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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. + diff --git a/Class_Work/nuce2101/final/latex/problem3.tex b/Class_Work/nuce2101/final/latex/problem3.tex index c9de8b7c4..0565a7354 100644 --- a/Class_Work/nuce2101/final/latex/problem3.tex +++ b/Class_Work/nuce2101/final/latex/problem3.tex @@ -13,7 +13,7 @@ negative reactivity to the system. + \frac{\partial \rho_{net}}{\partial Power} dPower \] -But with ingoring fuel temperature feedback and no boron effects, +But with ignoring fuel temperature feedback and no boron effects, \[\rho_{net} = -10 [\frac{\text{pcm}}{^\circ F}]dT + \frac{\partial \rho_{net}}{\partial H} dH diff --git a/Class_Work/nuce2101/final/latex/problem5.tex b/Class_Work/nuce2101/final/latex/problem5.tex index 3fe384276..f6884f3e6 100644 --- a/Class_Work/nuce2101/final/latex/problem5.tex +++ b/Class_Work/nuce2101/final/latex/problem5.tex @@ -1,5 +1,47 @@ \section*{Problem 5} \subsection*{Part A} +Xenon-135 is primarily produced in a reactor as a fission product or by the +decay of Iodine-135, which is a direct product of uranium fission. Xenon-135 +decays naturally over time or can absorb neutrons to become other isotopes with +a much smaller neutron cross section. + + + \subsection*{Part B} + +Xenon-135: + +The second reactor will have a lower concentration of xenon-135 in the reactor +core. Xenon-135 will be generated by the decay of iodine-135, but will also +decay naturally to other elements over time as well as being removed through +neutron absorption. If the reactors are shut down after a year, the xenon-135 in +both will spike with the full power reactor having a larger spike, and it will +take longer for the 100\% power reactor to be able to be restarted until the +extra xenon-135 has had time to decay away. + +Samarium-149: + +The second reactor will also have a lower concentration of samarium-149 for a +similar reason as the xenon-135 case. There will be less fissile products to +decay into samarium-149 in the low power reactor. If the reactors are powered +off, however, both reactors will develop additional samarium-149 concentrations +that will not decay away naturally as samarium-149 is stable. It will take more reactivity +to restart the full power reactor from shutdown as the samarium-149 +concentration will be much higher than the low power reactor. + \subsection*{Part C} + +Xenon is the more challenging poison to deal with. First, xenon has a much +larger cross section and has a more significant effect on reactivity than +samarium. Second, xenon-135 is not a stable element, and thus has decay effects +when the reactor is shut down. As such, the amount of time since shutdown is an +important factor to consider when restarting a reactor with consideration to +xenon. A miscalculation on xenon concentration can lead to a too-large insertion +of reactivity on start-up, which doesn't exist in the same way with samarium as +samarium doesn't decay during shutdown. Finally, xenon is a much faster response +dynamically compared to samarium. Xenon-135 can quickly change concentrations +in hours, while samarium-149 concentrations can take weeks to change +significantly. Human operators are much more likely to be able to cope with the +longer time constant poison versus misjudging the small time constant effects of +xenon. diff --git a/Class_Work/nuce2101/final/latex/problem6.tex b/Class_Work/nuce2101/final/latex/problem6.tex index e69de29bb..a07dba7b3 100644 --- a/Class_Work/nuce2101/final/latex/problem6.tex +++ b/Class_Work/nuce2101/final/latex/problem6.tex @@ -0,0 +1,22 @@ +\section*{Problem 6} +\subsection*{Part A} + +Withdrawing the control rods will move power production towards the upper half +of the core. Changing boron concentration will affect all parts of the core the +same amount, while increasing to full power does not provide enough information +to describe what's happening in the core. That can be accomplished through rod +position or through boron concentration. + +Assume