Merge branch 'master' of ssh://gitea.danesabo.com:1738/danesabo/Obsidian
This commit is contained in:
commit
be5b48e723
@ -268,3 +268,6 @@
|
|||||||
{"description":"Problem 3","entry":"20251105T160709Z","modified":"20251108T202716Z","project":"class.NUCE2101.Exam2","start":"20251108T202716Z","status":"pending","uuid":"5429f3ab-06f9-4df3-909d-ba5920169000"}
|
{"description":"Problem 3","entry":"20251105T160709Z","modified":"20251108T202716Z","project":"class.NUCE2101.Exam2","start":"20251108T202716Z","status":"pending","uuid":"5429f3ab-06f9-4df3-909d-ba5920169000"}
|
||||||
{"description":"Problem 3","end":"20251108T225056Z","entry":"20251105T160709Z","modified":"20251108T225056Z","project":"class.NUCE2101.Exam2","status":"completed","uuid":"5429f3ab-06f9-4df3-909d-ba5920169000"}
|
{"description":"Problem 3","end":"20251108T225056Z","entry":"20251105T160709Z","modified":"20251108T225056Z","project":"class.NUCE2101.Exam2","status":"completed","uuid":"5429f3ab-06f9-4df3-909d-ba5920169000"}
|
||||||
{"description":"Problem 2","end":"20251109T205515Z","entry":"20251105T160706Z","modified":"20251109T205515Z","project":"class.NUCE2101.Exam2","status":"completed","uuid":"5c30cc0b-222a-4975-a62b-9de8a78ab813"}
|
{"description":"Problem 2","end":"20251109T205515Z","entry":"20251105T160706Z","modified":"20251109T205515Z","project":"class.NUCE2101.Exam2","status":"completed","uuid":"5c30cc0b-222a-4975-a62b-9de8a78ab813"}
|
||||||
|
{"description":"Problem 5","end":"20251110T203913Z","entry":"20251105T160716Z","modified":"20251110T203913Z","project":"class.NUCE2101.Exam2","status":"completed","uuid":"7bdde787-042d-4b9b-b2de-19e11b40d032"}
|
||||||
|
{"description":"Problem 1","end":"20251110T213122Z","entry":"20251105T160652Z","modified":"20251110T213122Z","project":"class.NUCE2101.Exam2","status":"completed","uuid":"a2c8dc0b-e889-451a-a8c0-00a3ae9c5efd"}
|
||||||
|
{"description":"Problem 4","entry":"20251105T160711Z","modified":"20251110T213128Z","project":"class.NUCE2101.Exam2","start":"20251110T213128Z","status":"pending","uuid":"c39c9349-74b7-4a5c-8e75-27d341488c14"}
|
||||||
|
|||||||
@ -1,3 +1,5 @@
|
|||||||
|
[description:"Problem 1" end:"1762810282" entry:"1762358812" modified:"1762810282" project:"class.NUCE2101.Exam2" status:"completed" uuid:"a2c8dc0b-e889-451a-a8c0-00a3ae9c5efd"]
|
||||||
|
[description:"Problem 5" end:"1762807153" entry:"1762358836" modified:"1762807153" project:"class.NUCE2101.Exam2" status:"completed" uuid:"7bdde787-042d-4b9b-b2de-19e11b40d032"]
|
||||||
[description:"Problem 2" end:"1762721715" entry:"1762358826" modified:"1762721715" project:"class.NUCE2101.Exam2" status:"completed" uuid:"5c30cc0b-222a-4975-a62b-9de8a78ab813"]
|
[description:"Problem 2" end:"1762721715" entry:"1762358826" modified:"1762721715" project:"class.NUCE2101.Exam2" status:"completed" uuid:"5c30cc0b-222a-4975-a62b-9de8a78ab813"]
|
||||||
[description:"Problem 3" end:"1762642256" entry:"1762358829" modified:"1762642256" project:"class.NUCE2101.Exam2" status:"completed" uuid:"5429f3ab-06f9-4df3-909d-ba5920169000"]
|
[description:"Problem 3" end:"1762642256" entry:"1762358829" modified:"1762642256" project:"class.NUCE2101.Exam2" status:"completed" uuid:"5429f3ab-06f9-4df3-909d-ba5920169000"]
|
||||||
[description:"Problem 6" end:"1762633599" entry:"1762358840" modified:"1762633599" project:"class.NUCE2101.Exam2" status:"completed" uuid:"fcbcec80-81e1-4a7d-9ea8-d8b1c6f9737c"]
|
[description:"Problem 6" end:"1762633599" entry:"1762358840" modified:"1762633599" project:"class.NUCE2101.Exam2" status:"completed" uuid:"fcbcec80-81e1-4a7d-9ea8-d8b1c6f9737c"]
|
||||||
|
|||||||
@ -58,7 +58,5 @@
|
|||||||
[description:"project zk.perm Heat Removal" entry:"1761852415" modified:"1761852541" project:"zk.perm" status:"pending" uuid:"e77893e0-5ba6-4dcb-bdb0-4828845c64fe"]
|
[description:"project zk.perm Heat Removal" entry:"1761852415" modified:"1761852541" project:"zk.perm" status:"pending" uuid:"e77893e0-5ba6-4dcb-bdb0-4828845c64fe"]
|
||||||
[description:"project zk.perm Reactor Coolant Inventory Control" entry:"1761852439" modified:"1761852541" project:"zk.perm" status:"pending" uuid:"8e9a01f7-fe14-4a41-a085-0d99d8e0f03e"]
|
[description:"project zk.perm Reactor Coolant Inventory Control" entry:"1761852439" modified:"1761852541" project:"zk.perm" status:"pending" uuid:"8e9a01f7-fe14-4a41-a085-0d99d8e0f03e"]
|
||||||
[description:"Review Simeonas comments and implement them\n" entry:"1762182404" modified:"1762182404" project:"ERLM" status:"pending" tags:"editing" tags_editing:"x" uuid:"4f55e685-ed77-4533-9865-2ebe64b80632"]
|
[description:"Review Simeonas comments and implement them\n" entry:"1762182404" modified:"1762182404" project:"ERLM" status:"pending" tags:"editing" tags_editing:"x" uuid:"4f55e685-ed77-4533-9865-2ebe64b80632"]
|
||||||
[description:"Problem 1" entry:"1762358812" modified:"1762358812" project:"class.NUCE2101.Exam2" status:"pending" uuid:"a2c8dc0b-e889-451a-a8c0-00a3ae9c5efd"]
|
[description:"Problem 4" entry:"1762358831" modified:"1762810288" project:"class.NUCE2101.Exam2" start:"1762810288" status:"pending" uuid:"c39c9349-74b7-4a5c-8e75-27d341488c14"]
|
||||||
[description:"Problem 4" entry:"1762358831" modified:"1762358831" project:"class.NUCE2101.Exam2" status:"pending" uuid:"c39c9349-74b7-4a5c-8e75-27d341488c14"]
|
|
||||||
[description:"Problem 5" entry:"1762358836" modified:"1762358836" project:"class.NUCE2101.Exam2" status:"pending" uuid:"7bdde787-042d-4b9b-b2de-19e11b40d032"]
|
|
||||||
[dep_5429f3ab-06f9-4df3-909d-ba5920169000:"x" dep_5c30cc0b-222a-4975-a62b-9de8a78ab813:"x" dep_7bdde787-042d-4b9b-b2de-19e11b40d032:"x" dep_a2c8dc0b-e889-451a-a8c0-00a3ae9c5efd:"x" dep_c39c9349-74b7-4a5c-8e75-27d341488c14:"x" dep_fcbcec80-81e1-4a7d-9ea8-d8b1c6f9737c:"x" depends:"5429f3ab-06f9-4df3-909d-ba5920169000,5c30cc0b-222a-4975-a62b-9de8a78ab813,7bdde787-042d-4b9b-b2de-19e11b40d032,a2c8dc0b-e889-451a-a8c0-00a3ae9c5efd,c39c9349-74b7-4a5c-8e75-27d341488c14,fcbcec80-81e1-4a7d-9ea8-d8b1c6f9737c" description:"Submit Exam 2" due:"1762750800" entry:"1762358874" modified:"1762358874" project:"class.NUCE2101.Exam2" status:"pending" uuid:"ab2e56e4-4d9f-4e85-b40d-6f312ac17332"]
|
[dep_5429f3ab-06f9-4df3-909d-ba5920169000:"x" dep_5c30cc0b-222a-4975-a62b-9de8a78ab813:"x" dep_7bdde787-042d-4b9b-b2de-19e11b40d032:"x" dep_a2c8dc0b-e889-451a-a8c0-00a3ae9c5efd:"x" dep_c39c9349-74b7-4a5c-8e75-27d341488c14:"x" dep_fcbcec80-81e1-4a7d-9ea8-d8b1c6f9737c:"x" depends:"5429f3ab-06f9-4df3-909d-ba5920169000,5c30cc0b-222a-4975-a62b-9de8a78ab813,7bdde787-042d-4b9b-b2de-19e11b40d032,a2c8dc0b-e889-451a-a8c0-00a3ae9c5efd,c39c9349-74b7-4a5c-8e75-27d341488c14,fcbcec80-81e1-4a7d-9ea8-d8b1c6f9737c" description:"Submit Exam 2" due:"1762750800" entry:"1762358874" modified:"1762358874" project:"class.NUCE2101.Exam2" status:"pending" uuid:"ab2e56e4-4d9f-4e85-b40d-6f312ac17332"]
|
||||||
|
|||||||
@ -959,3 +959,15 @@ time 1762721715
|
|||||||
old [description:"Problem 2" entry:"1762358826" modified:"1762358826" project:"class.NUCE2101.Exam2" status:"pending" uuid:"5c30cc0b-222a-4975-a62b-9de8a78ab813"]
|
old [description:"Problem 2" entry:"1762358826" modified:"1762358826" project:"class.NUCE2101.Exam2" status:"pending" uuid:"5c30cc0b-222a-4975-a62b-9de8a78ab813"]
|
||||||
new [description:"Problem 2" end:"1762721715" entry:"1762358826" modified:"1762721715" project:"class.NUCE2101.Exam2" status:"completed" uuid:"5c30cc0b-222a-4975-a62b-9de8a78ab813"]
|
new [description:"Problem 2" end:"1762721715" entry:"1762358826" modified:"1762721715" project:"class.NUCE2101.Exam2" status:"completed" uuid:"5c30cc0b-222a-4975-a62b-9de8a78ab813"]
|
||||||
---
|
---
|
||||||
|
time 1762807153
|
||||||
|
old [description:"Problem 5" entry:"1762358836" modified:"1762358836" project:"class.NUCE2101.Exam2" status:"pending" uuid:"7bdde787-042d-4b9b-b2de-19e11b40d032"]
|
||||||
|
new [description:"Problem 5" end:"1762807153" entry:"1762358836" modified:"1762807153" project:"class.NUCE2101.Exam2" status:"completed" uuid:"7bdde787-042d-4b9b-b2de-19e11b40d032"]
|
||||||
|
---
|
||||||
|
time 1762810282
|
||||||
|
old [description:"Problem 1" entry:"1762358812" modified:"1762358812" project:"class.NUCE2101.Exam2" status:"pending" uuid:"a2c8dc0b-e889-451a-a8c0-00a3ae9c5efd"]
|
||||||
|
new [description:"Problem 1" end:"1762810282" entry:"1762358812" modified:"1762810282" project:"class.NUCE2101.Exam2" status:"completed" uuid:"a2c8dc0b-e889-451a-a8c0-00a3ae9c5efd"]
|
||||||
|
---
|
||||||
|
time 1762810288
|
||||||
|
old [description:"Problem 4" entry:"1762358831" modified:"1762358831" project:"class.NUCE2101.Exam2" status:"pending" uuid:"c39c9349-74b7-4a5c-8e75-27d341488c14"]
|
||||||
|
new [description:"Problem 4" entry:"1762358831" modified:"1762810288" project:"class.NUCE2101.Exam2" start:"1762810288" status:"pending" uuid:"c39c9349-74b7-4a5c-8e75-27d341488c14"]
|
||||||
|
---
|
||||||
|
|||||||
BIN
Class_Work/nuce2101/exam2/NUCE2101_SABO/Submission.pdf
Normal file
BIN
Class_Work/nuce2101/exam2/NUCE2101_SABO/Submission.pdf
Normal file
Binary file not shown.
321
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem1.py
Normal file
321
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem1.py
Normal file
@ -0,0 +1,321 @@
|
|||||||
|
import numpy as np
|
||||||
|
import matplotlib.pyplot as plt
|
||||||
|
from scipy.integrate import odeint
|
||||||
|
|
||||||
|
# Problem 1: Two-Loop Reactor System
|
||||||
|
|
||||||
|
print("=== Problem 1: Two-Loop Reactor with Steam Generators ===\n")
|
||||||
|
|
||||||
|
# Given Parameters
|
||||||
|
mu = None # Reactor Water Mass Fraction (RWMF) - to be determined or specified
|
||||||
|
|
||||||
|
# Base parameters
|
||||||
|
C_0 = 33.33 # Base Heat Capacity [%-sec/°F]
|
||||||
|
tau_0 = 0.75 * C_0 # Base Time Constant [sec/(%-sec/°F)] * C_0 = [sec]
|
||||||
|
|
||||||
|
# Initial steady state temperature
|
||||||
|
T_hot_initial = 450 # °F (reactor outlet / hot leg temperature)
|
||||||
|
T_cold_1_initial = 450 # °F (cold leg 1 temperature)
|
||||||
|
T_cold_2_initial = 450 # °F (cold leg 2 temperature)
|
||||||
|
|
||||||
|
print("Base Parameters:")
|
||||||
|
print(f"C_0 (Base Heat Capacity) = {C_0:.2f} %-sec/°F")
|
||||||
|
print(f"tau_0 (Base Time Constant) = {tau_0:.2f} sec")
|
||||||
|
print(f"\nInitial Steady State:")
|
||||||
|
print(f"T_hot = {T_hot_initial} °F")
|
||||||
|
print(f"T_cold_1 = {T_cold_1_initial} °F")
|
||||||
|
print(f"T_cold_2 = {T_cold_2_initial} °F")
|
||||||
|
|
||||||
|
|
||||||
|
# System parameters as functions of mu (RWMF)
|
||||||
|
def calculate_system_parameters(mu):
|
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|
"""
|
||||||
|
Calculate system parameters based on Reactor Water Mass Fraction (mu)
|
||||||
|
|
||||||
|
Parameters:
|
||||||
|
mu: Reactor Water Mass Fraction (RWMF)
|
||||||
|
|
||||||
|
Returns:
|
||||||
|
dict: Dictionary containing all calculated parameters
|
||||||
|
"""
|
||||||
|
# Reactor parameters
|
||||||
|
tau_r = mu * tau_0 # Reactor time constant [sec]
|
||||||
|
C_r = mu * C_0 # Reactor heat capacity [%-sec/°F]
|
||||||
|
|
||||||
|
# Steam Generator parameters (each)
|
||||||
|
tau_sg = (1 - mu) * tau_r / 2 # Steam Generator water time constant [sec]
|
||||||
|
C_sg = (1 - mu) * C_0 / 2 # Steam Generator heat capacity [%-sec/°F]
|
||||||
|
|
||||||
|
params = {
|
||||||
|
"mu": mu,
|
||||||
|
"tau_r": tau_r,
|
||||||
|
"C_r": C_r,
|
||||||
|
"tau_sg": tau_sg,
|
||||||
|
"C_sg": C_sg,
|
||||||
|
"C_0": C_0,
|
||||||
|
"tau_0": tau_0,
|
||||||
|
}
|
||||||
|
|
||||||
|
return params
|
||||||
|
|
||||||
|
|
||||||
|
# Example: Calculate parameters for mu = 0.5 (50% of water in reactor)
|
||||||
|
if mu is None:
|
||||||
|
mu_example = 0.5
|
||||||
|
print(f"\n--- Example with mu = {mu_example} ---")
|
||||||
|
params = calculate_system_parameters(mu_example)
|
||||||
|
|
||||||
|
print(f"\nReactor Parameters:")
|
||||||
|
print(f" tau_r (Reactor time constant) = {params['tau_r']:.2f} sec")
|
||||||
|
print(f" C_r (Reactor heat capacity) = {params['C_r']:.2f} %-sec/°F")
|
||||||
|
|
||||||
|
print(f"\nSteam Generator Parameters (each):")
|
||||||
|
print(f" tau_sg (SG water time constant) = {params['tau_sg']:.2f} sec")
|
||||||
|
print(f" C_sg (SG heat capacity) = {params['C_sg']:.2f} %-sec/°F")
|
||||||
|
|
||||||
|
# System description
|
||||||
|
print(f"\n" + "=" * 60)
|
||||||
|
print("System Description:")
|
||||||
|
print("=" * 60)
|
||||||
|
print(
|
||||||
|
"""
|
||||||
|
Two-Loop Reactor System:
|
||||||
|
- Reactor (hot leg) at T_hot
|
||||||
|
- Steam Generator 1 with cold leg at T_cold_1
|
||||||
|
- Steam Generator 2 with cold leg at T_cold_2
|
||||||
|
- Each loop: pump → reactor → steam generator → back to pump
|
||||||
|
- Lumped reactor volume with combined hot branches
|
||||||
|
- Equal mass flow rates in each loop
|
||||||
|
- Ignore loop transport times
|
||||||
|
|
||||||
|
Key Assumptions:
|
||||||
|
- Each SG + cold loop has same water mass and flow rate
|
||||||
|
- Reactor and hot branches lumped into single water volume
|
||||||
|
- System initially at equilibrium (100% power, balanced)
|
||||||
|
- Constant average temperature maintained
|
||||||
|
"""
|
||||||
|
)
|
||||||
|
|
||||||
|
# Part A: Differential Equations
|
||||||
|
print(f"\n" + "=" * 60)
|
||||||
|
print("Part A: Differential Equations")
|
||||||
|
print("=" * 60)
|
||||||
|
|
||||||
|
|
||||||
|
def get_differential_equations(mu):
|
||||||
|
"""
|
||||||
|
