Obsidian/Class_Work/nuce2101/final/latex/problem10.tex

\section*{Problem 10}
\subsection*{Part A}

To me, this looks like there was a major steam leak on the secondary side,
temporarily increasing steam demand by a lot. I think this was on the secondary
side because a primary side loss of coolant would be captured in the pressure
data.

\subsection*{Part B}

\subsubsection*{Subpart 1}

I'm going to assume a step change in reactivity, such that \(\dot \rho=0\).

Because we're considering the first few milliseconds of the transient, we can
ignore the contribution of precursors and delayed neutrons. This is an
appropriate assumption because the fastest precursor is on the order of 300ms
half-life. Much slower than our analysis.

\[\frac{dn(t)}{dt} = \frac{n(t)(\rho - \beta)}{\Lambda}\]
\[\frac{\dot n(t)}{n(t)} = \frac{(\rho - \beta)}{\Lambda}\]
\[\frac{781.88}{26.06} = \frac{(\rho - 0.000650)}{0.00026}\]
\[\rho = 0.008451\]

\subsubsection*{Subpart 2}
I would assume rise rate decreases so rapidly because fuel temperature would go
through the roof and Doppler broadening would add a lot of negative reactivity.
There is not enough time for any other physics to happen other than those at the
quantum level.

\subsection*{Part C}

At near steady state (around 800 seconds on the chart), the reactor power and
temperature have stabilized after the initial transient. We can estimate the
moderator temperature coefficient using the relationship between reactivity
change and temperature change.

Given:
\begin{itemize}
  \item Initial temperature: \(T_i = 600^\circ\)F
  \item Final temperature: \(T_f = 680^\circ\)F
  \item Temperature change: \(\Delta T = 80^\circ\)F
\end{itemize}

At near steady state, the reactor is critical, so the net reactivity is zero.
The reactivity balance is:

\[\rho_{net} = \rho_{inserted} + \alpha_m \Delta T = 0\]

From Part B.1, we found the inserted reactivity:

\[\rho_{inserted} = 0.008451 = 845.1 \text{ pcm}\]

Solving for the moderator temperature coefficient:

\[\alpha_m = -\frac{\rho_{inserted}}{\Delta T} = -\frac{845.1 \text{ pcm}}{80^\circ\text{F}}\]

\[\boxed{\alpha_m \approx -10.6 \text{ pcm}/^\circ\text{F}}\]

This value is reasonable for a PWR moderator temperature coefficient, typically
ranging from -10 to -40 pcm/\(^\circ\)F depending on core conditions and boron
concentration.
\subsection*{Part D}

Starting with the one delayed group prompt jump approximation:

\[n(t) = \frac{\lambda_{eff} C(t) \Lambda + S\Lambda}{\beta - \rho}\]

\[\dot C(t) = \frac{n(t)\beta}{\Lambda} - \lambda_{eff} C(t)\]

To find \(\dot n(t)\), we take the implicit derivative of the first equation.
Since both \(C(t)\) and \(\lambda_{eff}(t)\) can vary with time, we use the
product rule:

\[\dot n(t) = \frac{\Lambda}{\beta - \rho} \frac{d}{dt}[\lambda_{eff} C(t)] - \frac{\Lambda (\lambda_{eff} C + S)}{(\beta - \rho)^2} \dot \rho\]

\[\dot n(t) = \frac{\Lambda}{\beta - \rho} \left[\dot \lambda_{eff} C(t) + \lambda_{eff} \dot C(t)\right] - \frac{\Lambda (\lambda_{eff} C + S)}{(\beta - \rho)^2} \dot \rho\]

Substitute \(\dot C(t) = \frac{n(t)\beta}{\Lambda} - \lambda_{eff} C(t)\):

\[\dot n(t) = \frac{\Lambda}{\beta - \rho} \left[\dot \lambda_{eff} C(t) + \lambda_{eff}\left(\frac{n(t)\beta}{\Lambda} - \lambda_{eff} C(t)\right)\right] - \frac{n(t)(\beta - \rho) + S\Lambda}{(\beta - \rho)^2} \dot \rho\]

Simplify:

\[\dot n(t) = \frac{\Lambda \dot \lambda_{eff} C(t)}{\beta - \rho} + \frac{\beta n(t)}{\beta - \rho} - \frac{\Lambda \lambda_{eff}^2 C(t)}{\beta - \rho} - \frac{n(t)\dot \rho}{\beta - \rho} - \frac{S\Lambda \dot \rho}{(\beta - \rho)^2}\]

From the first equation: \(\lambda_{eff} C(t) \Lambda = n(t)(\beta - \rho) - S\Lambda\), so:

\[\Lambda C(t) = \frac{n(t)(\beta - \rho) - S\Lambda}{\lambda_{eff}}\]

After substitution and algebraic manipulation, we get:

\[\frac{\dot n(t)}{n(t)} = \frac{\dot \rho + \frac{\dot \lambda_{eff}}{\lambda_{eff}}\left[(\beta - \rho) - \frac{S\Lambda}{n(t)}\right] + \lambda_{eff}\left(\rho + \frac{S\Lambda}{n(t)}\right)}{(\beta - \rho) + \frac{S\Lambda}{n(t)}}\]

Power turning occurs when \(\dot n(t) = 0\):

\[\dot \rho + \frac{\dot \lambda_{eff}}{\lambda_{eff}}\left[(\beta - \rho) - \frac{S\Lambda}{n(t)}\right] + \lambda_{eff}\left(\rho + \frac{S\Lambda}{n(t)}\right) = 0\]

The key term \(\frac{\dot \lambda_{eff}}{\lambda_{eff}}\left[(\beta - \rho) - \frac{S\Lambda}{n(t)}\right]\) arises from the implicit derivative of the product \(\lambda_{eff} C(t)\). This accounts for the changing effective decay constant during the transient.

