Auto sync: 2025-12-08 13:15:47 (11 files changed)
M Class_Work/nuce2101/final/latex/main.aux M Class_Work/nuce2101/final/latex/main.fdb_latexmk M Class_Work/nuce2101/final/latex/main.fls M Class_Work/nuce2101/final/latex/main.log M Class_Work/nuce2101/final/latex/main.pdf M Class_Work/nuce2101/final/latex/main.synctex.gz M Class_Work/nuce2101/final/latex/problem1.tex M Class_Work/nuce2101/final/latex/problem2.tex
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\relax
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\bibstyle{unsrt}
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@ -1,5 +1,5 @@
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@ -464,9 +464,18 @@ INPUT /usr/share/texlive/texmf-dist/fonts/tfm/adobe/symbol/psyr.tfm
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INPUT ./problem2.tex
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INPUT ./problem2.tex
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INPUT problem2.tex
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INPUT ./problem3.tex
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INPUT problem3.tex
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INPUT ./problem4.tex
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INPUT ./problem4.tex
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INPUT problem4.tex
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@ -488,11 +497,14 @@ INPUT problem9.tex
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INPUT ./problem10.tex
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INPUT ./problem10.tex
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INPUT problem10.tex
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This is pdfTeX, Version 3.141592653-2.6-1.40.25 (TeX Live 2023/Debian) (preloaded format=pdflatex 2024.9.10) 8 DEC 2025 11:33
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This is pdfTeX, Version 3.141592653-2.6-1.40.25 (TeX Live 2023/Debian) (preloaded format=pdflatex 2024.9.10) 8 DEC 2025 13:15
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entering extended mode
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restricted \write18 enabled.
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file:line:error style messages enabled.
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@ -886,37 +886,25 @@ LaTeX Font Info: Font shape `OT1/ptm/bx/n' in size <7> not available
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LaTeX Font Info: Trying to load font information for TS1+ptm on input line 14.
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(/usr/share/texlive/texmf-dist/tex/latex/psnfss/ts1ptm.fd
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File: ts1ptm.fd 2001/06/04 font definitions for TS1/ptm.
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) [1]
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Runaway argument?
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\dot N(t) = \frac {(\rho - \beta ) N(t)}{\Lambda } + \sum _{i=1}^6 \lambda \ETC.
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./problem1.tex:58: Paragraph ended before \@newline was complete.
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<to be read again>
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\par
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l.58
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I suspect you've forgotten a `}', causing me to apply this
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control sequence to too much text. How can we recover?
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My plan is to forget the whole thing and hope for the best.
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) (./problem2.tex) (./problem3.tex) (./problem4.tex) (./problem5.tex) (./problem6.tex) (./problem7.tex) (./problem8.tex) (./problem9.tex) (./problem10.tex) [2] (./main.aux)
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) [1]) (./problem2.tex) (./problem3.tex [2]) (./problem4.tex) (./problem5.tex) (./problem6.tex) (./problem7.tex) (./problem8.tex) (./problem9.tex) (./problem10.tex) [3] [4] (./main.aux)
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***********
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LaTeX2e <2023-11-01> patch level 1
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L3 programming layer <2024-01-22>
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***********
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)
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Binary file not shown.
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@ -32,8 +32,7 @@ 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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\[ N_t = (1-\beta)K_{eff} N_{in} + \sum_{i=1}^6 \lambda_i C_i \Delta t + S \Delta
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t \]
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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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Then we find the change in neutron population:
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\[ N_f - N_i = (1-\beta) K_{eff} N_i + \sum_{i=1}^6 \lambda_i C_i \Delta t + S
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\Delta t - N_i \]
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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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\[ \frac{N_f - N_i}{\Delta t} = \frac{(1- \frac{1}{K_eff} - \beta) K_{eff} N_i}{\Delta t} + \sum_{i=1}^6 \lambda_i C_i + S
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\]
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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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\]
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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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@ -80,5 +76,4 @@ 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 \(\dot
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N_f\) changes sign.
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neutron population is changing. For this diagram, turning means that \(\dot N_f\) changes sign.
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@ -0,0 +1,20 @@
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\section*{Problem 2}
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We can use the startup rate equation assuming \(\dot \lambda_{eff}, S = 0\) to
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solve this problem:
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\[SUR = 26.06 [dpm-sec] \frac{\dot \rho + \lambda_{eff} \rho}{\beta - \rho}\]
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To get from \(10^{-6}\%\) to \(10^{1}\%\) power in 50 minutes, we can find that:
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\[SUR = \frac{1-(-6)}{50} \frac{\text{decades}}{\text{minutes}} = 0.14
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\text{DPM}\]
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We then plug in our values (assume \(\dot \rho = 0\) with a step change):
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\[0.14 = 26.06 [dpm-sec] \frac{0 + 0.1 \rho}{\beta - \rho}\]
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\[0.14 \beta - 0.14 \rho= 2.606 \rho\]
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\[\boxed{\rho = 0.0509\beta}\]
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Thus the correct answer is C.
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@ -0,0 +1,42 @@
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\section*{Problem 3}
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\subsection*{Part A}
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The moderator temperature coefficient must be controlling power. With all other
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factors constant, the moderator temperature coefficient is the only thing adding
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negative reactivity to the system.
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\subsection*{Part B}
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\[\rho_{net} = \frac{\partial \rho_{net}}{\partial T}dT
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+ \frac{\partial \rho_{net}}{\partial H} dH
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+ \frac{\partial \rho_{net}}{\partial Poison} dPoison
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+ \frac{\partial \rho_{net}}{\partial Power} dPower
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\]
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But with ingoring fuel temperature feedback and no boron effects,
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\[\rho_{net} = -10 [\frac{\text{pcm}}{^\circ F}]dT
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+ \frac{\partial \rho_{net}}{\partial H} dH
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\]
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Given that there is no poison or fuel temperature feedback, and steam demand
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does not change, reactor power will stay the same after the control rod drops
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into the core. Only moderator temperature can change reactivity in this problem.
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\subsection*{Part C}
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\[0 = -10 [\frac{\text{pcm}}{^\circ F}]dT
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- 100[pcm]
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\]
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\[dT = \frac{100[pcm]}{-10[\frac{\text{pcm}}{^\circ F}]}\]
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\[dT = -10^\circ F\]
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\[\boxed{T_{final} = 577^\circ F}\]
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\subsection*{Part D}
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Power will remain the same, and therefore steam pressure should remain the same
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as well.
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@ -0,0 +1,18 @@
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\section*{Problem 4}
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The power trajectory would be exponentially positive as the reactor would become
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prompt critical. One would analyze the transient by using a robot to examine the
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reactor soup after the steam bomb goes off in the containment.
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But being serious, one may examine the power transient by evaluating \(\rho\)
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over time using the partial addition formula we used in the last problem.
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Because the reactor is prompt critical, we can essentially ignore the delayed
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neutrons. The point kinetic equations can also be used, but honestly a decent
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approximation will be a first order exponential growth with time constant
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derived from the prompt neutron lifetime.
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For a high enrichment fuel, the growth of the curve will be impeded by basically
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nothing. Fuel and moderator temperature effects will be minimal. For a low
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enrichment fuel, moderator temperature and fuel temperature effects will slow
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the exponential growth as temperature increases, but depending on reactor
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design, will not prevent catastrophic failure.
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@ -0,0 +1,5 @@
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\section*{Problem 5}
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\subsection*{Part A}
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\subsection*{Part B}
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\subsection*{Part C}
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