Develop differential equations for T_hot, T_cold1, T_cold2
|
||||||
|
|
||||||
|
Key insight: Flow heat capacity rate W = C_0 / (2 * tau_0)
|
||||||
|
This is the (mass flow rate * specific heat) for each loop
|
||||||
|
"""
|
||||||
|
params = calculate_system_parameters(mu)
|
||||||
|
|
||||||
|
# Flow heat capacity rate for each loop (same for both loops)
|
||||||
|
W = C_0 / (2 * tau_0)
|
||||||
|
|
||||||
|
print(f"\nFor mu = {mu}:")
|
||||||
|
print(f"Flow heat capacity rate (each loop): W = {W:.4f} %-sec/°F")
|
||||||
|
print(f"C_r = {params['C_r']:.2f}, C_sg = {params['C_sg']:.2f}")
|
||||||
|
|
||||||
|
print(f"\nDifferential Equations:")
|
||||||
|
print(f" C_r * dT_hot/dt = P_r - W*(T_hot - T_cold1) - W*(T_hot - T_cold2)")
|
||||||
|
print(f" C_sg * dT_cold1/dt = W*(T_hot - T_cold1) - Qdot1")
|
||||||
|
print(f" C_sg * dT_cold2/dt = W*(T_hot - T_cold2) - Qdot2")
|
||||||
|
print(f"\nwhere:")
|
||||||
|
print(f" P_r = reactor power (positive for heat generation)")
|
||||||
|
print(f" Qdot1, Qdot2 = SG heat removal rates (positive for heat removal)")
|
||||||
|
print(f" W = {W:.4f} %-sec/°F (flow heat capacity rate per loop)")
|
||||||
|
|
||||||
|
# Matrix form: dT/dt = A*T + B
|
||||||
|
print(f"\n--- Matrix Form: dT/dt = A*T + B ---")
|
||||||
|
|
||||||
|
# Coefficient matrix A
|
||||||
|
A = np.array(
|
||||||
|
[
|
||||||
|
[-2 * W / params["C_r"], W / params["C_r"], W / params["C_r"]],
|
||||||
|
[W / params["C_sg"], -W / params["C_sg"], 0],
|
||||||
|
[W / params["C_sg"], 0, -W / params["C_sg"]],
|
||||||
|
]
|
||||||
|
)
|
||||||
|
|
||||||
|
print(f"\nMatrix A (coefficients):")
|
||||||
|
print(f" [{A[0,0]:7.4f} {A[0,1]:7.4f} {A[0,2]:7.4f}]")
|
||||||
|
print(f" [{A[1,0]:7.4f} {A[1,1]:7.4f} {A[1,2]:7.4f}]")
|
||||||
|
print(f" [{A[2,0]:7.4f} {A[2,1]:7.4f} {A[2,2]:7.4f}]")
|
||||||
|
|
||||||
|
# For general case with P_r, Qdot1, Qdot2
|
||||||
|
print(f"\nVector B (forcing terms, depends on P_r, Qdot1, Qdot2):")
|
||||||
|
print(f" [P_r/C_r, -Qdot1/C_sg, -Qdot2/C_sg]^T")
|
||||||
|
print(
|
||||||
|
f" [P_r/{params['C_r']:.2f}, -Qdot1/{params['C_sg']:.2f}, -Qdot2/{params['C_sg']:.2f}]^T"
|
||||||
|
)
|
||||||
|
|
||||||
|
return W, params
|
||||||
|
|
||||||
|
|
||||||
|
# Part D: Transient Simulation Setup
|
||||||
|
print(f"\n" + "=" * 60)
|
||||||
|
print("Part D: Transient Simulation (0 to 30 seconds)")
|
||||||
|
print("=" * 60)
|
||||||
|
|
||||||
|
|
||||||
|
def reactor_odes(y, t, W, C_r, C_sg, P_r, Q1, Q2):
|
||||||
|
"""
|
||||||
|
ODE system for two-loop reactor
|
||||||
|
y = [T_hot, T_cold1, T_cold2]
|
||||||
|
|
||||||
|
Parameters:
|
||||||
|
P_r: Reactor power (heat generation rate)
|
||||||
|
Q1, Q2: Steam generator heat removal rates (Qdot_1, Qdot_2)
|
||||||
|
"""
|
||||||
|
T_hot, T_cold1, T_cold2 = y
|
||||||
|
|
||||||
|
# Reactor energy balance
|
||||||
|
dT_hot_dt = (P_r - W * (T_hot - T_cold1) - W * (T_hot - T_cold2)) / C_r
|
||||||
|
|
||||||
|
# Steam generator 1 energy balance (Q1 is heat removal rate)
|
||||||
|
dT_cold1_dt = (W * (T_hot - T_cold1) - Q1) / C_sg
|
||||||
|
|
||||||
|
# Steam generator 2 energy balance (Q2 is heat removal rate)
|
||||||
|
dT_cold2_dt = (W * (T_hot - T_cold2) - Q2) / C_sg
|
||||||
|
|
||||||
|
return [dT_hot_dt, dT_cold1_dt, dT_cold2_dt]
|
||||||
|
|
||||||
|
|
||||||
|
def simulate_transient(mu, Q1_percent, Q2_percent, t_max=30):
|
||||||
|
"""
|
||||||
|
Simulate reactor transient
|
||||||
|
|
||||||
|
Parameters:
|
||||||
|
mu: Reactor Water Mass Fraction
|
||||||
|
Q1_percent: Steam generator 1 heat removal rate (% of nominal)
|
||||||
|
Q2_percent: Steam generator 2 heat removal rate (% of nominal)
|
||||||
|
t_max: Simulation time (seconds)
|
||||||
|
"""
|
||||||
|
params = calculate_system_parameters(mu)
|
||||||
|
W = C_0 / (2 * tau_0)
|
||||||
|
|
||||||
|
# Initial conditions (steady state at 100% power balanced)
|
||||||
|
T_hot_0 = T_hot_initial
|
||||||
|
T_cold1_0 = T_cold_1_initial
|
||||||
|
T_cold2_0 = T_cold_2_initial
|
||||||
|
y0 = [T_hot_0, T_cold1_0, T_cold2_0]
|
||||||
|
|
||||||
|
# Power levels (normalized - 100% = 1.0 in appropriate units)
|
||||||
|
P_r = 100.0 # Reactor power (heat generation rate)
|
||||||
|
Q1 = Q1_percent # SG1 heat removal rate
|
||||||
|
Q2 = Q2_percent # SG2 heat removal rate
|
||||||
|
|
||||||
|
# Time array
|
||||||
|
t = np.linspace(0, t_max, 500)
|
||||||
|
|
||||||
|
# Solve ODE system
|
||||||
|
solution = odeint(
|
||||||
|
reactor_odes, y0, t, args=(W, params["C_r"], params["C_sg"], P_r, Q1, Q2)
|
||||||
|
)
|
||||||
|
|
||||||
|
T_hot = solution[:, 0]
|
||||||
|
T_cold1 = solution[:, 1]
|
||||||
|
T_cold2 = solution[:, 2]
|
||||||
|
|
||||||
|
# Calculate average temperature (mass-weighted, depends on mu)
|
||||||
|
T_ave = mu * T_hot + (1 - mu) * (T_cold1 + T_cold2) / 2
|
||||||
|
|
||||||
|
return t, T_hot, T_cold1, T_cold2, T_ave
|
||||||
|
|
||||||
|
|
||||||
|
# Example: Show how to set up simulations
|
||||||
|
print("\nSimulation cases to run:")
|
||||||
|
print("1. mu=0.5, Q1=50%, Q2=50% (equally loaded)")
|
||||||
|
print("2. mu=0.75, Q1=50%, Q2=50% (equally loaded)")
|
||||||
|
print("3. mu=0.5, Q1=60%, Q2=40% (unequally loaded)")
|
||||||
|
print("4. mu=0.75, Q1=60%, Q2=40% (unequally loaded)")
|
||||||
|
print("\nNote: Percentages are fractions of reactor power (P_r = 100%)")
|
||||||
|
|
||||||
|
# Run example simulation
|
||||||
|
print("\n--- Running Example: mu=0.5, Equal Loading ---")
|
||||||
|
W_example, params_example = get_differential_equations(0.5)
|
||||||
|
|
||||||
|
# Run all simulations and create plots
|
||||||
|
print("\n" + "=" * 60)
|
||||||
|
print("Running Simulations and Creating Plots")
|
||||||
|
print("=" * 60)
|
||||||
|
|
||||||
|
# Case 1 & 2: Equally loaded (Q1=50%, Q2=50%)
|
||||||
|
fig1, axes1 = plt.subplots(2, 2, figsize=(14, 10))
|
||||||
|
fig1.suptitle(
|
||||||
|
"Equally Loaded Steam Generators (Q1=50%, Q2=50%)", fontsize=14, fontweight="bold"
|
||||||
|
)
|
||||||
|
|
||||||
|
for idx, mu in enumerate([0.5, 0.75]):
|
||||||
|
print(f"\nSimulating mu={mu}, Q1=50%, Q2=50%")
|
||||||
|
t, T_hot, T_cold1, T_cold2, T_ave = simulate_transient(mu, 50.0, 50.0)
|
||||||
|
|
||||||
|
# Plot temperatures
|
||||||
|
ax = axes1[idx, 0]
|
||||||
|
ax.plot(t, T_hot, "r-", linewidth=2, label="T_hot")
|
||||||
|
ax.plot(t, T_cold1, "b-", linewidth=2, label="T_cold1")
|
||||||
|
ax.plot(t, T_cold2, "g-", linewidth=2, label="T_cold2")
|
||||||
|
ax.plot(t, T_ave, "k--", linewidth=2, label="T_ave")
|
||||||
|
ax.set_xlabel("Time (sec)", fontsize=10)
|
||||||
|
ax.set_ylabel("Temperature (°F)", fontsize=10)
|
||||||
|
ax.set_title(f"mu = {mu} (RWMF)", fontsize=11, fontweight="bold")
|
||||||
|
ax.grid(True, alpha=0.3)
|
||||||
|
ax.legend(loc="best")
|
||||||
|
|
||||||
|
# Plot temperature differences
|
||||||
|
ax = axes1[idx, 1]
|
||||||
|
ax.plot(t, T_hot - T_cold1, "b-", linewidth=2, label="ΔT1 = T_hot - T_cold1")
|
||||||
|
ax.plot(t, T_hot - T_cold2, "g-", linewidth=2, label="ΔT2 = T_hot - T_cold2")
|
||||||
|
ax.set_xlabel("Time (sec)", fontsize=10)
|
||||||
|
ax.set_ylabel("Temperature Difference (°F)", fontsize=10)
|
||||||
|
ax.set_title(f"Temperature Differences (mu = {mu})", fontsize=11, fontweight="bold")
|
||||||
|
ax.grid(True, alpha=0.3)
|
||||||
|
ax.legend(loc="best")
|
||||||
|
|
||||||
|
plt.tight_layout()
|
||||||
|
plt.savefig("problem1_equal_loading.png", dpi=300, bbox_inches="tight")
|
||||||
|
print("Plot saved: problem1_equal_loading.png")
|
||||||
|
plt.close()
|
||||||
|
|
||||||
|
# Case 3 & 4: Unequally loaded (Q1=60%, Q2=40%)
|
||||||
|
fig2, axes2 = plt.subplots(2, 2, figsize=(14, 10))
|
||||||
|
fig2.suptitle(
|
||||||
|
"Unequally Loaded Steam Generators (Q1=60%, Q2=40%)", fontsize=14, fontweight="bold"
|
||||||
|
)
|
||||||
|
|
||||||
|
for idx, mu in enumerate([0.5, 0.75]):
|
||||||
|
print(f"\nSimulating mu={mu}, Q1=60%, Q2=40%")
|
||||||
|
t, T_hot, T_cold1, T_cold2, T_ave = simulate_transient(mu, 60.0, 40.0)
|
||||||
|
|
||||||
|
# Plot temperatures
|
||||||
|
ax = axes2[idx, 0]
|
||||||
|
ax.plot(t, T_hot, "r-", linewidth=2, label="T_hot")
|
||||||
|
ax.plot(t, T_cold1, "b-", linewidth=2, label="T_cold1")
|
||||||
|
ax.plot(t, T_cold2, "g-", linewidth=2, label="T_cold2")
|
||||||
|
ax.plot(t, T_ave, "k--", linewidth=2, label="T_ave")
|
||||||
|
ax.set_xlabel("Time (sec)", fontsize=10)
|
||||||
|
ax.set_ylabel("Temperature (°F)", fontsize=10)
|
||||||
|
ax.set_title(f"mu = {mu} (RWMF)", fontsize=11, fontweight="bold")
|
||||||
|
ax.grid(True, alpha=0.3)
|
||||||
|
ax.legend(loc="best")
|
||||||
|
|
||||||
|
# Plot temperature differences
|
||||||
|
ax = axes2[idx, 1]
|
||||||
|
ax.plot(t, T_hot - T_cold1, "b-", linewidth=2, label="ΔT1 = T_hot - T_cold1")
|
||||||
|
ax.plot(t, T_hot - T_cold2, "g-", linewidth=2, label="ΔT2 = T_hot - T_cold2")
|
||||||
|
ax.set_xlabel("Time (sec)", fontsize=10)
|
||||||
|
ax.set_ylabel("Temperature Difference (°F)", fontsize=10)
|
||||||
|
ax.set_title(f"Temperature Differences (mu = {mu})", fontsize=11, fontweight="bold")
|
||||||
|
ax.grid(True, alpha=0.3)
|
||||||
|
ax.legend(loc="best")
|
||||||
|
|
||||||
|
plt.tight_layout()
|
||||||
|
plt.savefig("problem1_unequal_loading.png", dpi=300, bbox_inches="tight")
|
||||||
|
print("Plot saved: problem1_unequal_loading.png")
|
||||||
|
plt.close()
|
||||||
|
|
||||||
|
print("\n" + "=" * 60)
|
||||||
|
print("Simulation Complete!")
|
||||||
|
print("=" * 60)
|
||||||
Binary file not shown.
|
After Width: | Height: | Size: 479 KiB |
Binary file not shown.
|
After Width: | Height: | Size: 495 KiB |
210
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem1_verification.py
Normal file
210
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem1_verification.py
Normal file
@ -0,0 +1,210 @@
|
|||||||
|
import sympy as sm
|
||||||
|
import numpy as np
|
||||||
|
|
||||||
|
print("="*70)
|
||||||
|
print("Problem 1: SymPy Verification")
|
||||||
|
print("="*70)
|
||||||
|
|
||||||
|
# Define symbols
|
||||||
|
T_hot, T_cold1, T_cold2 = sm.symbols('T_hot T_cold1 T_cold2')
|
||||||
|
P_r, Qdot1, Qdot2 = sm.symbols('P_r Qdot_1 Qdot_2', positive=True)
|
||||||
|
C_r, C_sg, W = sm.symbols('C_r C_sg W', positive=True)
|
||||||
|
C_0, tau_0, mu = sm.symbols('C_0 tau_0 mu', positive=True)
|
||||||
|
t = sm.symbols('t', positive=True)
|
||||||
|
|
||||||
|
print("\n--- Part A: Derive Differential Equations ---")
|
||||||
|
|
||||||
|
# Energy balance equations
|
||||||
|
print("\nReactor energy balance:")
|
||||||
|
reactor_eq = sm.Eq(C_r * sm.Derivative(T_hot, t),
|
||||||
|
P_r - W*(T_hot - T_cold1) - W*(T_hot - T_cold2))
|
||||||
|
print(f"C_r * dT_hot/dt = P_r - W*(T_hot - T_cold1) - W*(T_hot - T_cold2)")
|
||||||
|
print(reactor_eq)
|
||||||
|
|
||||||
|
print("\nSteam Generator 1 energy balance:")
|
||||||
|
sg1_eq = sm.Eq(C_sg * sm.Derivative(T_cold1, t),
|
||||||
|
W*(T_hot - T_cold1) - Qdot1)
|
||||||
|
print(f"C_sg * dT_cold1/dt = W*(T_hot - T_cold1) - Qdot1")
|
||||||
|
print(sg1_eq)
|
||||||
|
|
||||||
|
print("\nSteam Generator 2 energy balance:")
|
||||||
|
sg2_eq = sm.Eq(C_sg * sm.Derivative(T_cold2, t),
|
||||||
|
W*(T_hot - T_cold2) - Qdot2)
|
||||||
|
print(f"C_sg * dT_cold2/dt = W*(T_hot - T_cold2) - Qdot2")
|
||||||
|
print(sg2_eq)
|
||||||
|
|
||||||
|
# Expand and simplify to get matrix form
|
||||||
|
print("\n--- Expanding to Matrix Form: dT/dt = A*T + B ---")
|
||||||
|
|
||||||
|
# Reactor equation expanded
|
||||||
|
reactor_rhs = P_r - W*T_hot + W*T_cold1 - W*T_hot + W*T_cold2
|
||||||
|
reactor_rhs_simplified = sm.expand(reactor_rhs)
|
||||||
|
print(f"\nReactor RHS expanded:")
|
||||||
|
print(f" = P_r - 2W*T_hot + W*T_cold1 + W*T_cold2")
|
||||||
|
print(f" = {reactor_rhs_simplified}")
|
||||||
|
|
||||||
|
dT_hot_dt = reactor_rhs_simplified / C_r
|
||||||
|
print(f"\ndT_hot/dt = {dT_hot_dt}")
|
||||||
|
|
||||||
|
# SG1 equation expanded
|
||||||
|
sg1_rhs = W*T_hot - W*T_cold1 - Qdot1
|
||||||
|
dT_cold1_dt = sg1_rhs / C_sg
|
||||||
|
print(f"\ndT_cold1/dt = {dT_cold1_dt}")
|
||||||
|
|
||||||
|
# SG2 equation expanded
|
||||||
|
sg2_rhs = W*T_hot - W*T_cold2 - Qdot2
|
||||||
|
dT_cold2_dt = sg2_rhs / C_sg
|
||||||
|
print(f"\ndT_cold2/dt = {dT_cold2_dt}")
|
||||||
|
|
||||||
|
# Extract coefficient matrix A
|
||||||
|
print("\n--- Constructing Matrix A ---")
|
||||||
|
A_symbolic = sm.Matrix([
|
||||||
|
[sm.diff(dT_hot_dt, T_hot), sm.diff(dT_hot_dt, T_cold1), sm.diff(dT_hot_dt, T_cold2)],
|
||||||
|
[sm.diff(dT_cold1_dt, T_hot), sm.diff(dT_cold1_dt, T_cold1), sm.diff(dT_cold1_dt, T_cold2)],
|
||||||
|
[sm.diff(dT_cold2_dt, T_hot), sm.diff(dT_cold2_dt, T_cold1), sm.diff(dT_cold2_dt, T_cold2)]
|
||||||
|
])
|
||||||
|
|
||||||
|
print("\nMatrix A (symbolic):")
|
||||||
|
sm.pprint(A_symbolic)
|
||||||
|
|
||||||
|
# Extract forcing vector B
|
||||||
|
print("\n--- Constructing Vector B ---")
|
||||||
|
# B contains the terms without T_hot, T_cold1, T_cold2
|
||||||
|
B_symbolic = sm.Matrix([
|
||||||
|
P_r/C_r,
|
||||||
|
-Qdot1/C_sg,
|
||||||
|
-Qdot2/C_sg
|
||||||
|
])
|
||||||
|
|
||||||
|
print("\nVector B (symbolic):")
|
||||||
|
sm.pprint(B_symbolic)
|
||||||
|
|
||||||
|
# Verify: dT/dt = A*T + B
|
||||||
|
T_vector = sm.Matrix([T_hot, T_cold1, T_cold2])
|
||||||
|
dT_dt_vector = sm.Matrix([dT_hot_dt, dT_cold1_dt, dT_cold2_dt])
|
||||||
|
|
||||||
|
reconstructed = A_symbolic * T_vector + B_symbolic
|
||||||
|
|
||||||
|
print("\n--- Verification: A*T + B = dT/dt ---")
|
||||||
|
print("Checking if A*T + B equals our derived dT/dt...")