\subsection*{Part E}

Values estimated from graphs at t = 17.2s (power peak):

\begin{itemize}
  \item \(\Delta T_{ave} \approx 611 - 600 = 11^\circ\)F
  \item \(\frac{d(\Delta T_{ave})}{dt} \approx \frac{620 - 600}{25 - 5} = 1.0^\circ\)F/s
  \item \(\lambda_{eff} \approx 0.13\) s\(^{-1}\)
  \item \(\dot \lambda_{eff} \approx 0.081\) s\(^{-2}\)
  \item From Part B.1: \(\rho_0 = 0.008451\) (845.1 pcm)
  \item From Part C: \(\alpha_m \approx -10.6\) pcm/\(^\circ\)F
  \item \(\beta = 0.0065\) (650 pcm)
\end{itemize}

At power turning (\(\dot n(t) = 0\)):

\[\dot \rho + \frac{\dot \lambda_{eff}}{\lambda_{eff}}\left[(\beta - \rho) - \frac{S\Lambda}{n(t)}\right] + \lambda_{eff}\left(\rho + \frac{S\Lambda}{n(t)}\right) = 0\]

Assuming negligible source term \(S \approx 0\) at high power:

\[\dot \rho + \frac{\dot \lambda_{eff}}{\lambda_{eff}}(\beta - \rho) + \lambda_{eff} \rho = 0\]

The reactivity includes both moderator and fuel temperature feedback:

\[\rho(t) = \rho_0 + \alpha_m \Delta T_{ave} + \alpha_f \Delta T_{ave}\]

\[\dot \rho = (\alpha_m + \alpha_f) \frac{d(\Delta T_{ave})}{dt}\]

Substituting into the power turning equation:

\[(\alpha_m + \alpha_f) \frac{d(\Delta T_{ave})}{dt} + \frac{\dot \lambda_{eff}}{\lambda_{eff}}\left[\beta - \left(\rho_0 + \alpha_m \Delta T_{ave} + \alpha_f \Delta T_{ave}\right)\right] + \lambda_{eff} \left(\rho_0 + \alpha_m \Delta T_{ave} + \alpha_f \Delta T_{ave}\right) = 0\]

Solving numerically for \(\alpha_f\) using the given values:

\begin{align*}
\alpha_m &= -10.6 \times 10^{-5} \text{ per }^\circ\text{F} \\
\rho &= 0.008451 + (-10.6 \times 10^{-5})(11) + \alpha_f (11) \\
\dot \rho &= (-10.6 \times 10^{-5} + \alpha_f)(1.0)
\end{align*}

Using SymPy to solve:

\[\boxed{\alpha_f \approx 7.96 \text{ pcm}/^\circ\text{F}}\]

\textbf{Note:} This positive value is non-physical, as fuel temperature
coefficients should always be negative due to Doppler broadening. This suggests
the model assumptions may not be valid for this extreme transient.

\textbf{Physical interpretation:} The only physically reasonable explanation for
the continued increase in average coolant temperature even as power is turning
is that the fuel temperature is much hotter than the moderator temperature and
continues dumping heat into the coolant via conduction. This large temperature
gradient between fuel and moderator violates our assumption that both can be
approximated by \(\Delta T_{ave}\). The fuel has likely reached extreme
temperatures above normal operating
conditions while the moderator lags significantly behind. In a real scenario,
fuel damage or melting would likely have occurred.

\subsection*{Part F}

\textbf{Recommendation: Do NOT restart this plant without extensive inspection
and fuel integrity assessment.}

The analysis reveals several concerning indicators:

\textbf{Evidence of severe fuel damage:}

\begin{itemize}
  \item The non-physical positive fuel temperature coefficient from Part E
  indicates the normal reactor physics models have broken down

  \item The continued increase in average coolant temperature even as power was
  turning suggests fuel temperature was far exceeding moderator temperature

  \item The extreme reactivity insertion (845 pcm) combined with the rapid power
  excursion likely caused fuel temperatures to reach damage thresholds

  \item Power rose extremely rapidly in the first few seconds, indicating
  prompt-critical-like behavior that would cause severe thermal stress
\end{itemize}

The fuel almost certainly experienced temperatures well
beyond design limits. Cladding integrity is highly suspect. Restart should be
prohibited until a comprehensive inspection confirms the core is safe to
operate, and additional care should be paid to the chemistry of the coolant to
look for zirconium or zirconium-irradiated products.