\section*{Problem 7}
\subsection*{Part A}

Starting with the point kinetics equations with one delayed precursor group, using
the prompt jump approximation where prompt neutrons reach equilibrium instantly:

\[\frac{dN(t)}{dt} = \lambda C(t)\]

\[\frac{dC(t)}{dt} = \frac{\beta}{\Lambda}N(t) - \lambda C(t)\]

Using appropriate numerical values for a PWR with low enrichment U-235 fuel:

\begin{itemize}
  \item \(\beta = 0.0065\) (650 pcm)
  \item \(\Lambda = 5 \times 10^{-5}\) s (50 \(\mu\)s)
  \item \(\lambda = 0.08\) s\(^{-1}\)
  \item \(\rho = +0.0005\) (+50 pcm inserted)
\end{itemize}

The complete system of equations describing reactor power as a function of time is:

\[N(t) = \Lambda \frac{\lambda C(t)}{\beta -\rho} \]

\[\frac{dC(t)}{dt} = \frac{\beta}{\Lambda}N(t) - \lambda C(t)\]

Substituting N(t) into the precursor equation:

\[\frac{dC(t)}{dt} = \frac{\beta \lambda}{\beta -\rho} C(t) - \lambda C(t) = \lambda C(t) \left(\frac{\beta}{\beta - \rho} - 1\right)\]

Simplifying:

\[\frac{dC(t)}{dt} = \lambda C(t) \frac{\rho}{\beta - \rho}\]

With the given numerical values (\(\beta = 0.0065\), \(\rho = 0.0005\), \(\lambda = 0.08\) s\(^{-1}\), \(\Lambda = 5 \times 10^{-5}\) s):

\[\frac{dC(t)}{dt} = 0.08 \times C(t) \times \frac{0.0005}{0.006} = 6.67 \times 10^{-3} C(t)\]

\[N(t) = 5 \times 10^{-5} \times \frac{0.08 \times C(t)}{0.006} = 6.67 \times 10^{-4} C(t)\]

\subsection*{Part B}

The kinetics parameters chosen are justified as follows:

\begin{itemize}
  \item \(\beta = 0.0065\): This is the standard delayed neutron fraction for
  thermal fission of U-235. 

  \item \(\Lambda = 50 \mu\)s: This is the typical prompt neutron generation
  time for a pressurized water reactor. 

  \item \(\lambda = 0.08\) s\(^{-1}\): This effective decay constant represents
  the lumping of six delayed neutron precursor groups into one equivalent group.
  This gives an effective half-life of \(t_{1/2} = \ln(2)/\lambda \approx 8.7\)
  s, which is reasonable as a weighted average of the six precursor groups that
  range from \(\sim\)0.2 s to \(\sim\)80 s half-lives.

\end{itemize}

\subsection*{Part C}

Power will stop increasing when \(\frac{dN}{dt} = 0\), which occurs when
temperature feedback effects add sufficient negative reactivity to cancel the
+50 pcm reactivity insertion.

The reactor reaches a new equilibrium when:

\[\rho_{net} = \rho_{inserted} + \alpha_f \Delta T_f + \alpha_m \Delta T_m = 0\]

For \(\rho_{inserted} = +50\) pcm, temperature feedback must provide
\(-50\) pcm. 

\subsection*{Part D}

At end of life (EOL) with +200 pcm reactivity insertion, several key parameters
change:

\begin{itemize}
  \item \(\beta_{EOL} \approx 0.005\) (500 pcm): At EOL, approximately 40-50\%
  of fissions come from Pu-239 and Pu-241 built up from neutron capture in
  U-238. Since Pu-239 has \(\beta \approx 0.0021\) (much lower than U-235's
  0.0065), the effective \(\beta\) is a weighted average that decreases to
  around 0.005.
\end{itemize}

The time constant for the transient becomes:

\[\tau_{EOL} = \frac{\beta - \rho}{\lambda \rho} = \frac{0.0045 - 0.002}{0.08 \times 0.002} = 15.6 \text{ s}\]

compared to BOL:

\[\tau_{BOL} = \frac{0.0065 - 0.0005}{0.08 \times 0.0005} = 150 \text{ s}\]

The EOL transient is much faster (10x shorter period) because \(\beta - \rho\)
is smaller relative to \(\rho\). 

The spent core actually experiences a more dramatic transient per unit
reactivity insertion, but the stronger feedback from increased fuel temperature
effects provides self-limiting behavior that prevents excessive power excursion.
Plutonium-240 absorbs a lot of neutrons as 240Pu concentration builds up.