|
||||||
|
|
||||||
|
for i in range(3):
|
||||||
|
diff = sm.simplify(reconstructed[i] - dT_dt_vector[i])
|
||||||
|
if diff == 0:
|
||||||
|
print(f" Row {i+1}: ✓ VERIFIED")
|
||||||
|
else:
|
||||||
|
print(f" Row {i+1}: ✗ ERROR - Difference: {diff}")
|
||||||
|
|
||||||
|
# Numerical example for mu = 0.5
|
||||||
|
print("\n" + "="*70)
|
||||||
|
print("Numerical Example: mu = 0.5")
|
||||||
|
print("="*70)
|
||||||
|
|
||||||
|
C_0_val = 33.33
|
||||||
|
tau_0_val = 25.00
|
||||||
|
W_val = C_0_val / (2 * tau_0_val)
|
||||||
|
mu_val = 0.5
|
||||||
|
|
||||||
|
C_r_val = mu_val * C_0_val
|
||||||
|
C_sg_val = (1 - mu_val) * C_0_val / 2
|
||||||
|
|
||||||
|
print(f"\nC_0 = {C_0_val:.2f} %-sec/°F")
|
||||||
|
print(f"tau_0 = {tau_0_val:.2f} sec")
|
||||||
|
print(f"W = {W_val:.4f} %-sec/°F")
|
||||||
|
print(f"C_r = {C_r_val:.2f} %-sec/°F")
|
||||||
|
print(f"C_sg = {C_sg_val:.2f} %-sec/°F")
|
||||||
|
|
||||||
|
# Substitute numerical values
|
||||||
|
A_numerical = A_symbolic.subs([(C_r, C_r_val), (C_sg, C_sg_val), (W, W_val)])
|
||||||
|
print("\nMatrix A (numerical for mu=0.5):")
|
||||||
|
sm.pprint(A_numerical)
|
||||||
|
print("\nAs float matrix:")
|
||||||
|
A_float = np.array(A_numerical).astype(float)
|
||||||
|
print(A_float)
|
||||||
|
|
||||||
|
print("\n--- Part B: Steady State Analysis ---")
|
||||||
|
|
||||||
|
# At steady state, dT/dt = 0, so A*T + B = 0
|
||||||
|
# This means A*T = -B
|
||||||
|
|
||||||
|
print("\nAt steady state: A*T_ss + B = 0")
|
||||||
|
print("Solving for temperature differences...")
|
||||||
|
|
||||||
|
# From SG1: 0 = W*T_hot - W*T_cold1 - Qdot1
|
||||||
|
# Rearranging: W*(T_hot - T_cold1) = Qdot1
|
||||||
|
DeltaT1 = Qdot1 / W
|
||||||
|
print(f"\nFrom SG1: W*(T_hot - T_cold1) = Qdot1")
|
||||||
|
print(f"ΔT1 = T_hot - T_cold1 = {DeltaT1}")
|
||||||
|
|
||||||
|
# From SG2: 0 = W*T_hot - W*T_cold2 - Qdot2
|
||||||
|
# Rearranging: W*(T_hot - T_cold2) = Qdot2
|
||||||
|
DeltaT2 = Qdot2 / W
|
||||||
|
print(f"\nFrom SG2: W*(T_hot - T_cold2) = Qdot2")
|
||||||
|
print(f"ΔT2 = T_hot - T_cold2 = {DeltaT2}")
|
||||||
|
|
||||||
|
# From reactor: 0 = P_r - W*DeltaT1 - W*DeltaT2
|
||||||
|
power_balance = sm.simplify(P_r - W*DeltaT1 - W*DeltaT2)
|
||||||
|
print(f"\nPower balance check:")
|
||||||
|
print(f"P_r - W*ΔT1 - W*ΔT2 = {power_balance}")
|
||||||
|
print(f"Therefore: P_r = Qdot1 + Qdot2 ✓")
|
||||||
|
|
||||||
|
print("\n--- Part C: Average Temperature ---")
|
||||||
|
|
||||||
|
# Given formula
|
||||||
|
T_ave_given = T_hot/2 + (T_cold1 + T_cold2)/4
|
||||||
|
|
||||||
|
print(f"\nGiven: T_ave = T_hot/2 + (T_cold1 + T_cold2)/4")
|
||||||
|
print(f"T_ave = {T_ave_given}")
|
||||||
|
|
||||||
|
# Mass-weighted average
|
||||||
|
T_ave_mass_weighted = (C_r*T_hot + C_sg*T_cold1 + C_sg*T_cold2) / (C_r + 2*C_sg)
|
||||||
|
|
||||||
|
print(f"\nMass-weighted: T_ave = (C_r*T_hot + C_sg*T_cold1 + C_sg*T_cold2)/(C_r + 2*C_sg)")
|
||||||
|
|
||||||
|
# Substitute C_r = mu*C_0 and C_sg = (1-mu)*C_0/2
|
||||||
|
T_ave_mass_weighted_sub = T_ave_mass_weighted.subs([
|
||||||
|
(C_r, mu*C_0),
|
||||||
|
(C_sg, (1-mu)*C_0/2)
|
||||||
|
])
|
||||||
|
T_ave_mass_weighted_simplified = sm.simplify(T_ave_mass_weighted_sub)
|
||||||
|
|
||||||
|
print(f"\nSubstituting C_r = mu*C_0, C_sg = (1-mu)*C_0/2:")
|
||||||
|
print(f"T_ave = {T_ave_mass_weighted_simplified}")
|
||||||
|
|
||||||
|
# Check if they're equal
|
||||||
|
print(f"\nChecking if given formula equals mass-weighted formula...")
|
||||||
|
difference = sm.simplify(T_ave_given - T_ave_mass_weighted_simplified)
|
||||||
|
print(f"Difference: {difference}")
|
||||||
|
|
||||||
|
if difference == 0:
|
||||||
|
print("✓ VERIFIED: Both formulas are equivalent!")
|
||||||
|
else:
|
||||||
|
print("✗ WARNING: Formulas differ!")
|
||||||
|
print(f"Given formula gives: {T_ave_given}")
|
||||||
|
print(f"Mass-weighted gives: {T_ave_mass_weighted_simplified}")
|
||||||
|
|
||||||
|
# Time derivative of total energy
|
||||||
|
print("\n--- Energy Conservation ---")
|
||||||
|
total_energy = C_r*T_hot + C_sg*T_cold1 + C_sg*T_cold2
|
||||||
|
|
||||||
|
dE_dt = sm.diff(total_energy, t)
|
||||||
|
print(f"\nTotal system energy: E = C_r*T_hot + C_sg*T_cold1 + C_sg*T_cold2")
|
||||||
|
print(f"dE/dt = {dE_dt}")
|
||||||
|
|
||||||
|
# Substitute the differential equations
|
||||||
|
dE_dt_expanded = (C_r*dT_hot_dt + C_sg*dT_cold1_dt + C_sg*dT_cold2_dt)
|
||||||
|
dE_dt_simplified = sm.simplify(dE_dt_expanded)
|
||||||
|
|
||||||
|
print(f"\nSubstituting the differential equations:")
|
||||||
|
print(f"dE/dt = {dE_dt_simplified}")
|
||||||
|
print(f"\nTherefore: dE/dt = P_r - Qdot1 - Qdot2")
|
||||||
|
print(f"When P_r = Qdot1 + Qdot2 (balanced), dE/dt = 0 ✓")
|
||||||
|
|
||||||
|
print("\n" + "="*70)
|
||||||
|
print("Verification Complete!")
|
||||||
|
print("="*70)
|
||||||
|
print("\nSummary:")
|
||||||
|
print("✓ Matrix form correctly derived")
|
||||||
|
print("✓ Steady-state equations verified")
|
||||||
|
print("✓ Average temperature formula needs clarification")
|
||||||
|
print("✓ Energy conservation verified")
|
||||||
217
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem2.py
Normal file
217
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem2.py
Normal file
@ -0,0 +1,217 @@
|
|||||||
|
import numpy as np
|
||||||
|
|
||||||
|
# Two-group cross-section data
|
||||||
|
cross_sections = {
|
||||||
|
"fast": {
|
||||||
|
"D": 1.4, # Diffusion constant [cm]
|
||||||
|
"Sigma_a": 0.010, # Absorption [cm^-1]
|
||||||
|
"Sigma_s": 0.050, # Scattering from fast to thermal [cm^-1]
|
||||||
|
"nu_Sigma_f": 0.000, # Neutrons per fission times fission cross section [cm^-1]
|
||||||
|
"chi": 1, # Neutron group distribution (fission spectrum)
|
||||||
|
"v": 1.8e7, # Average group velocity [cm/sec]
|
||||||
|
},
|
||||||
|
"thermal": {
|
||||||
|
"D": 0.35, # Diffusion constant [cm]
|
||||||
|
"Sigma_a": 0.080, # Absorption [cm^-1]
|
||||||
|
"Sigma_s": 0.0, # Scattering from thermal to fast [cm^-1]
|
||||||
|
"nu_Sigma_f": 0.125, # Neutrons per fission times fission cross section [cm^-1]
|
||||||
|
"chi": 0, # Neutron group distribution (fission spectrum)
|
||||||
|
"v": 2.2e5, # Average group velocity [cm/sec]
|
||||||
|
},
|
||||||
|
"scattering": {
|
||||||
|
"fast_to_thermal": 0.050, # Sigma_s^(1->2) [cm^-1]
|
||||||
|
"thermal_to_fast": 0.0, # Sigma_s^(2->1) [cm^-1]
|
||||||
|
},
|
||||||
|
}
|
||||||
|
|
||||||
|
# Easy access shortcuts
|
||||||
|
D_fast = cross_sections["fast"]["D"]
|
||||||
|
D_thermal = cross_sections["thermal"]["D"]
|
||||||
|
Sigma_a_fast = cross_sections["fast"]["Sigma_a"]
|
||||||
|
Sigma_a_thermal = cross_sections["thermal"]["Sigma_a"]
|
||||||
|
Sigma_s_fast = cross_sections["fast"]["Sigma_s"] # Scattering from fast to thermal
|
||||||
|
Sigma_s_thermal = cross_sections["thermal"][
|
||||||
|
"Sigma_s"
|
||||||
|
] # Scattering from thermal to fast
|
||||||
|
nu_Sigma_f_fast = cross_sections["fast"]["nu_Sigma_f"]
|
||||||
|
nu_Sigma_f_thermal = cross_sections["thermal"]["nu_Sigma_f"]
|
||||||
|
chi_fast = cross_sections["fast"]["chi"]
|
||||||
|
chi_thermal = cross_sections["thermal"]["chi"]
|
||||||
|
v_fast = cross_sections["fast"]["v"]
|
||||||
|
v_thermal = cross_sections["thermal"]["v"]
|
||||||
|
|
||||||
|
print("Two-Group Cross-Section Data Loaded")
|
||||||
|
print(f"Fast Group: D={D_fast} cm, Sigma_a={Sigma_a_fast} cm^-1, v={v_fast:.2e} cm/s")
|
||||||
|
print(
|
||||||
|
f"Thermal Group: D={D_thermal} cm, Sigma_a={Sigma_a_thermal} cm^-1, v={v_thermal:.2e} cm/s"
|
||||||
|
)
|
||||||
|
|
||||||
|
# Problem 2A: Calculate k_inf using Four-Factor Formula
|
||||||
|
# k_inf = epsilon * p * f * eta
|
||||||
|
# where:
|
||||||
|
# epsilon = fast fission factor (neutrons from fast fission per thermal fission)
|
||||||
|
# p = resonance escape probability (fraction of fast neutrons reaching thermal)
|
||||||
|
# f = thermal utilization factor (thermal neutrons absorbed in fuel vs total)
|
||||||
|
# eta = reproduction factor (neutrons produced per thermal neutron absorbed in fuel)
|
||||||
|
|
||||||
|
# Fast fission factor: epsilon = 1 (no fast fissions since nu_Sigma_f_fast = 0)
|
||||||
|
epsilon = 1.0
|
||||||
|
|
||||||
|
# Resonance escape probability: fraction of fast neutrons that scatter to thermal
|
||||||
|
# p = Sigma_s_fast / (Sigma_a_fast + Sigma_s_fast)
|
||||||
|
p = Sigma_s_fast / (Sigma_a_fast + Sigma_s_fast)
|
||||||
|
|
||||||
|
# Thermal utilization factor: f = 1 (single-region, all absorptions in "fuel")
|
||||||
|
f = 1.0
|
||||||
|
|
||||||
|
# Reproduction factor: eta = nu*Sigma_f / Sigma_a (thermal group)
|
||||||
|
eta = nu_Sigma_f_thermal / Sigma_a_thermal
|
||||||
|
|
||||||
|
# Four-Factor Formula
|
||||||
|
k_inf = epsilon * p * f * eta
|
||||||
|
|
||||||
|
print(f"\n=== Problem 2A: k_infinity (Four-Factor Formula) ===")
|
||||||
|
print(f"epsilon (fast fission factor) = {epsilon:.4f}")
|
||||||
|
print(f"p (resonance escape probability) = {p:.4f}")
|
||||||
|
print(f"f (thermal utilization) = {f:.4f}")
|
||||||
|
print(f"eta (reproduction factor) = {eta:.4f}")
|
||||||
|
print(f"k_inf = epsilon * p * f * eta = {k_inf:.4f}")
|
||||||
|
|
||||||
|
# Problem 2B: Calculate diffusion lengths
|
||||||
|
import sympy as sm
|
||||||
|
|
||||||
|
# Define buckling as a symbol
|
||||||
|
B_squared = sm.Symbol("B^2", positive=True)
|
||||||
|
|
||||||
|
# Calculate diffusion lengths
|
||||||
|
# L^2_fast = D_fast / Sigma_total_fast
|
||||||
|
# where Sigma_total_fast = Sigma_a_fast + Sigma_s_fast (removal from fast group)
|
||||||
|
Sigma_total_fast = Sigma_a_fast + Sigma_s_fast
|
||||||
|
L_squared_fast = D_fast / Sigma_total_fast
|
||||||
|
|
||||||
|
# L^2_th = D_th / Sigma_a_th
|
||||||
|
L_squared_th = D_thermal / Sigma_a_thermal
|
||||||
|
|
||||||
|
print(f"\n=== Problem 2B: Diffusion Lengths ===")
|
||||||
|
print(f"L_fast = {np.sqrt(L_squared_fast):.3f} cm")
|
||||||
|
print(f"L_thermal = {np.sqrt(L_squared_th):.3f} cm")
|
||||||
|
print(f"\nL^2_fast = {L_squared_fast:.3f} cm^2")
|
||||||
|
print(f"L^2_thermal = {L_squared_th:.3f} cm^2")
|
||||||
|
|
||||||
|
# k_eff equation as function of buckling (for reference):
|
||||||
|
# k_eff = k_inf / [(L^2_fast * B^2 + 1)(L^2_th * B^2 + 1)]
|
||||||
|
# At criticality: k_eff = 1
|
||||||
|
k_eff_expr = k_inf / ((L_squared_fast * B_squared + 1) * (L_squared_th * B_squared + 1))
|
||||||
|
|
||||||
|
print(f"\nCriticality relation:")
|
||||||
|
print(f"k_eff = k_inf / [(L^2_fast * B^2 + 1)(L^2_thermal * B^2 + 1)]")
|
||||||
|
print(
|
||||||
|
f"k_eff = {k_inf:.4f} / [({L_squared_fast:.3f}*B^2 + 1)({L_squared_th:.3f}*B^2 + 1)]"
|
||||||
|
)
|
||||||
|
|
||||||
|
# Problem 2C: Geometric buckling for rectangular solid
|
||||||
|
print(f"\n=== Problem 2C: Geometric Buckling ===")
|
||||||
|
print(f"For a rectangular solid with dimensions L_x, L_y, L_z")
|
||||||
|
print(f"with flux going to zero at the boundaries (bare reactor):")
|
||||||
|
print(f"\nB^2 = (pi/L_x)^2 + (pi/L_y)^2 + (pi/L_z)^2")
|
||||||
|
print(f"\nThis is the geometric buckling for the fundamental mode.")
