Dane Sabo
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43 lines
1.2 KiB
TeX
43 lines
1.2 KiB
TeX
\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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