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80 lines
2.6 KiB
TeX
80 lines
2.6 KiB
TeX
\section*{Problem 1}
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\subsection*{Part A}
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\(K_{eff}\) is the effective growth or decay of the neutron population based on
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the prompt reactions. This block can be represented by basic physical quantities
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by using the six factor formula:
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\[ k_{eff} = \eta f p \epsilon P_{FNL} P_{TNL} \]
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where:
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\begin{itemize}
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\item \(\eta\) is the thermal neutron fission factor
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\item \(f\) is the thermal neutron absorption factor
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\item \(p\) is the resonance escape probability
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\item \(\epsilon\) is the fast fission neutron production factor
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\item \(P_{FNL}\) is the fast non-leakage probability
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\item \(P_{TNL}\) is the thermal non-leakage probability
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\end{itemize}
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\subsection*{Part B}
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We start with a set of equations representing our neutron populations:
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\[ N_{in} = N_{f} \]
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\[ N_{f} = N_{p} + N_{decay} + S \Delta t \]
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\[ N_{t} = N_{d} + N_{p} \]
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\[ N_t = K_{eff}N_{in} \]
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Now, how are these related? Well, the total number of neutrons is split between
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the neutrons that are prompt neutrons and those that are destined to become
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delayed neutrons. We can represent the fraction between the two as:
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\[ N_t = (1-\beta)K_{eff} N_{in} + \sum_{i=1}^6 \lambda_i C_i \Delta t + S \Delta t \]
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where
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\[ N_d = \sum_{i=1}^6 K_{eff} \beta_i N_{in} \]
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and
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\[ N_p = (1-\beta) K_{eff} N_{in} \]
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Then we find the change in neutron population:
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\[ N_f - N_i = (1-\beta) K_{eff} N_i + \sum_{i=1}^6 \lambda_i C_i \Delta t + S \Delta t - N_i \]
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and take the 'derivative':
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\[ \frac{N_f - N_i}{\Delta t} = \frac{(1- \frac{1}{K_eff} - \beta) K_{eff} N_i}{\Delta t} + \sum_{i=1}^6 \lambda_i C_i + S \]
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then after a little more substitution found in Fundamental Kinetics Ideas:
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\[ \dot N(t) = \frac{(\rho - \beta) N(t)}{\Lambda} + \sum_{i=1}^6 \lambda_i C_i + S \]
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and not forgetting our precursors:
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\[\dot C_i(t) = \frac{\beta_i}{\Lambda}N(t) - \lambda_i C_i\]
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\subsection*{Part C}
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The prompt jump assumption assumes that the speed of the prompt cycle is so fast
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that we can effectively ignore the dynamics of the prompt neutron effects. It is
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equivalent to us only using the outside loop, and not considering additions from
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\(N_p\).
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\subsection*{Part D}
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First, the number of prompt neutrons would increase dramatically. This would
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happen over the first hundreds of milliseconds. Then, over a much longer time
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horizon (seconds, minutes), the delayed number of neutrons would increase as the
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precursors catch up to the reactivity increase.
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\subsection*{Part E}
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Power turning has to do with start-up rate and the rate at which the total
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neutron population is changing. For this diagram, turning means that \(\dot N_f\) changes sign.
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