
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numbers=left,
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numberstyle=\tiny,
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frame=single,
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breaklines=true]
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from scipy.integrate import odeint
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# Define ODE system
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def xenon_ode(y, t, power_func):
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I, X = y
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t_hours = t / 3600
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rho = power_func(t_hours)
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dI_dt = -lambda_I * I + rho * gamma_I * P0
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dX_dt = -lambda_Xe * X - rho * R_max * X + lambda_I * I + rho * gamma_Xe * P0
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return [dI_dt, dX_dt]
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# Initial conditions at full power equilibrium
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I0 = gamma_I * P0 / lambda_I
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X0 = abs(Xe_eq_reactivity) / K
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# Solve ODE over time period
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t_hours = np.linspace(0, 150, 2000)
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t_seconds = t_hours * 3600
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solution = odeint(xenon_ode, [I0, X0], t_seconds, args=(get_power,))
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# Find peak after first shutdown (5-15 hours)
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X_transient = solution[:, 1]
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Xe_reactivity = -K * X_transient
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mask = (t_hours >= 5) & (t_hours <= 15)
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peak_idx = np.argmin(Xe_reactivity[mask])
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\end{lstlisting}
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\subsubsection*{Solution}
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After the first shutdown at t = 5 hours (shutdown period: 5-15 hours), xenon-135 concentration increases due to:
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\begin{enumerate}
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\item Continued decay of I-135 inventory into Xe-135
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\item Elimination of xenon burnout (no neutron flux)
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\end{enumerate}
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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.
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\textbf{Given parameters:}
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\begin{itemize}
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\item $\gamma_I = 0.057$ (I-135 fission yield)
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\item $\gamma_{Xe} = 0.003$ (Xe-135 fission yield)
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\item $\lambda_I = 2.87 \times 10^{-5}$ sec$^{-1}$ (I-135 decay, $t_{1/2}$ = 6.7 hr)
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\item $\lambda_{Xe} = 2.09 \times 10^{-5}$ sec$^{-1}$ (Xe-135 decay, $t_{1/2}$ = 9.2 hr)
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\item $R^{Max} = 7.34 \times 10^{-5}$ sec$^{-1}$ (full power burnout)
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\item $K = 4.56$ pcm$\cdot$sec$^{-1}$
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\item Initial Xe reactivity = -2900 pcm (at 100\% power)
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\end{itemize}
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\textbf{Initial equilibrium concentrations (100\% power):}
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At equilibrium with $\rho = 1.0$:
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\[I_{eq} = \frac{\gamma_I P_0}{\lambda_I} = 1985.12 \text{ [arb. units]}\]
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\[X_{eq} = \frac{|\text{Xe reactivity}|}{K} = \frac{2900}{4.56} = 635.96 \text{ [arb. units]}\]
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\textbf{Results from numerical integration:}
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\[\boxed{\text{Time of peak: } t = 13.36 \text{ hours}}\]
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\[\boxed{\text{Time after shutdown: } \Delta t = 8.36 \text{ hours}}\]
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\[\boxed{\text{Peak xenon reactivity: } -5261 \text{ pcm}}\]
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\textbf{Interpretation:}
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\begin{itemize}
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\item The xenon reactivity becomes 2361 pcm more negative than equilibrium
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\item Peak occurs approximately 8.4 hours after shutdown
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\item This represents a significant reactivity penalty that must be overcome to restart
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\item If reactivity worth is insufficient, the reactor cannot be restarted until xenon decays
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\item At t = 15 hours, when power returns to 100\%, xenon has already started to decay from peak
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\end{itemize}
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\textbf{Physical insight:}
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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:
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\[\frac{dX}{dt} = \lambda_I I(t) - \lambda_{Xe} X(t) = 0\]
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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.
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