|
||||||
|
|
||||||
|
# Example calculation with symbolic dimensions
|
||||||
|
L_x, L_y, L_z = sm.symbols("L_x L_y L_z", positive=True)
|
||||||
|
B_squared_geometric = (sm.pi / L_x) ** 2 + (sm.pi / L_y) ** 2 + (sm.pi / L_z) ** 2
|
||||||
|
print(f"\nSymbolic expression:")
|
||||||
|
print(f"B^2 = {B_squared_geometric}")
|
||||||
|
|
||||||
|
# Problem 2D: Critical height of trough
|
||||||
|
print(f"\n=== Problem 2D: Critical Height of Trough ===")
|
||||||
|
print(f"Given: Width L_x = 150 cm, Length L_y = 200 cm")
|
||||||
|
print(f"Find: Height L_z for criticality (k_eff = 1)")
|
||||||
|
|
||||||
|
# Given dimensions
|
||||||
|
L_x_val = 150 # cm
|
||||||
|
L_y_val = 200 # cm
|
||||||
|
|
||||||
|
# At criticality: k_eff = 1 = k_inf / [(L²_fast * B² + 1)(L²_thermal * B² + 1)]
|
||||||
|
# Rearranging: (L²_fast * B² + 1)(L²_thermal * B² + 1) = k_inf
|
||||||
|
# where: B² = (π/L_x)² + (π/L_y)² + (π/L_z)²
|
||||||
|
|
||||||
|
# Define L_z as unknown
|
||||||
|
L_z_sym = sm.Symbol("L_z", positive=True)
|
||||||
|
|
||||||
|
# Buckling with known dimensions and unknown height
|
||||||
|
B_sq = (sm.pi / L_x_val) ** 2 + (sm.pi / L_y_val) ** 2 + (sm.pi / L_z_sym) ** 2
|
||||||
|
|
||||||
|
# Criticality equation: k_inf = (L²_fast * B² + 1)(L²_thermal * B² + 1)
|
||||||
|
criticality_eq = (L_squared_fast * B_sq + 1) * (L_squared_th * B_sq + 1) - k_inf
|
||||||
|
|
||||||
|
# Solve for L_z
|
||||||
|
L_z_solutions = sm.solve(criticality_eq, L_z_sym)
|
||||||
|
|
||||||
|
# Filter for positive real solutions
|
||||||
|
L_z_critical = None
|
||||||
|
for sol in L_z_solutions:
|
||||||
|
if sol.is_real and sol > 0:
|
||||||
|
L_z_critical = float(sol)
|
||||||
|
break
|
||||||
|
|
||||||
|
if L_z_critical:
|
||||||
|
print(f"\nCritical height L_z = {L_z_critical:.2f} cm")
|
||||||
|
|
||||||
|
# Verify by calculating k_eff
|
||||||
|
B_sq_check = (
|
||||||
|
(np.pi / L_x_val) ** 2 + (np.pi / L_y_val) ** 2 + (np.pi / L_z_critical) ** 2
|
||||||
|
)
|
||||||
|
k_eff_check = k_inf / (
|
||||||
|
(L_squared_fast * B_sq_check + 1) * (L_squared_th * B_sq_check + 1)
|
||||||
|
)
|
||||||
|
print(f"\nVerification: k_eff = {k_eff_check:.6f} (should be 1.0)")
|
||||||
|
print(f"Geometric buckling B^2 = {B_sq_check:.6f} cm^-2")
|
||||||
|
else:
|
||||||
|
print("No positive real solution found!")
|
||||||
|
|
||||||
|
# Problem 2E: Prompt criticality
|
||||||
|
print(f"\n=== Problem 2E: Prompt Critical Height ===")
|
||||||
|
BETA = 640e-5 # Delayed neutron fraction
|
||||||
|
print(f"Beta (delayed neutron fraction) = {BETA:.6f}")
|
||||||
|
|
||||||
|
# Prompt criticality occurs when k_eff = 1/(1-beta)
|
||||||
|
k_eff_prompt = 1 / (1 - BETA)
|
||||||
|
print(f"Prompt critical k_eff = 1/(1-beta) = {k_eff_prompt:.6f}")
|
||||||
|
|
||||||
|
# Solve for height at prompt criticality
|
||||||
|
# k_eff_prompt = k_inf / [(L²_fast * B² + 1)(L²_thermal * B² + 1)]
|
||||||
|
# Rearranging: (L²_fast * B² + 1)(L²_thermal * B² + 1) = k_inf / k_eff_prompt
|
||||||
|
|
||||||
|
L_z_prompt_sym = sm.Symbol("L_z_prompt", positive=True)
|
||||||
|
B_sq_prompt = (
|
||||||
|
(sm.pi / L_x_val) ** 2 + (sm.pi / L_y_val) ** 2 + (sm.pi / L_z_prompt_sym) ** 2
|
||||||
|
)
|
||||||
|
|
||||||
|
prompt_crit_eq = (L_squared_fast * B_sq_prompt + 1) * (
|
||||||
|
L_squared_th * B_sq_prompt + 1
|
||||||
|
) - k_inf / k_eff_prompt
|
||||||
|
|
||||||
|
L_z_prompt_solutions = sm.solve(prompt_crit_eq, L_z_prompt_sym)
|
||||||
|
|
||||||
|
# Filter for positive real solutions
|
||||||
|
L_z_prompt = None
|
||||||
|
for sol in L_z_prompt_solutions:
|
||||||
|
if sol.is_real and sol > 0:
|
||||||
|
L_z_prompt = float(sol)
|
||||||
|
break
|
||||||
|
|
||||||
|
if L_z_prompt:
|
||||||
|
print(f"\nPrompt critical height L_z = {L_z_prompt:.2f} cm")
|
||||||
|
|
||||||
|
# Verify
|
||||||
|
B_sq_prompt_check = (
|
||||||
|
(np.pi / L_x_val) ** 2 + (np.pi / L_y_val) ** 2 + (np.pi / L_z_prompt) ** 2
|
||||||
|
)
|
||||||
|
k_eff_prompt_check = k_inf / (
|
||||||
|
(L_squared_fast * B_sq_prompt_check + 1)
|
||||||
|
* (L_squared_th * B_sq_prompt_check + 1)
|
||||||
|
)
|
||||||
|
print(f"\nVerification: k_eff = {k_eff_prompt_check:.6f}")
|
||||||
|
print(f"Difference from delayed critical: {L_z_prompt - L_z_critical:.2f} cm")
|
||||||
|
else:
|
||||||
|
print("No positive real solution found!")
|
||||||
63
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem3.py
Normal file
63
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem3.py
Normal file
@ -0,0 +1,63 @@
|
|||||||
|
import numpy as np
|
||||||
|
import sympy as sm
|
||||||
|
|
||||||
|
# Problem 3A
|
||||||
|
## Using formulas from Fundamental Kinetics Ideas R17 Page 51
|
||||||
|
DRW = 10 # pcm/step
|
||||||
|
STEPS = 8
|
||||||
|
LAMBDA_EFF = 0.1 # hz
|
||||||
|
|
||||||
|
# ASSUMING AFTER ROD PULL COMPLETE, RHO_DOT = 0
|
||||||
|
RHO_DOT = 0
|
||||||
|
BETA = 640 # pcm
|
||||||
|
|
||||||
|
# FIND RHO AFTER ROD PULL
|
||||||
|
rho = DRW * STEPS # pcm
|
||||||
|
|
||||||
|
sur = 26.06 * (RHO_DOT + LAMBDA_EFF * rho) / (BETA - rho)
|
||||||
|
|
||||||
|
print(f"The Start Up Rate is: {sur:.3f}")
|
||||||
|
|
||||||
|
# Problem 3C
|
||||||
|
# rho_net = rho_T + rho_rods + rho_poison + rho_fuel + ...
|
||||||
|
# rho_net = alpha_w * (4 degrees) + rho (from above) + 0 + alpha_f * 2.5
|
||||||
|
|
||||||
|
D_POWER = 2.5 # %
|
||||||
|
D_T_AVG = 4 # degrees
|
||||||
|
HEAT_UP_RATE = 0.15 # F/s
|
||||||
|
ALPHA_F = -10 # pcm/%power (negative feedback)
|
||||||
|
|
||||||
|
rho_rod = rho
|
||||||
|
|
||||||
|
# The heat up rate introduces a rho_dot, so SUR becomes 0 at the peak power.
|
||||||
|
alpha_w_sym = sm.Symbol("alpha_w")
|
||||||
|
rho_dot = alpha_w_sym * HEAT_UP_RATE
|
||||||
|
rho_net = alpha_w_sym * D_T_AVG + rho_rod + ALPHA_F * D_POWER
|
||||||
|
|
||||||
|
# At peak power, SUR = 0, which means: rho_dot + lambda_eff * rho_net = 0
|
||||||
|
# (the numerator must be zero)
|
||||||
|
equation = rho_dot + LAMBDA_EFF * rho_net
|
||||||
|
|
||||||
|
# Solve for alpha_w
|
||||||
|
alpha_w_solution = sm.solve(equation, alpha_w_sym)[0]
|
||||||
|
alpha_w = float(alpha_w_solution)
|
||||||
|
|
||||||
|
print(f"\nThe water temperature reactivity coefficient is: {alpha_w:.3f} pcm/°F")
|
||||||
|
|
||||||
|
# Problem 3D
|
||||||
|
# At final equilibrium: temperature stops changing (rho_dot = 0) and rho_net = 0
|
||||||
|
# This means: alpha_w * T_final + rho_rod + alpha_f * P_final = 0
|
||||||
|
#
|
||||||
|
# However, without knowing the heat removal characteristics (relationship between
|
||||||
|
# power and temperature at equilibrium), we cannot solve for exact values.
|
||||||
|
#
|
||||||
|
# Qualitative analysis:
|
||||||
|
# - At peak (4°F, 2.5%): SUR = 0 but temperature still rising
|
||||||
|
# - After peak: Temperature continues to rise → more negative reactivity → power decreases
|
||||||
|
# - At final equilibrium: Temperature plateaus at T_final > 4°F, Power at P_final < 2.5%
|
||||||
|
|
||||||
|
print(f"\nPart D - Qualitative Answer:")
|
||||||
|
print(f"At peak power: ΔT = {D_T_AVG}°F, ΔP = {D_POWER}%")
|
||||||
|
print(f"At final equilibrium:")
|
||||||
|
print(f" - Temperature: T_final > {D_T_AVG}°F (continues rising after peak)")
|
||||||
|
print(f" - Power: P_final < {D_POWER}% (decreases from peak as T rises further)")
|
||||||
99
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem4.py
Normal file
99
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem4.py
Normal file
@ -0,0 +1,99 @@
|
|||||||
|
import numpy as np
|
||||||
|
import matplotlib.pyplot as plt
|
||||||
|
|
||||||
|
# Problem 4: PWR Power Increase Transient (25% → 50% Steam Load)
|
||||||
|
|
||||||
|
print("=" * 70)
|
||||||
|
print("Problem 4: PWR Power Increase Transient")
|
||||||
|
print("=" * 70)
|
||||||
|
|
||||||
|
# Given Initial Conditions (25% power steady state)
|
||||||
|
T_ave_initial = 500 # °F - Average temperature
|
||||||
|
T_h_initial = 510 # °F - Hot leg temperature
|
||||||
|
T_c_initial = 490 # °F - Cold leg temperature
|
||||||
|
P_initial = 25 # % - Initial power level
|
||||||
|
P_final = 50 # % - Final power level
|
||||||
|
|
||||||
|
print("\n--- Initial Conditions (25% Power) ---")
|
||||||
|
print(f"T_ave = {T_ave_initial} °F")
|
||||||
|
print(f"T_h = {T_h_initial} °F")
|
||||||
|
print(f"T_c = {T_c_initial} °F")
|
||||||
|
print(f"Power = {P_initial}%")
|
||||||
|
|
||||||
|
# Calculate initial ΔT across core
|
||||||
|
DeltaT_initial = T_h_initial - T_c_initial
|
||||||
|
print(f"ΔT across core = {DeltaT_initial} °F")
|
||||||
|
|
||||||
|
print("\n" + "=" * 70)
|
||||||
|
print("Part A: Chronological Physical Impacts")
|
||||||
|
print("=" * 70)
|
||||||
|
|
||||||
|
print("\n" + "=" * 70)
|
||||||
|
print("Part B: Final State Comparison")
|
||||||
|
print("=" * 70)
|
||||||
|
|
||||||
|
# For a PWR at steady state:
|
||||||
|
# Power ∝ mass flow rate × Cp × ΔT
|
||||||
|
# If flow rate is constant (no pump speed change):
|
||||||
|
# P_new / P_old = ΔT_new / ΔT_old
|
||||||
|
|
||||||
|
# Calculate new ΔT
|
||||||
|
DeltaT_final = DeltaT_initial * (P_final / P_initial)
|
||||||
|
print(f"\nΔT calculation:")
|
||||||
|
print(f"ΔT_initial = {DeltaT_initial} °F at {P_initial}% power")
|
||||||
|
print(f"ΔT_final = ΔT_initial × (P_final/P_initial)")
|
||||||
|
print(f"ΔT_final = {DeltaT_initial} × ({P_final}/{P_initial}) = {DeltaT_final} °F")
|
||||||
|
|
||||||
|
# Estimate final temperatures
|
||||||
|
# Assume T_ave returns to approximately initial value (no automatic control,
|
||||||
|
# but moderator feedback will drive it back toward equilibrium)
|
||||||
|
# This is a simplification - in reality, T_ave might be slightly different
|
||||||
|
|
||||||
|
# Method 1: Assume T_ave = constant (strong moderator feedback)
|
||||||
|
T_ave_final_method1 = T_ave_initial
|
||||||
|
T_h_final_method1 = T_ave_final_method1 + DeltaT_final / 2
|
||||||
|
T_c_final_method1 = T_ave_final_method1 - DeltaT_final / 2
|
||||||
|
|
||||||
|
print(f"\n--- Method 1: Assume T_ave constant (strong moderator feedback) ---")
|
||||||
|
print(f"T_ave_final = {T_ave_final_method1} °F")
|
||||||
|
print(
|
||||||
|
f"T_h_final = T_ave + ΔT/2 = {T_ave_final_method1} + {DeltaT_final/2} = {T_h_final_method1} °F"
|
||||||
|
)
|
||||||
|
print(
|
||||||
|
f"T_c_final = T_ave - ΔT/2 = {T_ave_final_method1} - {DeltaT_final/2} = {T_c_final_method1} °F"
|
||||||
|
)
|
||||||
|
|
||||||
|
# Method 2: More realistic - Tc drops somewhat, Th rises more
|
||||||
|
# The SG cooling will lower Tc, but not back to initial Tc
|
||||||
|
# Let's estimate Tc drops by ~5-10°F, then Th rises to accommodate ΔT
|
||||||
|
T_c_drop = 5 # °F - estimated drop in Tc due to increased SG cooling
|
||||||
|
T_c_final_method2 = T_c_initial - T_c_drop
|
||||||
|
T_h_final_method2 = T_c_final_method2 + DeltaT_final
|
||||||
|
T_ave_final_method2 = (T_h_final_method2 + T_c_final_method2) / 2
|
||||||
|
|
||||||
|
print(f"\n--- Method 2: Estimate Tc drop due to SG cooling ---")
|
||||||
|
print(f"Assume Tc drops by ~{T_c_drop}°F due to increased steam flow")
|
||||||
|
print(f"T_c_final = {T_c_final_method2} °F")
|
||||||
|
print(
|
||||||
|
f"T_h_final = T_c + ΔT = {T_c_final_method2} + {DeltaT_final} = {T_h_final_method2} °F"
|
||||||
|
)
|
||||||
|
print(f"T_ave_final = {T_ave_final_method2} °F")
|
||||||
|
|
||||||
|
print(f"\n--- Comparison Table ---")
|
||||||
|
print(
|
||||||
|
f"{'Parameter':<20} {'Initial (25%)':<20} {'Final (50%) M1':<20} {'Final (50%) M2':<20}"
|
||||||
|
)
|
||||||
|
print(f"{'-'*80}")
|
||||||
|
print(f"{'Power':<20} {P_initial:<20} {P_final:<20} {P_final:<20}")
|
||||||
|
print(
|
||||||
|
f"{'T_h (°F)':<20} {T_h_initial:<20} {T_h_final_method1:<20.1f} {T_h_final_method2:<20.1f}"
|
||||||
|
)
|
||||||
|
print(
|
||||||
|
f"{'T_c (°F)':<20} {T_c_initial:<20} {T_c_final_method1:<20.1f} {T_c_final_method2:<20.1f}"
|
||||||
|
)
|
||||||
|
print(
|
||||||
|
f"{'T_ave (°F)':<20} {T_ave_initial:<20} {T_ave_final_method1:<20.1f} {T_ave_final_method2:<20.1f}"
|
||||||
|
)
|
||||||
|
print(
|
||||||
|
f"{'ΔT (°F)':<20} {DeltaT_initial:<20} {DeltaT_final:<20.1f} {DeltaT_final:<20.1f}"
|
||||||
|
)
|
||||||
187
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem5.py
Normal file
187
Class_Work/nuce2101/exam2/NUCE2101_SABO/problem5.py
Normal file
@ -0,0 +1,187 @@
|
|||||||
|
import numpy as np
|
||||||
|
import matplotlib.pyplot as plt
|
||||||
|
from scipy.integrate import odeint
|
||||||
|
|
||||||
|
# Problem 5: Xenon-135 Transient Analysis
|
||||||
|
|
||||||
|
# Given parameters
|
||||||
|
gamma_I = 0.057 # I-135 fission yield (5.7%)
|
||||||
|
gamma_Xe = 0.003 # Xe-135 fission yield (0.3%)
|
||||||
|
lambda_I = 2.87e-5 # I-135 decay constant [sec^-1] (t_1/2 = 6.7 hr)
|
||||||
|
lambda_Xe = 2.09e-5 # Xe-135 decay constant [sec^-1] (t_1/2 = 9.2 hr)
|
||||||
|
R_max = 7.34e-5 # Full power burnout factor [sec^-1]
|
||||||
|
K = 4.56 # Power constant [pcm*sec^-1]
|
||||||
|
Xe_eq_reactivity = -2900 # Full power equilibrium Xe reactivity [pcm]
|
||||||
|
|
||||||
|
print("=== Problem 5: Xenon-135 Transient ===")
|
||||||
|
print(f"\nGiven Parameters:")
|
||||||
|
print(f"gamma_I = {gamma_I} (I-135 fission yield)")
|
||||||
|
print(f"gamma_Xe = {gamma_Xe} (Xe-135 fission yield)")
|
||||||
|
print(f"lambda_I = {lambda_I:.2e} sec^-1 (t_1/2 = 6.7 hr)")
|
||||||
|
print(f"lambda_Xe = {lambda_Xe:.2e} sec^-1 (t_1/2 = 9.2 hr)")
|
||||||
|
print(f"R_max = {R_max:.2e} sec^-1 (full power burnout)")
|
||||||
|
print(f"K = {K} pcm*sec^-1")
|
||||||
|
print(f"Initial Xe reactivity = {Xe_eq_reactivity} pcm (at 100% power)")
|
||||||
|
|
||||||
|
# Power history (normalized to 1.0 = 100% power)
|
||||||
|
# Time ranges in hours
|
||||||
|
power_history = [
|
||||||
|
(0, 5, 1.0), # 0-5 hr: 100% power
|
||||||
|
(5, 15, 0.0), # 5-15 hr: Shutdown
|
||||||
|
(15, 50, 1.0), # 15-50 hr: 100% power
|
||||||
|
(50, 80, 0.4), # 50-80 hr: 40% power
|
||||||
|
(80, 100, 0.0), # 80-100 hr: Shutdown
|
||||||
|
(100, 150, 1.0), # 100-150 hr: 100% power
|
||||||
|
]
|
||||||
|
|
||||||
|
def get_power(t_hours):
|
||||||
|
"""Return normalized power at time t (in hours)"""
|
||||||
|
for t_start, t_end, power in power_history:
|
||||||
|
if t_start <= t_hours < t_end:
|
||||||
|
return power
|
||||||
|
return 0.0 # Default to shutdown
|
||||||
|
|
||||||
|
# Calculate equilibrium I and Xe at full power
|
||||||
|
# At equilibrium: dI/dt = 0, dX/dt = 0, rho = 1.0
|
||||||
|
# I_eq = gamma_I * Sigma_f * phi / lambda_I
|
||||||
|
# X_eq = (lambda_I * I_eq + gamma_Xe * Sigma_f * phi) / (lambda_Xe + sigma_a^Xe * phi)
|
||||||
|
|
||||||
|
# From the reactivity equation and given parameters, we can compute equilibrium
|
||||||
|
# Xe_reactivity = -K * X / rho
|
||||||
|
# At full power equilibrium: -2900 = -K * X_eq / 1.0
|
||||||
|
# X_eq = 2900 / K (in relative units)
|
||||||
|
|
||||||
|
# But we need to be careful about units. Let's work in terms of production rates.
|
||||||
|
# Production rate at full power: P_I = gamma_I * (production source)
|
||||||
|
# At equilibrium: I_eq = P_I / lambda_I
|
||||||
|
# At equilibrium: X_eq = (lambda_I * I_eq + gamma_Xe * P_source) / (lambda_Xe + R_max)
|
||||||
|
|
||||||
|
# Let's define production in terms that give correct equilibrium
|
||||||
|
# From B = rho*K*[gamma_I, gamma_Xe], production scales with power
|
||||||
|
|
||||||
|
# For now, let's compute initial conditions at full power equilibrium
|
||||||
|
rho_initial = 1.0 # Full power
|
||||||
|
|
||||||
|
# At equilibrium with rho=1:
|
||||||
|
# dI/dt = 0 = -lambda_I * I + rho * gamma_I * P0
|
||||||
|
# dX/dt = 0 = -lambda_Xe * X - rho * R_max * X + lambda_I * I + rho * gamma_Xe * P0
|
||||||
|
|
||||||
|
# Let P0 be a production constant we need to determine
|
||||||
|
# From equilibrium: I_eq = rho * gamma_I * P0 / lambda_I
|
||||||
|
# X_eq = (lambda_I * I_eq + rho * gamma_Xe * P0) / (lambda_Xe + rho * R_max)
|
||||||
|
|
||||||
|
# We know Xe_reactivity = -K * X at full power
|
||||||
|
# So X_eq = 2900 / K at rho = 1
|
||||||
|
|
||||||
|
X_eq_fullpower = abs(Xe_eq_reactivity) / K # Equilibrium Xe "concentration" at full power
|
||||||
|
print(f"\nEquilibrium Xe concentration (full power) = {X_eq_fullpower:.2f} [arbitrary units]")
|
||||||
|
|
||||||
|
# From equilibrium equation at full power (rho=1):
|
||||||
|
# X_eq = (lambda_I * I_eq + gamma_Xe * P0) / (lambda_Xe + R_max)
|
||||||
|
# And: I_eq = gamma_I * P0 / lambda_I
|
||||||
|
|
||||||
|
# Substituting:
|
||||||
|
# X_eq = (lambda_I * gamma_I * P0 / lambda_I + gamma_Xe * P0) / (lambda_Xe + R_max)
|
||||||
|
# X_eq = P0 * (gamma_I + gamma_Xe) / (lambda_Xe + R_max)
|
||||||
|
|
||||||
|
# Solving for P0:
|
||||||
|
P0 = X_eq_fullpower * (lambda_Xe + R_max) / (gamma_I + gamma_Xe)
|
||||||
|
I_eq_fullpower = gamma_I * P0 / lambda_I
|
||||||
|
|
||||||
|
print(f"Equilibrium I-135 concentration (full power) = {I_eq_fullpower:.2f} [arbitrary units]")
|
||||||
|
print(f"Production constant P0 = {P0:.2e}")
|
||||||
|
|
||||||
|
# Define ODE system for I-135 and Xe-135
|
||||||
|
def xenon_ode(y, t, power_func):
|
||||||
|
"""
|
||||||
|
ODE system for I-135 and Xe-135
|
||||||
|
y = [I, X] where I is I-135, X is Xe-135
|
||||||
|
t is time in seconds
|
||||||
|
"""
|
||||||
|
I, X = y
|
||||||
|
t_hours = t / 3600 # Convert to hours for power lookup
|
||||||
|
rho = power_func(t_hours)
|
||||||
|
|
||||||
|
# dI/dt = -lambda_I * I + rho * gamma_I * P0
|
||||||
|
dI_dt = -lambda_I * I + rho * gamma_I * P0
|
||||||
|
|
||||||
|
# dX/dt = -lambda_Xe * X - rho * R_max * X + lambda_I * I + rho * gamma_Xe * P0
|
||||||
|
dX_dt = -lambda_Xe * X - rho * R_max * X + lambda_I * I + rho * gamma_Xe * P0
|
||||||
|
|
||||||
|
return [dI_dt, dX_dt]
|
||||||
|
|
||||||
|
# Initial conditions at t=0 (full power equilibrium)
|
||||||
|
I0 = I_eq_fullpower
|
||||||
|
X0 = X_eq_fullpower
|
||||||
|
y0 = [I0, X0]
|
||||||
|
|
||||||
|
print(f"\nInitial conditions at t=0 (full power equilibrium):")
|
||||||
|
print(f"I-135 = {I0:.2f}")
|
||||||
|
print(f"Xe-135 = {X0:.2f}")
|
||||||
|
print(f"Xe reactivity = {-K * X0:.1f} pcm")
|
||||||
|
|
||||||
|
# Part A: Solve and sketch the xenon transient
|
||||||
|
print(f"\n=== Part A: Xenon Transient Sketch ===")
|
||||||
|
|
||||||
|
# Time array: 0 to 150 hours
|
||||||
|
t_hours = np.linspace(0, 150, 2000)
|
||||||
|
t_seconds = t_hours * 3600
|
||||||
|
|
||||||
|
# Solve ODE
|
||||||
|
solution = odeint(xenon_ode, y0, t_seconds, args=(get_power,))
|
||||||
|
I_transient = solution[:, 0]
|
||||||
|
X_transient = solution[:, 1]
|
||||||
|
|
||||||
|
# Calculate xenon reactivity
|
||||||
|
Xe_reactivity_transient = -K * X_transient
|
||||||
|
|
||||||
|
# Plot results
|
||||||
|
fig, (ax1, ax2, ax3) = plt.subplots(3, 1, figsize=(12, 10))
|
||||||
|
|
||||||
|
# Plot power history
|
||||||
|
power_values = [get_power(t) for t in t_hours]
|
||||||
|
ax1.plot(t_hours, power_values, 'b-', linewidth=2)
|
||||||
|
ax1.set_ylabel('Normalized Power', fontsize=12)
|
||||||
|
ax1.set_title('Power History', fontsize=14, fontweight='bold')
|
||||||
|
ax1.grid(True, alpha=0.3)
|
||||||
|
ax1.set_ylim(-0.1, 1.2)
|
||||||
|
|
||||||
|
# Plot I-135
|
||||||
|
ax2.plot(t_hours, I_transient, 'g-', linewidth=2, label='I-135')
|
||||||
|
ax2.set_ylabel('I-135 Concentration', fontsize=12)
|
||||||
|
ax2.set_title('I-135 Transient', fontsize=14, fontweight='bold')
|
||||||
|
ax2.grid(True, alpha=0.3)
|
||||||
|
ax2.legend()
|
||||||
|
|
||||||
|
# Plot Xe reactivity
|
||||||
|
ax3.plot(t_hours, Xe_reactivity_transient, 'r-', linewidth=2)
|
||||||
|
ax3.axhline(y=-2900, color='k', linestyle='--', alpha=0.5, label='Initial Eq (-2900 pcm)')
|
||||||
|
ax3.set_xlabel('Time (hours)', fontsize=12)
|
||||||
|
ax3.set_ylabel('Xenon Reactivity (pcm)', fontsize=12)
|
||||||
|
ax3.set_title('Xenon Reactivity Transient', fontsize=14, fontweight='bold')
|
||||||
|
ax3.grid(True, alpha=0.3)
|
||||||
|
ax3.legend()
|
||||||
|
|
||||||
|
plt.tight_layout()
|
||||||
|
plt.savefig('problem5_xenon_transient.png', dpi=300, bbox_inches='tight')
|
||||||
|
print("Xenon transient plot saved as 'problem5_xenon_transient.png'")
|
||||||
|
plt.close()
|
||||||
|
|
||||||
|
# Part B: Find the peak xenon after first shutdown (at t=5 hours)
|
||||||
|
print(f"\n=== Part B: First Xenon Peak Analysis ===")
|
||||||
|
|
||||||
|
# Focus on the first shutdown period (5 to 15 hours)
|
||||||
|
mask = (t_hours >= 5) & (t_hours <= 15)
|
||||||
|
t_peak_range = t_hours[mask]
|
||||||
|
Xe_react_peak_range = Xe_reactivity_transient[mask]
|
||||||
|
|
||||||
|
# Find the peak (most negative reactivity)
|
||||||
|
peak_idx = np.argmin(Xe_react_peak_range)
|
||||||
|
t_peak = t_peak_range[peak_idx]
|
||||||
|
Xe_peak_reactivity = Xe_react_peak_range[peak_idx]
|
||||||
|
|
||||||
|
print(f"\nFirst shutdown: t = 5 hours to t = 15 hours")
|
||||||
|
print(f"Peak xenon occurs at t = {t_peak:.2f} hours")
|
||||||
|
print(f"Time after shutdown = {t_peak - 5:.2f} hours")
|
||||||
|
print(f"Peak xenon reactivity = {Xe_peak_reactivity:.1f} pcm")
|
||||||
|
print(f"\nChange from equilibrium: {Xe_peak_reactivity - Xe_eq_reactivity:.1f} pcm")
|
||||||
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Class_Work/nuce2101/exam2/NUCE2101_SABO_EXAM2.pdf
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@ -1,4 +1,4 @@
|
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\relax
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\relax
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This is pdfTeX, Version 3.141592653-2.6-1.40.25 (TeX Live 2023/Debian) (preloaded format=pdflatex 2024.9.11) 9 NOV 2025 16:05
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This is pdfTeX, Version 3.141592653-2.6-1.40.25 (TeX Live 2023/Debian) (preloaded format=pdflatex 2024.9.11) 10 NOV 2025 17:53
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{/var/lib/texmf/fonts/map/pdftex/updmap/pdftex.map}{/usr/share/texlive/texmf-dist/fonts/enc/dvips/base/8r.enc}] (./problem1.tex) (./problem2.tex (/usr/share/texlive/texmf-dist/tex/latex/listings/lstlang1.sty
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) [2] [3] [4] [5] [6] [7]) (./problem3.tex [8] [9] [10]) (./problem4.tex) (./problem5.tex [11]
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<../python/problem5_xenon_transient.png, id=78, 858.5676pt x 713.7867pt>
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./problem1.tex:16: Undefined control sequence.
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l.16 ...eat capacity: $C_0 = 33.33$ \%-sec/\degree
|
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F
|
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The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
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|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
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|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
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|
and I'll forget about whatever was undefined.
|
||||||
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./problem1.tex:18: Undefined control sequence.
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l.18 ...ot} = T_{cold1} = T_{cold2} = 450$ \degree
|
||||||
|
F
|
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|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem1.tex:25: Undefined control sequence.
|
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|
<argument> \%-sec/\degree
|
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|
F
|
||||||
|
l.25 ...s 25.00} = 0.6667 \text{ \%-sec/\degree F}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem1.tex:25: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \%-sec/\degree
|
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|
F
|
||||||
|
l.25 ...s 25.00} = 0.6667 \text{ \%-sec/\degree F}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem1.tex:25: Undefined control sequence.
|
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|
<argument> \firstchoice@false \%-sec/\degree
|
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|
F
|
||||||
|
l.25 ...s 25.00} = 0.6667 \text{ \%-sec/\degree F}
|
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|
\]
|
||||||
|
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|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem1.tex:25: Undefined control sequence.
|
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|
<argument> \firstchoice@false \%-sec/\degree
|
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|
F
|
||||||
|
l.25 ...s 25.00} = 0.6667 \text{ \%-sec/\degree F}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
(/usr/share/texlive/texmf-dist/tex/latex/listings/lstlang1.sty
|
||||||
|
File: lstlang1.sty 2023/02/27 1.9 listings language file
|
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|
) [1]
|
||||||
|
./problem1.tex:93: Undefined control sequence.
|
||||||
|
l.93 ...rac{C_0}{2\tau_0} = 0.6667$ \%-sec/\degree
|
||||||
|
F = flow heat capacity ra...
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
[2]
|
||||||
|
./problem1.tex:121: Undefined control sequence.
|
||||||
|
l.121 ...$\mu = 0.5$: $C_r = 16.66$ \%-sec/\degree
|
||||||
|
F, $C_{sg} = 8.33$ \%-sec...
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem1.tex:121: Undefined control sequence.
|
||||||
|
l.121 ...\degree F, $C_{sg} = 8.33$ \%-sec/\degree
|
||||||
|
F, $W = 0.6667$ \%-sec/\d...
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem1.tex:121: Undefined control sequence.
|
||||||
|
l.121 ...ec/\degree F, $W = 0.6667$ \%-sec/\degree
|
||||||
|
F
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
[3] [4]
|
||||||
|
Overfull \hbox (25.26381pt too wide) in paragraph at lines 275--276
|
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|
\OT1/ptm/m/n/12 Transient sim-u-la-tions were per-formed for four cases us-ing nu-mer-i-cal in-te-gra-tion (scipy.integrate.odeint):
|
||||||
|
[]
|
||||||
|
|
||||||
|
./problem1.tex:285: Undefined control sequence.
|
||||||
|
l.285 ...t} = T_{cold1} = T_{cold2} = 450$ \degree
|
||||||
|
F).
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
[5]
|
||||||
|
<../python/problem1_equal_loading.png, id=58, 1004.553pt x 711.8595pt>
|
||||||
|
File: ../python/problem1_equal_loading.png Graphic file (type png)
|
||||||
|
<use ../python/problem1_equal_loading.png>
|
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<../python/problem1_unequal_loading.png, id=60, 1004.553pt x 711.8595pt>
|
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|
||||||
|
<use ../python/problem1_unequal_loading.png>
|
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Package pdftex.def Info: ../python/problem1_unequal_loading.png used on input line 298.
|
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|
(pdftex.def) Requested size: 446.26582pt x 316.2388pt.
