Obsidian/Class_Work/nuce2101/exam2/latex/problem3.tex

\section*{Problem 3}

\subsection*{Part A}

\subsubsection*{Python Code}
\begin{lstlisting}[language=Python,
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import numpy as np

# Problem 3A
## Using formulas from Fundamental Kinetics Ideas R17 Page 51
DRW = 10  # pcm/step
STEPS = 8
LAMBDA_EFF = 0.1  # hz

# ASSUMING AFTER ROD PULL COMPLETE, RHO_DOT = 0
RHO_DOT = 0
BETA = 640  # pcm

# FIND RHO AFTER ROD PULL
rho = DRW * STEPS  # pcm

sur = 26.06 * (RHO_DOT + LAMBDA_EFF * rho) / (BETA - rho)

print(f"The Start Up Rate is: {sur:.3f}")
\end{lstlisting}

\subsubsection*{Solution}
Given:
\begin{itemize}
    \item Differential Rod Worth (DRW) = 10 pcm/step
    \item Number of steps = 8
    \item $\lambda_{eff}$ = 0.1 Hz
    \item $\dot{\rho}$ = 0 (after rod pull complete)
    \item $\beta$ = 640 pcm
\end{itemize}

Reactivity after rod pull:
\[\rho = \text{DRW} \times \text{STEPS} = 10 \times 8 = 80 \text{ pcm}\]

Start-up rate calculation:
\[\text{SUR} = \frac{26.06 \times (\dot{\rho} + \lambda_{eff} \times \rho)}{\beta - \rho} = \frac{26.06 \times (0 + 0.1 \times 80)}{640 - 80}\]

\[\boxed{\text{Start Up Rate} = 0.373 \text{ DPM}}\]

\subsection*{Part B}

Negative reactivity feedback due to temperature would cause this power level
off. I would expect that the average reactor temperature would have increased
from the low power state significantly. I would also expect xenon concentration
would have increased, but would not have been the culprit in power leveling off. 

\subsection*{Part C}

\subsubsection*{Python Code}
\begin{lstlisting}[language=Python,
    basicstyle=\ttfamily\small,
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# Problem 3C
import sympy as sm

D_POWER = 2.5  # %
D_T_AVG = 4  # degrees
HEAT_UP_RATE = 0.15  # F/s
ALPHA_F = 10  # pcm/%power

rho_rod = rho

# The heat up rate introduces a rho_dot, so SUR becomes 0 at the peak power.
alpha_w_sym = sm.Symbol("alpha_w")
rho_dot = alpha_w_sym * HEAT_UP_RATE
rho_net = alpha_w_sym * D_T_AVG + rho_rod + ALPHA_F * D_POWER

# At peak power, SUR = 0, which means: rho_dot + lambda_eff * rho_net = 0
# (the numerator must be zero)
equation = rho_dot + LAMBDA_EFF * rho_net

# Solve for alpha_w
alpha_w_solution = sm.solve(equation, alpha_w_sym)[0]
alpha_w = float(alpha_w_solution)

print(f"The water temperature reactivity coefficient is: {alpha_w:.3f} pcm/F")
\end{lstlisting}

\subsubsection*{Solution}
Given:
\begin{itemize}
    \item Power change at peak: $\Delta P$ = 2.5\%
    \item Average temperature change at peak: $\Delta T_{avg}$ = 4$^\circ$F
    \item Heat-up rate: $\dot{T}$ = 0.15 $^\circ$F/s
    \item Fuel temperature coefficient: $\alpha_f$ = 10 pcm/\%power
    \item Rod reactivity: $\rho_{rod}$ = 80 pcm (from Part A)
    \item $\lambda_{eff}$ = 0.1 Hz
\end{itemize}

At peak power, the start-up rate becomes zero (SUR = 0), but temperature is still rising. This is the key insight: the numerator of the SUR equation must equal zero:

\[\dot{\rho} + \lambda_{eff} \times \rho_{net} = 0\]

The temperature rise creates a reactivity change rate:
\[\dot{\rho} = \alpha_w \times \dot{T} = \alpha_w \times 0.15\]

The net reactivity at the peak is:
\[\rho_{net} = \alpha_w \Delta T_{avg} + \rho_{rod} + \alpha_f \Delta P = \alpha_w \times 4 + 80 + 10 \times 2.5\]

Substituting into the SUR = 0 condition:
\[\alpha_w \times 0.15 + 0.1 \times (\alpha_w \times 4 + 80 + 25) = 0\]

\[0.15\alpha_w + 0.4\alpha_w + 10.5 = 0\]

\[0.55\alpha_w = -10.5\]

\[\boxed{\alpha_w = -19.091 \text{ pcm/}^\circ\text{F}}\]

\subsection*{Part D}

\subsubsection*{Solution}

At final equilibrium when the transient is complete:
\begin{itemize}
    \item Temperature stops changing: $\dot{T} = 0 \Rightarrow \dot{\rho} = 0$
    \item Start-up rate returns to zero: SUR = 0
    \item Net reactivity must be zero: $\rho_{net} = 0$
\end{itemize}

Since $\dot{\rho} = 0$ at equilibrium, the SUR equation requires:
\[\text{SUR} = \frac{26.06 \times (0 + \lambda_{eff} \times \rho_{net})}{\beta - \rho_{net}} = 0\]

This is satisfied when $\rho_{net} = 0$:
\[\alpha_w T_{final} + \rho_{rod} + \alpha_f P_{final} = 0\]

However, without knowing the heat removal characteristics (i.e., the relationship between power generation and temperature at thermal equilibrium with ambient losses), we cannot solve for exact values of $T_{final}$ and $P_{final}$.

\textbf{Qualitative Analysis:}

The transient behavior proceeds as follows:
\begin{enumerate}
    \item \textbf{At the peak} ($\Delta T = 4^\circ$F, $\Delta P = 2.5\%$): SUR = 0, but temperature is still rising at 0.15 $^\circ$F/s
    \item \textbf{After the peak}: Temperature continues to rise $\Rightarrow$ more negative reactivity is added $\Rightarrow$ power decreases from its maximum
    \item \textbf{At final equilibrium}: Temperature plateaus when heat generation equals ambient heat removal
\end{enumerate}

Therefore:
\[\boxed{T_{final} > 4^\circ\text{F} \quad \text{and} \quad P_{final} < 2.5\%}\]

The final power is lower than the peak power, but the final temperature is higher than the peak temperature. The peak power at 2.5\% is a transient maximum, not the steady-state equilibrium value.