rods remove through the top of the core (basically all PWRs), moving the +rods out will allow more flux higher in the core and thus move the average +production upwards. + +\subsection*{Part B} + +This happens because the moderator is the coolant, and the +coolant is colder as it enters the core at the bottom of the core. A low +enrichment core using boron does not have as pronounced of an effect because at +the same time because the boron is a poison. As boron enters the core, it is +more effective compared to as it travels through the core and is 'used up'. This +is the opposite effect of the moderator temperature, and as such they balance +each other out to a degree. diff --git a/Class_Work/nuce2101/final/latex/problem7.tex b/Class_Work/nuce2101/final/latex/problem7.tex index e69de29bb..0b38b9fd5 100644 --- a/Class_Work/nuce2101/final/latex/problem7.tex +++ b/Class_Work/nuce2101/final/latex/problem7.tex @@ -0,0 +1,99 @@ +\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. diff --git a/Class_Work/nuce2101/final/latex/problem8.tex b/Class_Work/nuce2101/final/latex/problem8.tex index e69de29bb..65f274d18 100644 --- a/Class_Work/nuce2101/final/latex/problem8.tex +++ b/Class_Work/nuce2101/final/latex/problem8.tex @@ -0,0 +1,70 @@ +\section*{Problem 8} + +\subsection*{Part A} + +The one delayed-group model assumes all six precursor groups can be lumped into +a single effective group with average parameters. This introduces errors: + +\begin{itemize} + \item Each group has different decay constants (\(\lambda_i\)) ranging from + 0.012 to 3.01 s\(^{-1}\), corresponding to half-lives from 0.2 to 80 seconds + + \item The one-group model cannot capture the multi-timescale behavior - early + time dynamics are dominated by fast-decaying groups, late time by slow groups + + \item Effective parameters (\(\beta_{eff}\), \(\lambda_{eff}\)) are only + approximate averages that work reasonably for long-term behavior but miss + short-term details +\end{itemize} + +This weakness matters most for short transients where individual group dynamics +are important, less so for long-term steady-state calculations. + +\subsection*{Part B} + +For six delayed groups with prompt jump approximation: + +Power is given by: + +\[N(t) = \frac{\Lambda}{\beta - \rho} \sum_{i=1}^{6} \lambda_i C_i(t)\] + +Precursor concentrations evolve as: + +\[\frac{dC_i(t)}{dt} = \frac{\beta_i}{\Lambda}N(t) - \lambda_i C_i(t), \quad i = 1,\ldots,6\] + +In matrix form, define the state vector: + +\[\mathbf{C}(t) = \begin{bmatrix} C_1(t) \\ C_2(t) \\ C_3(t) \\ C_4(t) \\ C_5(t) \\ C_6(t) \end{bmatrix}\] + +The precursor equation becomes: + +\[\frac{d\mathbf{C}}{dt} = \mathbf{A} \mathbf{C}(t)\] + +where the matrix \(\mathbf{A}\) is: + +{\tiny +\[\mathbf{A} = \frac{1}{\beta - \rho} \begin{bmatrix} +\beta_1 \lambda_1 - \lambda_1(\beta - \rho) & \beta_1 \lambda_2 & \beta_1 \lambda_3 & \beta_1 \lambda_4 & \beta_1 \lambda_5 & \beta_1 \lambda_6 \\ +\beta_2 \lambda_1 & \beta_2 \lambda_2 - \lambda_2(\beta - \rho) & \beta_2 \lambda_3 & \beta_2 \lambda_4 & \beta_2 \lambda_5 & \beta_2 \lambda_6 \\ +\beta_3 \lambda_1 & \beta_3 \lambda_2 & \beta_3 \lambda_3 - \lambda_3(\beta - \rho) & \beta_3 \lambda_4 & \beta_3 \lambda_5 & \beta_3 \lambda_6 \\ +\beta_4 \lambda_1 & \beta_4 \lambda_2 & \beta_4 \lambda_3 & \beta_4 \lambda_4 - \lambda_4(\beta - \rho) & \beta_4 \lambda_5 & \beta_4 \lambda_6 \\ +\beta_5 \lambda_1 & \beta_5 \lambda_2 & \beta_5 \lambda_3 & \beta_5 \lambda_4 & \beta_5 \lambda_5 - \lambda_5(\beta - \rho) & \beta_5 \lambda_6 \\ +\beta_6 \lambda_1 & \beta_6 \lambda_2 & \beta_6 \lambda_3 & \beta_6 \lambda_4 & \beta_6 \lambda_5 & \beta_6 \lambda_6 - \lambda_6(\beta - \rho) +\end{bmatrix}\] +} + +\subsection*{Part C} + +The prompt jump approximation error is likely smaller than the one-group error. + +The prompt jump assumes prompt neutrons equilibrate instantly (valid when +\(\Lambda \ll\) timescales of interest). For a 50 \(\mu\)s generation time and +transients on the scale of seconds, this is excellent. + +The one-group