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|
[6 <../python/problem1_equal_loading.png>]) (./problem2.tex [7 <../python/problem1_unequal_loading.png>] [8] [9] [10] [11] [12] [13]) (./problem3.tex [14] [15] [16]) (./problem4.tex
|
||||||
|
./problem4.tex:9: Undefined control sequence.
|
||||||
|
l.9 \item $T_{ave} = 500$ \degree
|
||||||
|
F
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
[17]
|
||||||
|
./problem4.tex:10: Undefined control sequence.
|
||||||
|
l.10 \item $T_h = 510$ \degree
|
||||||
|
F (hot leg)
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:11: Undefined control sequence.
|
||||||
|
l.11 \item $T_c = 490$ \degree
|
||||||
|
F (cold leg)
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
[18] [19]
|
||||||
|
./problem4.tex:155: Undefined control sequence.
|
||||||
|
l.155 ...al} = T_h - T_c = 510 - 490 = 20$ \degree
|
||||||
|
F
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:159: Undefined control sequence.
|
||||||
|
<argument> \degree
|
||||||
|
F
|
||||||
|
l.159 ...ac{50}{25} = \boxed{40 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:159: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.159 ...ac{50}{25} = \boxed{40 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:159: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.159 ...ac{50}{25} = \boxed{40 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:159: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.159 ...ac{50}{25} = \boxed{40 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:162: Undefined control sequence.
|
||||||
|
<argument> \degree
|
||||||
|
F
|
||||||
|
l.162 ...{ave,final} \approx 500 \text{ \degree F}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:162: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.162 ...{ave,final} \approx 500 \text{ \degree F}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:162: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.162 ...{ave,final} \approx 500 \text{ \degree F}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:162: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.162 ...{ave,final} \approx 500 \text{ \degree F}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:164: Undefined control sequence.
|
||||||
|
<argument> \degree
|
||||||
|
F
|
||||||
|
l.164 ... 500 + 20 = \boxed{520 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:164: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.164 ... 500 + 20 = \boxed{520 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:164: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.164 ... 500 + 20 = \boxed{520 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:164: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.164 ... 500 + 20 = \boxed{520 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:166: Undefined control sequence.
|
||||||
|
<argument> \degree
|
||||||
|
F
|
||||||
|
l.166 ... 500 - 20 = \boxed{480 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:166: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.166 ... 500 - 20 = \boxed{480 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:166: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.166 ... 500 - 20 = \boxed{480 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:166: Undefined control sequence.
|
||||||
|
<argument> \firstchoice@false \degree
|
||||||
|
F
|
||||||
|
l.166 ... 500 - 20 = \boxed{480 \text{ \degree F}}
|
||||||
|
\]
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:176: Undefined control sequence.
|
||||||
|
l.176 $T_h$ (\degree
|
||||||
|
F) & 510 & 520 & $+10$ \degree F \\
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:176: Undefined control sequence.
|
||||||
|
l.176 ...$ (\degree F) & 510 & 520 & $+10$ \degree
|
||||||
|
F \\
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:177: Undefined control sequence.
|
||||||
|
l.177 $T_c$ (\degree
|
||||||
|
F) & 490 & 480 & $-10$ \degree F \\
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:177: Undefined control sequence.
|
||||||
|
l.177 ...$ (\degree F) & 490 & 480 & $-10$ \degree
|
||||||
|
F \\
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:178: Undefined control sequence.
|
||||||
|
l.178 $T_{ave}$ (\degree
|
||||||
|
F) & 500 & $\sim$500 & $\sim$0 \degree F \\
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:178: Undefined control sequence.
|
||||||
|
l.178 ...ee F) & 500 & $\sim$500 & $\sim$0 \degree
|
||||||
|
F \\
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:179: Undefined control sequence.
|
||||||
|
l.179 $\Delta T$ (\degree
|
||||||
|
F) & 20 & 40 & $+20$ \degree F \\
|
||||||
|
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|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
./problem4.tex:179: Undefined control sequence.
|
||||||
|
l.179 ... T$ (\degree F) & 20 & 40 & $+20$ \degree
|
||||||
|
F \\
|
||||||
|
The control sequence at the end of the top line
|
||||||
|
of your error message was never \def'ed. If you have
|
||||||
|
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||||
|
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||||
|
and I'll forget about whatever was undefined.
|
||||||
|
|
||||||
|
[20]) (./problem5.tex [21]
|
||||||
|
<../python/problem5_xenon_transient.png, id=115, 858.5676pt x 713.7867pt>
|
||||||
File: ../python/problem5_xenon_transient.png Graphic file (type png)
|
File: ../python/problem5_xenon_transient.png Graphic file (type png)
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||||||
<use ../python/problem5_xenon_transient.png>
|
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||||||
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(pdftex.def) Requested size: 446.26582pt x 371.00784pt.
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[12] [13 <../python/problem5_xenon_transient.png>] [14]) (./problem6.tex) [15] (./main.aux)
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[22 <../python/problem5_xenon_transient.png>] [23]) (./problem6.tex) [24] (./main.aux)
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LaTeX2e <2023-11-01> patch level 1
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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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L3 programming layer <2024-01-22>
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\section*{Problem 1}
|
||||||
|
|
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|
\subsection*{System Description}
|
||||||
|
|
||||||
|
A two-loop reactor system with:
|
||||||
|
\begin{itemize}
|
||||||
|
\item Reactor (lumped hot leg) at temperature $T_{hot}$
|
||||||
|
\item Steam Generator 1 with cold leg at $T_{cold1}$
|
||||||
|
\item Steam Generator 2 with cold leg at $T_{cold2}$
|
||||||
|
\item Equal mass flow rates and water masses in each loop
|
||||||
|
\item Reactor Water Mass Fraction (RWMF): $\mu$
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\textbf{Given parameters:}
|
||||||
|
\begin{itemize}
|
||||||
|
\item Base heat capacity: $C_0 = 33.33$ \%-sec/\degree F
|
||||||
|
\item Base time constant: $\tau_0 = 0.75 \times C_0 = 25.00$ sec
|
||||||
|
\item Initial temperature: $T_{hot} = T_{cold1} = T_{cold2} = 450$ \degree F
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\textbf{Derived parameter:}
|
||||||
|
|
||||||
|
The flow heat capacity rate per loop is \textit{not directly given} in the problem statement but can be derived from the relationship between time constants and heat capacities:
|
||||||
|
|
||||||
|
\[W = \frac{C_0}{2\tau_0} = \frac{33.33}{2 \times 25.00} = 0.6667 \text{ \%-sec/\degree F}\]
|
||||||
|
|
||||||
|
This represents the (mass flow rate $\times$ specific heat) for each loop.
|
||||||
|
|
||||||
|
\textbf{System parameters as functions of $\mu$:}
|
||||||
|
\begin{itemize}
|
||||||
|
\item Reactor time constant: $\tau_r = \mu \tau_0$
|
||||||
|
\item Reactor heat capacity: $C_r = \mu C_0$
|
||||||
|
\item Steam generator time constant (each): $\tau_{sg} = \frac{(1-\mu)\tau_r}{2}$
|
||||||
|
\item Steam generator heat capacity (each): $C_{sg} = \frac{(1-\mu)C_0}{2}$
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\subsection*{Part A}
|
||||||
|
|
||||||
|
\subsubsection*{Python Code}
|
||||||
|
\begin{lstlisting}[language=Python,
|
||||||
|
basicstyle=\ttfamily\small,
|
||||||
|
keywordstyle=\color{blue},
|
||||||
|
commentstyle=\color{gray},
|
||||||
|
stringstyle=\color{red},
|
||||||
|
showstringspaces=false,
|
||||||
|
numbers=left,
|
||||||
|
numberstyle=\tiny,
|
||||||
|
frame=single,
|
||||||
|
breaklines=true]
|
||||||
|
import numpy as np
|
||||||
|
import sympy as sm
|
||||||
|
|
||||||
|
# Given parameters
|
||||||
|
C_0 = 33.33 # Base heat capacity [%-sec/degF]
|
||||||
|
tau_0 = 25.00 # Base time constant [sec]
|
||||||
|
W = C_0 / (2 * tau_0) # Flow heat capacity rate per loop
|
||||||
|
|
||||||
|
def calculate_system_parameters(mu):
|
||||||
|
C_r = mu * C_0 # Reactor heat capacity
|
||||||
|
C_sg = (1 - mu) * C_0 / 2 # Steam generator heat capacity (each)
|
||||||
|
return C_r, C_sg
|
||||||
|
|
||||||
|
# Matrix form: dT/dt = A*T + B
|
||||||
|
def get_matrix_A(C_r, C_sg, W):
|
||||||
|
A = np.array([
|
||||||
|
[-2*W/C_r, W/C_r, W/C_r],
|
||||||
|
[W/C_sg, -W/C_sg, 0],
|
||||||
|
[W/C_sg, 0, -W/C_sg]
|
||||||
|
])
|
||||||
|
return A
|
||||||
|
|
||||||
|
# Vector B (forcing terms):
|
||||||
|
# B = [P_r/C_r, -Qdot1/C_sg, -Qdot2/C_sg]^T
|
||||||
|
\end{lstlisting}
|
||||||
|
|
||||||
|
\subsubsection*{Solution}
|
||||||
|
|
||||||
|
The energy balance for each component yields differential equations:
|
||||||
|
|
||||||
|
\textbf{Reactor energy balance:}
|
||||||
|
\[C_r \frac{dT_{hot}}{dt} = P_r - W(T_{hot} - T_{cold1}) - W(T_{hot} - T_{cold2})\]
|
||||||
|
|
||||||
|
\textbf{Steam Generator 1 energy balance:}
|
||||||
|
\[C_{sg} \frac{dT_{cold1}}{dt} = W(T_{hot} - T_{cold1}) - \dot{Q}_1\]
|
||||||
|
|
||||||
|
\textbf{Steam Generator 2 energy balance:}
|
||||||
|
\[C_{sg} \frac{dT_{cold2}}{dt} = W(T_{hot} - T_{cold2}) - \dot{Q}_2\]
|
||||||
|
|
||||||
|
where:
|
||||||
|
\begin{itemize}
|
||||||
|
\item $P_r$ = reactor power (positive for heat generation)
|
||||||
|
\item $\dot{Q}_1, \dot{Q}_2$ = steam generator heat removal rates (positive for heat removal)
|
||||||
|
\item $W = \frac{C_0}{2\tau_0} = 0.6667$ \%-sec/\degree F = flow heat capacity rate per loop
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\textbf{Matrix Form:}
|
||||||
|
|
||||||
|
Define the temperature vector: $\mathbf{T} = \begin{bmatrix} T_{hot} \\ T_{cold1} \\ T_{cold2} \end{bmatrix}$
|
||||||
|
|
||||||
|
The system can be written as:
|
||||||
|
\[\boxed{\frac{d\mathbf{T}}{dt} = \mathbf{A} \mathbf{T} + \mathbf{B}}\]
|
||||||
|
|
||||||
|
where the coefficient matrix $\mathbf{A}$ is:
|
||||||
|
|
||||||
|
\[\mathbf{A} = \begin{bmatrix}
|
||||||
|
-\frac{2W}{C_r} & \frac{W}{C_r} & \frac{W}{C_r} \\[0.3em]
|
||||||
|
\frac{W}{C_{sg}} & -\frac{W}{C_{sg}} & 0 \\[0.3em]
|
||||||
|
\frac{W}{C_{sg}} & 0 & -\frac{W}{C_{sg}}
|
||||||
|
\end{bmatrix}\]
|
||||||
|
|
||||||
|
and the forcing vector $\mathbf{B}$ is:
|
||||||
|
|
||||||
|
\[\mathbf{B} = \begin{bmatrix}
|
||||||
|
\frac{P_r}{C_r} \\[0.3em]
|
||||||
|
-\frac{\dot{Q}_1}{C_{sg}} \\[0.3em]
|
||||||
|
-\frac{\dot{Q}_2}{C_{sg}}
|
||||||
|
\end{bmatrix}\]
|
||||||
|
|
||||||
|
\textbf{Numerical example for $\mu = 0.5$:}
|
||||||
|
|
||||||
|
With $\mu = 0.5$: $C_r = 16.66$ \%-sec/\degree F, $C_{sg} = 8.33$ \%-sec/\degree F, $W = 0.6667$ \%-sec/\degree F
|
||||||
|
|
||||||
|
\[\boxed{\mathbf{A} = \begin{bmatrix}
|
||||||
|
-0.0800 & 0.0400 & 0.0400 \\
|
||||||
|
0.0800 & -0.0800 & 0 \\
|
||||||
|
0.0800 & 0 & -0.0800
|
||||||
|
\end{bmatrix}}\]
|
||||||
|
|
||||||
|
\subsection*{Part B}
|
||||||
|
|
||||||
|
\subsubsection*{Python Code}
|
||||||
|
\begin{lstlisting}[language=Python,
|
||||||
|
basicstyle=\ttfamily\small,
|
||||||
|
keywordstyle=\color{blue},
|
||||||
|
commentstyle=\color{gray},
|
||||||
|
stringstyle=\color{red},
|
||||||
|
showstringspaces=false,
|
||||||
|
numbers=left,
|
||||||
|
numberstyle=\tiny,
|
||||||
|
frame=single,
|
||||||
|
breaklines=true]
|
||||||
|
# At steady state, dT/dt = 0
|
||||||
|
# From SG1: 0 = W*(T_hot - T_cold1) - Qdot1
|
||||||
|
DeltaT1 = Qdot1 / W
|
||||||
|
|
||||||
|
# From SG2: 0 = W*(T_hot - T_cold2) - Qdot2
|
||||||
|
DeltaT2 = Qdot2 / W
|
||||||
|
|
||||||
|
# From reactor: P_r = W*DeltaT1 + W*DeltaT2
|
||||||
|
power_balance = Qdot1 + Qdot2
|
||||||
|
\end{lstlisting}
|
||||||
|
|
||||||
|
\subsubsection*{Solution}
|
||||||
|
|
||||||
|
At steady state, all time derivatives are zero ($\frac{dT}{dt} = 0$).
|
||||||
|
|
||||||
|
\textbf{From Steam Generator 1 equation:}
|
||||||
|
\[0 = W(T_{hot} - T_{cold1}) - \dot{Q}_1\]
|
||||||
|
\[\boxed{\Delta T_1 = T_{hot} - T_{cold1} = \frac{\dot{Q}_1}{W}}\]
|
||||||
|
|
||||||
|
\textbf{From Steam Generator 2 equation:}
|
||||||
|
\[0 = W(T_{hot} - T_{cold2}) - \dot{Q}_2\]
|
||||||
|
\[\boxed{\Delta T_2 = T_{hot} - T_{cold2} = \frac{\dot{Q}_2}{W}}\]
|
||||||
|
|
||||||
|
\textbf{From Reactor equation:}
|
||||||
|
\[0 = P_r - W\Delta T_1 - W\Delta T_2\]
|
||||||
|
\[P_r = W\Delta T_1 + W\Delta T_2 = \dot{Q}_1 + \dot{Q}_2\]
|
||||||
|
|
||||||
|
This confirms the power balance: reactor power equals total steam generator heat removal rates.
|
||||||
|
|
||||||
|
\subsection*{Part C}
|
||||||
|
|
||||||
|
\subsubsection*{Python Code}
|
||||||
|
\begin{lstlisting}[language=Python,
|
||||||
|
basicstyle=\ttfamily\small,
|
||||||
|
keywordstyle=\color{blue},
|
||||||
|
commentstyle=\color{gray},
|
||||||
|
stringstyle=\color{red},
|
||||||
|
showstringspaces=false,
|
||||||
|
numbers=left,
|
||||||
|
numberstyle=\tiny,
|
||||||
|
frame=single,
|
||||||
|
breaklines=true]
|
||||||
|
# Average reactor temperature (mass-weighted)
|
||||||
|
def calculate_T_ave(T_hot, T_cold1, T_cold2, mu, C_0):
|
||||||
|
C_r = mu * C_0
|
||||||
|
C_sg = (1 - mu) * C_0 / 2
|
||||||
|
|
||||||
|
# Mass-weighted average
|
||||||
|
T_ave = (C_r*T_hot + C_sg*T_cold1 + C_sg*T_cold2) / C_0
|
||||||
|
|
||||||
|
# Simplified form
|
||||||
|
T_ave_simplified = mu*T_hot + (1-mu)*(T_cold1 + T_cold2)/2
|
||||||
|
|
||||||
|
return T_ave_simplified
|
||||||
|
|
||||||
|
# For mu = 0.5:
|
||||||
|
T_ave_50 = 0.5*T_hot + 0.25*T_cold1 + 0.25*T_cold2
|
||||||
|
|
||||||
|
# For mu = 0.75:
|
||||||
|
T_ave_75 = 0.75*T_hot + 0.125*T_cold1 + 0.125*T_cold2
|
||||||
|
\end{lstlisting}
|
||||||
|
|
||||||
|
\subsubsection*{Solution}
|
||||||
|
|
||||||
|
The average reactor temperature must be calculated as a \textbf{mass-weighted average}:
|
||||||
|
|
||||||
|
\[T_{ave} = \frac{C_r T_{hot} + C_{sg} T_{cold1} + C_{sg} T_{cold2}}{C_r + 2C_{sg}}\]
|
||||||
|
|
||||||
|
Since $C_r + 2C_{sg} = \mu C_0 + 2 \cdot \frac{(1-\mu)C_0}{2} = C_0$, this simplifies to:
|
||||||
|
|
||||||
|
\[T_{ave} = \frac{C_r T_{hot} + C_{sg} T_{cold1} + C_{sg} T_{cold2}}{C_0}\]
|
||||||
|
|
||||||
|
Substituting $C_r = \mu C_0$ and $C_{sg} = \frac{(1-\mu)C_0}{2}$:
|
||||||
|
|
||||||
|
\[\boxed{T_{ave} = \mu T_{hot} + \frac{(1-\mu)}{2}(T_{cold1} + T_{cold2})}\]
|
||||||
|
|
||||||
|
\textbf{Important:} This formula \textit{depends on $\mu$}!
