Dane Sabo
df8f28fce3
M .sessions/nvim_config.vim M .task/backlog.data M .task/completed.data M .task/pending.data M .task/undo.data A Class_Work/nuce2101/exam2/2101_Exam_2_2025.pdf A "Class_Work/nuce2101/exam2/Fundamental Kinetics Ideas_Rev_17.pdf" A "Class_Work/nuce2101/exam2/Simplified Parallel Coupled Reactors Rev 8.pdf"
64 lines
2.1 KiB
Python
64 lines
2.1 KiB
Python
import numpy as np
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import sympy as sm
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# Problem 3A
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## Using formulas from Fundamental Kinetics Ideas R17 Page 51
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DRW = 10 # pcm/step
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STEPS = 8
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LAMBDA_EFF = 0.1 # hz
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# ASSUMING AFTER ROD PULL COMPLETE, RHO_DOT = 0
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RHO_DOT = 0
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BETA = 640 # pcm
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# FIND RHO AFTER ROD PULL
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rho = DRW * STEPS # pcm
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sur = 26.06 * (RHO_DOT + LAMBDA_EFF * rho) / (BETA - rho)
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print(f"The Start Up Rate is: {sur:.3f}")
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# Problem 3C
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# rho_net = rho_T + rho_rods + rho_poison + rho_fuel + ...
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# rho_net = alpha_w * (4 degrees) + rho (from above) + 0 + alpha_f * 2.5
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D_POWER = 2.5 # %
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D_T_AVG = 4 # degrees
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HEAT_UP_RATE = 0.15 # F/s
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ALPHA_F = 10 # pcm/%power
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rho_rod = rho
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# The heat up rate introduces a rho_dot, so SUR becomes 0 at the peak power.
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alpha_w_sym = sm.Symbol("alpha_w")
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rho_dot = alpha_w_sym * HEAT_UP_RATE
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rho_net = alpha_w_sym * D_T_AVG + rho_rod + ALPHA_F * D_POWER
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# At peak power, SUR = 0, which means: rho_dot + lambda_eff * rho_net = 0
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# (the numerator must be zero)
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equation = rho_dot + LAMBDA_EFF * rho_net
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# Solve for alpha_w
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alpha_w_solution = sm.solve(equation, alpha_w_sym)[0]
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alpha_w = float(alpha_w_solution)
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print(f"\nThe water temperature reactivity coefficient is: {alpha_w:.3f} pcm/°F")
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# Problem 3D
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# At final equilibrium: temperature stops changing (rho_dot = 0) and rho_net = 0
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# This means: alpha_w * T_final + rho_rod + alpha_f * P_final = 0
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#
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# However, without knowing the heat removal characteristics (relationship between
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# power and temperature at equilibrium), we cannot solve for exact values.
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#
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# Qualitative analysis:
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# - At peak (4°F, 2.5%): SUR = 0 but temperature still rising
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# - After peak: Temperature continues to rise → more negative reactivity → power decreases
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# - At final equilibrium: Temperature plateaus at T_final > 4°F, Power at P_final < 2.5%
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print(f"\nPart D - Qualitative Answer:")
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print(f"At peak power: ΔT = {D_T_AVG}°F, ΔP = {D_POWER}%")
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print(f"At final equilibrium:")
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print(f" - Temperature: T_final > {D_T_AVG}°F (continues rising after peak)")
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print(f" - Power: P_final < {D_POWER}% (decreases from peak as T rises further)")
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