approximation loses the multi-timescale structure of the six +groups, which significantly affects transient shape, especially in the first +10-20 seconds where fast groups dominate. + +For this problem (low reactivity, second-scale transient), prompt jump +introduces \(<1\%\) error while one-group can introduce 10-20\% errors in peak +timing and shape. diff --git a/Class_Work/nuce2101/final/latex/problem9.tex b/Class_Work/nuce2101/final/latex/problem9.tex index e69de29bb..30dbfc72e 100644 --- a/Class_Work/nuce2101/final/latex/problem9.tex +++ b/Class_Work/nuce2101/final/latex/problem9.tex @@ -0,0 +1,23 @@ +\section*{Problem 9} +\subsection*{Part A} + +The boron concentration must be less than 1200ppm. + +The plant startup within 10 hours of full power shutdown must have a higher +xenon-135 concentration than a plant that has been shut down for 25 days. As a +result, boron must be diluted to compensate for the xenon poisoning effect. + +\subsection*{Part B} + +At first boron must be concentrated to reduce power. At this point, xenon will +increase in the core as xenon burnup reduces. To compensate with this additional +poison as power falls, boron must be diluted to maintain 50\% power. Then, the +xenon concentration will fall off as it decays (over days), so boron will need to be +concentrated as the xenon negative reactivity diminishes and settles at 50\% +power. The boron concentration will be HIGHER at the end of the transient than +at the start. An example of the xenon concentration can be found on page 11 of +'Xenon Transient Information Rev 2024-2'. + +Samarium on the other hand will not have a significant effect as the amount of +time spent on maintenance will be too short for significant samarium effects to +build up. diff --git a/Class_Work/nuce2101/final/python/part_e_solver.py b/Class_Work/nuce2101/final/python/part_e_solver.py new file mode 100644 index 000000000..4f1e3cc4f --- /dev/null +++ b/Class_Work/nuce2101/final/python/part_e_solver.py @@ -0,0 +1,54 @@ +import sympy as sp + +# Define symbolic variable +alpha_f = sp.Symbol('alpha_f', real=True) + +# Given values from graphs at t = 17.2s +Delta_T_ave = 11.0 # degrees F +d_Delta_T_dt = 1.0 # degrees F/s +lambda_eff = 0.13 # s^-1 +dot_lambda_eff = 0.081 # s^-2 + +# Known parameters +rho_0 = 0.008451 # initial reactivity (845.1 pcm) +alpha_m = -10.6e-5 # in absolute units (-10.6 pcm/F = -10.6e-5 per F) +beta = 0.0065 # delayed neutron fraction + +print(f"alpha_m = {alpha_m} per degF = {alpha_m*1e5} pcm/degF") +print() + +# Reactivity at t = 17.2s (including BOTH moderator and fuel feedback) +rho = rho_0 + alpha_m * Delta_T_ave + alpha_f * Delta_T_ave + +# Rate of change of reactivity +dot_rho = (alpha_m + alpha_f) * d_Delta_T_dt + +# Power turning condition (S ≈ 0 at high power): +# dot_rho + (dot_lambda_eff / lambda_eff) * (beta - rho) + lambda_eff * rho = 0 + +power_turning_eq = dot_rho + (dot_lambda_eff / lambda_eff) * (beta - rho) + lambda_eff * rho + +# Solve for alpha_f +solution = sp.solve(power_turning_eq, alpha_f) + +print("Power turning equation:") +print(f"dot_rho + (dot_lambda_eff/lambda_eff)*(beta - rho) + lambda_eff*rho = 0") +print() +print("Solving for alpha_f...") +print() + +for sol in solution: + alpha_f_value = float(sol) + alpha_f_pcm = alpha_f_value * 1e5 # Convert to pcm/degF + print(f"alpha_f = {alpha_f_value:.6e} per degF") + print(f"alpha_f = {alpha_f_pcm:.2f} pcm/degF") + print() + + # Verify the solution + rho_check = rho_0 + alpha_m * Delta_T_ave + alpha_f_value * Delta_T_ave + dot_rho_check = (alpha_m + alpha_f_value) * d_Delta_T_dt + lhs = dot_rho_check + (dot_lambda_eff/lambda_eff)*(beta - rho_check) + lambda_eff*rho_check + + print(f"Verification:") + print(f"rho at t=17.2s = {rho_check:.6f} ({rho_check*1e5:.1f} pcm)") + print(f"Power turning equation LHS = {lhs:.6e} (should be ~0)") diff --git a/Writing/CV/SABO_CV_OCT25.pdf b/Writing/CV/SABO_CV_OCT25.pdf new file mode 100644 index 000000000..8be8fb790 Binary files /dev/null and b/Writing/CV/SABO_CV_OCT25.pdf differ