|
||||||
|
|
||||||
|
\textbf{For $\mu = 0.5$:}
|
||||||
|
\[T_{ave} = \frac{T_{hot}}{2} + \frac{T_{cold1} + T_{cold2}}{4}\]
|
||||||
|
|
||||||
|
\textbf{For $\mu = 0.75$:}
|
||||||
|
\[T_{ave} = \frac{3T_{hot}}{4} + \frac{T_{cold1} + T_{cold2}}{8}\]
|
||||||
|
|
||||||
|
\textbf{Note:} When the system is balanced ($P_r = \dot{Q}_1 + \dot{Q}_2$), the mass-weighted average temperature remains constant even during transients, since $\frac{d(C_0 \cdot T_{ave})}{dt} = P_r - \dot{Q}_1 - \dot{Q}_2 = 0$.
|
||||||
|
|
||||||
|
\subsection*{Part D}
|
||||||
|
|
||||||
|
\subsubsection*{Python Code}
|
||||||
|
\begin{lstlisting}[language=Python,
|
||||||
|
basicstyle=\ttfamily\small,
|
||||||
|
keywordstyle=\color{blue},
|
||||||
|
commentstyle=\color{gray},
|
||||||
|
stringstyle=\color{red},
|
||||||
|
showstringspaces=false,
|
||||||
|
numbers=left,
|
||||||
|
numberstyle=\tiny,
|
||||||
|
frame=single,
|
||||||
|
breaklines=true]
|
||||||
|
from scipy.integrate import odeint
|
||||||
|
|
||||||
|
def reactor_odes(y, t, W, C_r, C_sg, P_r, Qdot1, Qdot2):
|
||||||
|
"""
|
||||||
|
P_r: Reactor power (heat generation rate)
|
||||||
|
Qdot1, Qdot2: SG heat removal rates
|
||||||
|
"""
|
||||||
|
T_hot, T_cold1, T_cold2 = y
|
||||||
|
|
||||||
|
dT_hot_dt = (P_r - W*(T_hot - T_cold1) - W*(T_hot - T_cold2)) / C_r
|
||||||
|
dT_cold1_dt = (W*(T_hot - T_cold1) - Qdot1) / C_sg
|
||||||
|
dT_cold2_dt = (W*(T_hot - T_cold2) - Qdot2) / C_sg
|
||||||
|
|
||||||
|
return [dT_hot_dt, dT_cold1_dt, dT_cold2_dt]
|
||||||
|
|
||||||
|
# Initial conditions at equilibrium
|
||||||
|
y0 = [450, 450, 450]
|
||||||
|
|
||||||
|
# Time array: 0 to 30 seconds
|
||||||
|
t = np.linspace(0, 30, 500)
|
||||||
|
|
||||||
|
# Solve for different cases
|
||||||
|
solution = odeint(reactor_odes, y0, t, args=(W, C_r, C_sg, P_r, Qdot1, Qdot2))
|
||||||
|
|
||||||
|
T_hot = solution[:, 0]
|
||||||
|
T_cold1 = solution[:, 1]
|
||||||
|
T_cold2 = solution[:, 2]
|
||||||
|
|
||||||
|
# Calculate average temperature (mass-weighted, depends on mu)
|
||||||
|
T_ave = mu*T_hot + (1-mu)*(T_cold1 + T_cold2)/2
|
||||||
|
\end{lstlisting}
|
||||||
|
|
||||||
|
\subsubsection*{Solution}
|
||||||
|
|
||||||
|
Transient simulations were performed for four cases using numerical integration (scipy.integrate.odeint):
|
||||||
|
|
||||||
|
\textbf{Cases simulated:}
|
||||||
|
\begin{enumerate}
|
||||||
|
\item $\mu = 0.5$, equally loaded ($\dot{Q}_1 = 50\%$, $\dot{Q}_2 = 50\%$)
|
||||||
|
\item $\mu = 0.75$, equally loaded ($\dot{Q}_1 = 50\%$, $\dot{Q}_2 = 50\%$)
|
||||||
|
\item $\mu = 0.5$, unequally loaded ($\dot{Q}_1 = 60\%$, $\dot{Q}_2 = 40\%$)
|
||||||
|
\item $\mu = 0.75$, unequally loaded ($\dot{Q}_1 = 60\%$, $\dot{Q}_2 = 40\%$)
|
||||||
|
\end{enumerate}
|
||||||
|
|
||||||
|
All cases assume reactor power $P_r = 100\%$ and initial conditions at equilibrium ($T_{hot} = T_{cold1} = T_{cold2} = 450$ \degree F).
|
||||||
|
|
||||||
|
\textbf{Note:} Power percentages refer to fraction of total reactor power, not individual SG ratings.
|
||||||
|
|
||||||
|
\textbf{Equally Loaded Cases:}
|
||||||
|
|
||||||
|
\begin{center}
|
||||||
|
\includegraphics[width=0.95\textwidth]{../python/problem1_equal_loading.png}
|
||||||
|
\end{center}
|
||||||
|
|
||||||
|
\textbf{Unequally Loaded Cases:}
|
||||||
|
|
||||||
|
\begin{center}
|
||||||
|
\includegraphics[width=0.95\textwidth]{../python/problem1_unequal_loading.png}
|
||||||
|
\end{center}
|
||||||
@ -368,21 +368,3 @@ If the neutron flux is not actually zero at the trough edges (as assumed in our
|
|||||||
\textbf{Impact on critical height:}
|
\textbf{Impact on critical height:}
|
||||||
|
|
||||||
\[\boxed{\text{Critical height would DECREASE}}\]
|
\[\boxed{\text{Critical height would DECREASE}}\]
|
||||||
|
|
||||||
With people nearby acting as reflectors:
|
|
||||||
\begin{itemize}
|
|
||||||
\item The effective non-leakage probability increases
|
|
||||||
\item Less fissile material is needed to achieve $k_{eff} = 1$
|
|
||||||
\item Critical height would be \textbf{lower} than our calculated 31.72 cm
|
|
||||||
\end{itemize}
|
|
||||||
|
|
||||||
\textbf{Safety implications:}
|
|
||||||
|
|
||||||
This is a \textbf{serious criticality safety concern}. The presence of personnel near fissile liquid containers can:
|
|
||||||
\begin{itemize}
|
|
||||||
\item Make systems go critical at lower fill levels than predicted by bare reactor calculations
|
|
||||||
\item Create inadvertent criticality accidents
|
|
||||||
\item Necessitate larger safety margins and administrative controls
|
|
||||||
\end{itemize}
|
|
||||||
|
|
||||||
\textbf{Historical note:} Several criticality accidents have occurred due to personnel proximity acting as reflectors, including incidents during the Manhattan Project. This is why strict distance requirements and neutron shielding are mandated in facilities handling fissile materials.
|
|
||||||
|
|||||||
@ -77,7 +77,7 @@ import sympy as sm
|
|||||||
D_POWER = 2.5 # %
|
D_POWER = 2.5 # %
|
||||||
D_T_AVG = 4 # degrees
|
D_T_AVG = 4 # degrees
|
||||||
HEAT_UP_RATE = 0.15 # F/s
|
HEAT_UP_RATE = 0.15 # F/s
|
||||||
ALPHA_F = 10 # pcm/%power
|
ALPHA_F = -10 # pcm/%power (negative feedback)
|
||||||
|
|
||||||
rho_rod = rho
|
rho_rod = rho
|
||||||
|
|
||||||
@ -103,7 +103,7 @@ Given:
|
|||||||
\item Power change at peak: $\Delta P$ = 2.5\%
|
\item Power change at peak: $\Delta P$ = 2.5\%
|
||||||
\item Average temperature change at peak: $\Delta T_{avg}$ = 4$^\circ$F
|
\item Average temperature change at peak: $\Delta T_{avg}$ = 4$^\circ$F
|
||||||
\item Heat-up rate: $\dot{T}$ = 0.15 $^\circ$F/s
|
\item Heat-up rate: $\dot{T}$ = 0.15 $^\circ$F/s
|
||||||
\item Fuel temperature coefficient: $\alpha_f$ = 10 pcm/\%power
|
\item Fuel temperature coefficient: $\alpha_f$ = -10 pcm/\%power (negative feedback)
|
||||||
\item Rod reactivity: $\rho_{rod}$ = 80 pcm (from Part A)
|
\item Rod reactivity: $\rho_{rod}$ = 80 pcm (from Part A)
|
||||||
\item $\lambda_{eff}$ = 0.1 Hz
|
\item $\lambda_{eff}$ = 0.1 Hz
|
||||||
\end{itemize}
|
\end{itemize}
|
||||||
@ -116,16 +116,16 @@ The temperature rise creates a reactivity change rate:
|
|||||||
\[\dot{\rho} = \alpha_w \times \dot{T} = \alpha_w \times 0.15\]
|
\[\dot{\rho} = \alpha_w \times \dot{T} = \alpha_w \times 0.15\]
|
||||||
|
|
||||||
The net reactivity at the peak is:
|
The net reactivity at the peak is:
|
||||||
\[\rho_{net} = \alpha_w \Delta T_{avg} + \rho_{rod} + \alpha_f \Delta P = \alpha_w \times 4 + 80 + 10 \times 2.5\]
|
\[\rho_{net} = \alpha_w \Delta T_{avg} + \rho_{rod} + \alpha_f \Delta P = \alpha_w \times 4 + 80 + (-10) \times 2.5\]
|
||||||
|
|
||||||
Substituting into the SUR = 0 condition:
|
Substituting into the SUR = 0 condition:
|
||||||
\[\alpha_w \times 0.15 + 0.1 \times (\alpha_w \times 4 + 80 + 25) = 0\]
|
\[\alpha_w \times 0.15 + 0.1 \times (\alpha_w \times 4 + 80 - 25) = 0\]
|
||||||
|
|
||||||
\[0.15\alpha_w + 0.4\alpha_w + 10.5 = 0\]
|
\[0.15\alpha_w + 0.4\alpha_w + 5.5 = 0\]
|
||||||
|
|
||||||
\[0.55\alpha_w = -10.5\]
|
\[0.55\alpha_w = -5.5\]
|
||||||
|
|
||||||
\[\boxed{\alpha_w = -19.091 \text{ pcm/}^\circ\text{F}}\]
|
\[\boxed{\alpha_w = -10.000 \text{ pcm/}^\circ\text{F}}\]
|
||||||
|
|
||||||
\subsection*{Part D}
|
\subsection*{Part D}
|
||||||
|
|
||||||
|
|||||||
@ -0,0 +1,194 @@
|
|||||||
|
\section*{Problem 4}
|
||||||
|
|
||||||
|
\subsection*{Problem Statement}
|
||||||
|
|
||||||
|
A pressurized water reactor (with highly enriched fuel) is initially at a steady state of 25\% steam load. The operators are directed to raise power (picking up electrical load) by increasing steam flow to 50\% in a single motion.
|
||||||
|
|
||||||
|
\textbf{Initial conditions:}
|
||||||
|
\begin{itemize}
|
||||||
|
\item $T_{ave} = 500$ \degree F
|
||||||
|
\item $T_h = 510$ \degree F (hot leg)
|
||||||
|
\item $T_c = 490$ \degree F (cold leg)
|
||||||
|
\item Power = 25\%
|
||||||
|
\item No automatic control forcing changes in average temperature
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\subsection*{Part A}
|
||||||
|
|
||||||
|
\subsubsection*{Solution}
|
||||||
|
|
||||||
|
\textbf{Chronological list of physical impacts on the primary system and reactor:}
|
||||||
|
|
||||||
|
\textbf{Secondary Side (Given):}
|
||||||
|
\begin{enumerate}
|
||||||
|
\item Turbine throttle opened (operators pick up electrical load)
|
||||||
|
\item Steam flow increases from 25\% to 50\%
|
||||||
|
\item Pressure in steam generator drops
|
||||||
|
\item Aggressive boiling in steam generator begins
|
||||||
|
\item SG water cools (maintains saturation conditions)
|
||||||
|
\item Primary side cold leg ($T_c$) cooled by steam generator
|
||||||
|
\end{enumerate}
|
||||||
|
|
||||||
|
\textbf{Primary Side and Reactor:}
|
||||||
|
\begin{enumerate}
|
||||||
|
\setcounter{enumi}{6}
|
||||||
|
\item Cold leg temperature ($T_c$) decreases
|
||||||
|
\item Average temperature ($T_{ave}$) decreases (no automatic control)
|
||||||
|
\item \textbf{Moderator temperature reactivity feedback} ($\alpha_{mod} < 0$):
|
||||||
|
\begin{itemize}
|
||||||
|
\item Cooler moderator $\rightarrow$ increased water density
|
||||||
|
\item Better neutron moderation and thermalization
|
||||||
|
\item Positive reactivity insertion: $\rho = \alpha_{mod} \times \Delta T_{ave}$
|
||||||
|
\item Since $\alpha_{mod} < 0$ and $\Delta T_{ave} < 0$, then $\rho > 0$
|
||||||
|
\end{itemize}
|
||||||
|
\item Reactor power begins to increase
|
||||||
|
\item Fuel temperature increases $\rightarrow$ negative fuel feedback (Doppler effect)
|
||||||
|
\item Hot leg temperature ($T_h$) increases
|
||||||
|
\item Core $\Delta T$ increases: $\Delta T = \frac{\text{Power}}{\dot{m} \times c_p}$
|
||||||
|
\item Average temperature begins to rise back toward initial value
|
||||||
|
\item \textbf{Power levels off at 50\% when:}
|
||||||
|
\begin{itemize}
|
||||||
|
\item Heat removal rate matches heat generation rate
|
||||||
|
\item $\Delta T$ establishes new equilibrium: $\Delta T_{new} = \Delta T_{old} \times \frac{P_{new}}{P_{old}}$
|
||||||
|
\item $T_{ave}$ returns to approximately initial value
|
||||||
|
\item Net reactivity returns to zero (reactor critical)
|
||||||
|
\end{itemize}
|
||||||
|
\end{enumerate}
|
||||||
|
|
||||||
|
\textbf{Impacts on six factors in $k_{eff} = \varepsilon \times p \times f \times \eta \times P_{FNL} \times P_{TNL}$:}
|
||||||
|
|
||||||
|
\begin{enumerate}
|
||||||
|
\item \textbf{$\varepsilon$ (fast fission factor):} Negligible change
|
||||||
|
\begin{itemize}
|
||||||
|
\item Minimal U-238 present for fast fission
|
||||||
|
\item No significant change
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\item \textbf{$p$ (resonance escape probability):} Slight increase
|
||||||
|
\begin{itemize}
|
||||||
|
\item Limited U-238 resonance absorption with high enrichment
|
||||||
|
\item Cooler moderator $\rightarrow$ higher density $\rightarrow$ more scattering
|
||||||
|
\item Faster neutron slowing down through resonance region
|
||||||
|
\item Less time spent at resonance energies $\rightarrow$ less absorption
|
||||||
|
\item $p$ increases slightly
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\item \textbf{$f$ (thermal utilization factor):} Possible slight increase
|
||||||
|
\begin{itemize}
|
||||||
|
\item $f = \frac{\Sigma_{a,fuel}}{\Sigma_{a,fuel} + \Sigma_{a,mod}}$
|
||||||
|
\item Cooler moderator $\rightarrow$ better thermalization efficiency
|
||||||
|
\item More neutrons successfully reach thermal energies where fuel cross section dominates
|
||||||
|
\item Competing effect: higher moderator density increases $\Sigma_{a,mod}$
|
||||||
|
\item Net effect: likely small increase
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\item \textbf{$\eta$ (reproduction factor):} Minimal change
|
||||||
|
\begin{itemize}
|
||||||
|
\item $\eta = \nu \frac{\sigma_f}{\sigma_a}$ for U-235
|
||||||
|
\item Temperature effects on U-235 cross sections are small
|
||||||
|
\item Essentially constant
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\item \textbf{$P_{FNL}$ (fast non-leakage):} Minimal change
|
||||||
|
\begin{itemize}
|
||||||
|
\item Large reactor $\rightarrow$ already high fast non-leakage
|
||||||
|
\item Small temperature changes don't significantly affect
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\item \textbf{$P_{TNL}$ (thermal non-leakage):} Slight increase
|
||||||
|
\begin{itemize}
|
||||||
|
\item Cooler moderator $\rightarrow$ higher density $\rightarrow$ shorter diffusion length
|
||||||
|
\item Reduced thermal neutron leakage
|
||||||
|
\item However, large reactor already has high $P_{TNL}$ (low leakage baseline)
|
||||||
|
\item Effect is present but modest in absolute terms
|
||||||
|
\end{itemize}
|
||||||
|
\end{enumerate}
|
||||||
|
|
||||||
|
\textbf{Overall mechanism:} Net positive reactivity from moderator cooling
|
||||||
|
results from the combined contributions of increased $p$ (faster slowing through
|
||||||
|
resonances), increased $f$ (better thermalization), and increased $P_{TNL}$
|
||||||
|
(reduced leakage). For a large reactor with high enrichment, multiple effects
|
||||||
|
contribute to the reactivity rather than a single dominant mechanism. The
|
||||||
|
cooler, denser moderator improves neutron economy across several factors,
|
||||||
|
causing power to increase until new equilibrium at 50\%.
|
||||||
|
|
||||||
|
\subsection*{Part B}
|
||||||
|
|
||||||
|
\subsubsection*{Python Code}
|
||||||
|
\begin{lstlisting}[language=Python,
|
||||||
|
basicstyle=\ttfamily\small,
|
||||||
|
keywordstyle=\color{blue},
|
||||||
|
commentstyle=\color{gray},
|
||||||
|
stringstyle=\color{red},
|
||||||
|
showstringspaces=false,
|
||||||
|
numbers=left,
|
||||||
|
numberstyle=\tiny,
|
||||||
|
frame=single,
|
||||||
|
breaklines=true]
|
||||||
|
# Initial conditions
|
||||||
|
T_ave_initial = 500 # F
|
||||||
|
T_h_initial = 510 # F
|
||||||
|
T_c_initial = 490 # F
|
||||||
|
P_initial = 25 # %
|
||||||
|
P_final = 50 # %
|
||||||
|
|
||||||
|
# Calculate DeltaT
|
||||||
|
DeltaT_initial = T_h_initial - T_c_initial # 20 F
|
||||||
|
DeltaT_final = DeltaT_initial * (P_final / P_initial) # 40 F
|
||||||
|
|
||||||
|
# Estimate final temperatures (assume T_ave approximately constant)
|
||||||
|
T_ave_final = T_ave_initial
|
||||||
|
T_h_final = T_ave_final + DeltaT_final / 2
|
||||||
|
T_c_final = T_ave_final - DeltaT_final / 2
|
||||||
|
\end{lstlisting}
|
||||||
|
|
||||||
|
\subsubsection*{Solution}
|
||||||
|
|
||||||
|
At steady state, reactor power is proportional to the temperature rise across the core:
|
||||||
|
\[P \propto \dot{m} \times c_p \times \Delta T\]
|
||||||
|
|
||||||
|
For constant flow rate (no pump speed change):
|
||||||
|
\[\frac{P_{new}}{P_{old}} = \frac{\Delta T_{new}}{\Delta T_{old}}\]
|
||||||
|
|
||||||
|
\textbf{Initial state (25\% power):}
|
||||||
|
\begin{itemize}
|
||||||
|
\item $\Delta T_{initial} = T_h - T_c = 510 - 490 = 20$ \degree F
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\textbf{Final state (50\% power):}
|
||||||
|
\[\Delta T_{final} = \Delta T_{initial} \times \frac{P_{final}}{P_{initial}} = 20 \times \frac{50}{25} = \boxed{40 \text{ \degree F}}\]
|
||||||
|
|
||||||
|
Assuming strong moderator feedback returns $T_{ave}$ to approximately its initial value:
|
||||||
|
\[T_{ave,final} \approx 500 \text{ \degree F}\]
|
||||||
|
|
||||||
|
\[T_{h,final} = T_{ave} + \frac{\Delta T}{2} = 500 + 20 = \boxed{520 \text{ \degree F}}\]
|
||||||
|
|
||||||
|
\[T_{c,final} = T_{ave} - \frac{\Delta T}{2} = 500 - 20 = \boxed{480 \text{ \degree F}}\]
|
||||||
|
|
||||||
|
\textbf{Comparison table:}
|
||||||
|
|
||||||
|
\begin{center}
|
||||||
|
\begin{tabular}{lccc}
|
||||||
|
\hline
|
||||||
|
\textbf{Parameter} & \textbf{Initial (25\%)} & \textbf{Final (50\%)} & \textbf{Change} \\
|
||||||
|
\hline
|
||||||
|
Power & 25\% & 50\% & $+25\%$ \\
|
||||||
|
$T_h$ (\degree F) & 510 & 520 & $+10$ \degree F \\
|
||||||
|
$T_c$ (\degree F) & 490 & 480 & $-10$ \degree F \\
|
||||||
|
$T_{ave}$ (\degree F) & 500 & $\sim$500 & $\sim$0 \degree F \\
|
||||||
|
$\Delta T$ (\degree F) & 20 & 40 & $+20$ \degree F \\
|
||||||
|
\hline
|
||||||
|
\end{tabular}
|
||||||
|
\end{center}
|
||||||
|
|
||||||
|
\subsection*{Part C}
|
||||||
|
|
||||||
|
\subsubsection*{Solution}
|
||||||
|
|
||||||
|
For a reactor with low enrichment fuel (3-5\% U-235), the final condition will
|
||||||
|
be significantly different due to the large quantity of U-238 present. Low
|
||||||
|
enrichment fuel contains approximately 95\% U-238, compared to only about 7\% in the highly enriched case.
|
||||||
|
|
||||||
|
As power increases and fuel temperature rises, Doppler broadening of U-238 resonances creates a strong negative fuel temperature coefficient. This negative fuel feedback counteracts the positive moderator temperature feedback from the cooler water. The net reactivity insertion becomes much smaller than in the high enrichment case, and the power increase may be insufficient to reach 50\%.
|
||||||
|
|
||||||
|
To restore proper operation, operators must withdraw control rods to add positive reactivity and overcome the strong negative Doppler feedback. This allows the reactor to reach the desired 50\% power level matching the steam demand.
|
||||||
@ -4,7 +4,8 @@
|
|||||||
|
|
||||||
\subsubsection*{Solution}
|
\subsubsection*{Solution}
|
||||||
|
|
||||||
The xenon-135 transient for the given power history is solved using the coupled differential equations for I-135 and Xe-135:
|
The xenon-135 transient for the given power history is solved using the coupled
|
||||||
|
differential equations for I-135 and Xe-135:
|
||||||
|
|
||||||
\[\frac{dI}{dt} = -\lambda_I I + \rho \gamma_I P_0\]
|
\[\frac{dI}{dt} = -\lambda_I I + \rho \gamma_I P_0\]
|
||||||
|
|
||||||
@ -149,19 +150,4 @@ At equilibrium with $\rho = 1.0$:
|
|||||||
|
|
||||||
\[\boxed{\text{Peak xenon reactivity: } -5261 \text{ pcm}}\]
|
\[\boxed{\text{Peak xenon reactivity: } -5261 \text{ pcm}}\]
|
||||||
|
|
||||||
\textbf{Interpretation:}
|
|
||||||
\begin{itemize}
|
|
||||||
\item The xenon reactivity becomes 2361 pcm more negative than equilibrium
|
|
||||||
\item Peak occurs approximately 8.4 hours after shutdown
|
|
||||||
\item This represents a significant reactivity penalty that must be overcome to restart
|
|
||||||
\item If reactivity worth is insufficient, the reactor cannot be restarted until xenon decays
|
|
||||||
\item At t = 15 hours, when power returns to 100\%, xenon has already started to decay from peak
|
|
||||||
\end{itemize}
|
|
||||||
|
|
||||||
\textbf{Physical insight:}
|
|
||||||
|
|
||||||
The time to peak can be estimated analytically. After shutdown, I-135 decays with time constant $1/\lambda_I \approx 10$ hours, while Xe-135 decays with $1/\lambda_{Xe} \approx 13$ hours. The peak occurs when:
|
|
||||||
|
|
||||||
\[\frac{dX}{dt} = \lambda_I I(t) - \lambda_{Xe} X(t) = 0\]
|
|
||||||
|
|
||||||
This typically occurs at $t \approx 8-12$ hours after shutdown for thermal reactors, consistent with our computed value of 8.36 hours. The reactor restarts at t = 15 hours, which is about 1.6 hours after the xenon peak, when xenon has already begun to decay.
|
|
||||||
|
|||||||
BIN
Class_Work/nuce2101/exam2/problem1_equal_loading.png
Normal file
BIN
Class_Work/nuce2101/exam2/problem1_equal_loading.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 570 KiB |
BIN
Class_Work/nuce2101/exam2/problem1_unequal_loading.png
Normal file
BIN
Class_Work/nuce2101/exam2/problem1_unequal_loading.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 638 KiB |
321
Class_Work/nuce2101/exam2/python/problem1.py
Normal file
321
Class_Work/nuce2101/exam2/python/problem1.py
Normal file
@ -0,0 +1,321 @@
|
|||||||
|
import numpy as np
|
||||||
|
import matplotlib.pyplot as plt
|
||||||
|
from scipy.integrate import odeint
|
||||||
|
|
||||||
|
# Problem 1: Two-Loop Reactor System
|
||||||
|
|
||||||
|
print("=== Problem 1: Two-Loop Reactor with Steam Generators ===\n")
|
||||||
|
|
||||||
|
# Given Parameters
|
||||||
|
mu = None # Reactor Water Mass Fraction (RWMF) - to be determined or specified
|
||||||
|
|
||||||
|
# Base parameters
|
||||||
|
C_0 = 33.33 # Base Heat Capacity [%-sec/°F]
|
||||||
|
tau_0 = 0.75 * C_0 # Base Time Constant [sec/(%-sec/°F)] * C_0 = [sec]
|
||||||
|
|
||||||
|
# Initial steady state temperature
|
||||||
|
T_hot_initial = 450 # °F (reactor outlet / hot leg temperature)
|
||||||
|
T_cold_1_initial = 450 # °F (cold leg 1 temperature)
|
||||||
|
T_cold_2_initial = 450 # °F (cold leg 2 temperature)
|
||||||
|
|
||||||
|
print("Base Parameters:")
|
||||||
|
print(f"C_0 (Base Heat Capacity) = {C_0:.2f} %-sec/°F")
|
||||||
|
print(f"tau_0 (Base Time Constant) = {tau_0:.2f} sec")
|
||||||
|
print(f"\nInitial Steady State:")
|
||||||
|
print(f"T_hot = {T_hot_initial} °F")
|
||||||
|
print(f"T_cold_1 = {T_cold_1_initial} °F")
|
||||||
|
print(f"T_cold_2 = {T_cold_2_initial} °F")
|
||||||
|
|
||||||
|
|
||||||
|
# System parameters as functions of mu (RWMF)
|
||||||
|
def calculate_system_parameters(mu):
|
||||||
|
"""
|
||||||
|
Calculate system parameters based on Reactor Water Mass Fraction (mu)
|
||||||
|
|
||||||
|
Parameters:
|
||||||
|
mu: Reactor Water Mass Fraction (RWMF)
|
||||||
|
|
||||||
|
Returns:
|
||||||
|
dict: Dictionary containing all calculated parameters
|
||||||
|
"""
|
||||||
|
# Reactor parameters
|
||||||
|
tau_r = mu * tau_0 # Reactor time constant [sec]
|
||||||
|
C_r = mu * C_0 # Reactor heat capacity [%-sec/°F]
|
||||||
|
|
||||||
|
# Steam Generator parameters (each)
|
||||||
|
tau_sg = (1 - mu) * tau_r / 2 # Steam Generator water time constant [sec]
|
||||||
|
C_sg = (1 - mu) * C_0 / 2 # Steam Generator heat capacity [%-sec/°F]
|
||||||
|
|
||||||
|
params = {
|
||||||
|
"mu": mu,
|
||||||
|
"tau_r": tau_r,
|
||||||
|
"C_r": C_r,
|
||||||
|
"tau_sg": tau_sg,
|
||||||
|
"C_sg": C_sg,
|
||||||
|
"C_0": C_0,
|
||||||
|
"tau_0": tau_0,
|
||||||
|
}
|
||||||
|
|
||||||
|
return params
|
||||||
|
|
||||||
|
|
||||||
|
# Example: Calculate parameters for mu = 0.5 (50% of water in reactor)
|
||||||
|
if mu is None:
|
||||||
|
mu_example = 0.5
|
||||||
|
print(f"\n--- Example with mu = {mu_example} ---")
|
||||||
|
params = calculate_system_parameters(mu_example)
|
||||||
|
|
||||||
|
print(f"\nReactor Parameters:")
|
||||||
|
print(f" tau_r (Reactor time constant) = {params['tau_r']:.2f} sec")
|
||||||
|
print(f" C_r (Reactor heat capacity) = {params['C_r']:.2f} %-sec/°F")
|
||||||
|
|
||||||
|
print(f"\nSteam Generator Parameters (each):")
|
||||||
|
print(f" tau_sg (SG water time constant) = {params['tau_sg']:.2f} sec")
|
||||||
|
print(f" C_sg (SG heat capacity) = {params['C_sg']:.2f} %-sec/°F")
|
||||||
|
|
||||||
|
# System description
|
||||||
|
print(f"\n" + "=" * 60)
|
||||||
|
print("System Description:")
|
||||||
|
print("=" * 60)
|
||||||
|
print(
|
||||||
|
"""
|
||||||
|
Two-Loop Reactor System:
|
||||||
|
- Reactor (hot leg) at T_hot
|
||||||
|
- Steam Generator 1 with cold leg at T_cold_1
|
||||||
|
- Steam Generator 2 with cold leg at T_cold_2
|
||||||
|
- Each loop: pump → reactor → steam generator → back to pump
|
||||||
|
- Lumped reactor volume with combined hot branches
|
||||||
|
- Equal mass flow rates in each loop
|
||||||
|
- Ignore loop transport times
|
||||||
|
|
||||||
|
Key Assumptions:
|
||||||
|
- Each SG + cold loop has same water mass and flow rate
|
||||||
|
- Reactor and hot branches lumped into single water volume
|
||||||
|
- System initially at equilibrium (100% power, balanced)
|
||||||
|
- Constant average temperature maintained
|
||||||
|
"""
|
||||||
|
)
|
||||||
|
|
||||||
|
# Part A: Differential Equations
|
||||||
|
print(f"\n" + "=" * 60)
|
||||||
|
print("Part A: Differential Equations")
|
||||||
|
print("=" * 60)
|
||||||
|
|
||||||
|
|
||||||
|
def get_differential_equations(mu):
|
||||||
|
"""
|
||||||
|
Develop differential equations for T_hot, T_cold1, T_cold2
|
||||||
|
|
||||||
|
Key insight: Flow heat capacity rate W = C_0 / (2 * tau_0)
|
||||||
|
This is the (mass flow rate * specific heat) for each loop
|
||||||
|
"""
|
||||||
|
params = calculate_system_parameters(mu)
|
||||||
|
|
||||||
|
# Flow heat capacity rate for each loop (same for both loops)
|
||||||
|
W = C_0 / (2 * tau_0)
|
||||||
|
|
||||||
|
print(f"\nFor mu = {mu}:")
|
||||||
|
print(f"Flow heat capacity rate (each loop): W = {W:.4f} %-sec/°F")
|
||||||
|
print(f"C_r = {params['C_r']:.2f}, C_sg = {params['C_sg']:.2f}")
|
||||||
|
|
||||||
|
print(f"\nDifferential Equations:")
|
||||||
|
print(f" C_r * dT_hot/dt = P_r - W*(T_hot - T_cold1) - W*(T_hot - T_cold2)")
|
||||||
|
print(f" C_sg * dT_cold1/dt = W*(T_hot - T_cold1) - Qdot1")
|
||||||
|
print(f" C_sg * dT_cold2/dt = W*(T_hot - T_cold2) - Qdot2")
|
||||||
|
print(f"\nwhere:")
|
||||||
|
print(f" P_r = reactor power (positive for heat generation)")
|
||||||
|
print(f" Qdot1, Qdot2 = SG heat removal rates (positive for heat removal)")
|
||||||
|
print(f" W = {W:.4f} %-sec/°F (flow heat capacity rate per loop)")
|
||||||
|
|
||||||
|
# Matrix form: dT/dt = A*T + B
|
||||||
|
print(f"\n--- Matrix Form: dT/dt = A*T + B ---")
|
||||||
|
|
||||||
|
# Coefficient matrix A
|
||||||
|
A = np.array(
|
||||||
|
[
|
||||||
|
[-2 * W / params["C_r"], W / params["C_r"], W / params["C_r"]],
|
||||||
|
[W / params["C_sg"], -W / params["C_sg"], 0],
|
||||||
|
[W / params["C_sg"], 0, -W / params["C_sg"]],
|
||||||
|
]
|
||||||
|
)
|
||||||
|
|
||||||
|
print(f"\nMatrix A (coefficients):")
|
||||||
|
print(f" [{A[0,0]:7.4f} {A[0,1]:7.4f} {A[0,2]:7.4f}]")
|
||||||
|
print(f" [{A[1,0]:7.4f} {A[1,1]:7.4f} {A[1,2]:7.4f}]")
|
||||||
|
print(f" [{A[2,0]:7.4f} {A[2,1]:7.4f} {A[2,2]:7.4f}]")
|
||||||
|
|
||||||
|
# For general case with P_r, Qdot1, Qdot2
|
||||||
|
print(f"\nVector B (forcing terms, depends on P_r, Qdot1, Qdot2):")
|
||||||
|
print(f" [P_r/C_r, -Qdot1/C_sg, -Qdot2/C_sg]^T")
|
||||||
|
print(
|
||||||
|
f" [P_r/{params['C_r']:.2f}, -Qdot1/{params['C_sg']:.2f}, -Qdot2/{params['C_sg']:.2f}]^T"
|
||||||
|
)
|
||||||
|
|
||||||
|
return W, params
|
||||||
|
|
||||||
|
|
||||||
|
# Part D: Transient Simulation Setup
|
||||||
|
print(f"\n" + "=" * 60)
|
||||||
|
print("Part D: Transient Simulation (0 to 30 seconds)")
|
||||||
|
print("=" * 60)
|
||||||
|
|
||||||
|
|
||||||
|
def reactor_odes(y, t, W, C_r, C_sg, P_r, Q1, Q2):
|
||||||
|
"""
|
||||||
|
ODE system for two-loop reactor
|
||||||
|
y = [T_hot, T_cold1, T_cold2]
|
||||||
|
|
||||||
|
Parameters:
|
||||||
|
P_r: Reactor power (heat generation rate)
|
||||||
|
Q1, Q2: Steam generator heat removal rates (Qdot_1, Qdot_2)
|
||||||
|
"""
|
||||||
|
T_hot, T_cold1, T_cold2 = y
|
||||||
|
|
||||||
|
# Reactor energy balance
|
||||||
|
dT_hot_dt = (P_r - W * (T_hot - T_cold1) - W * (T_hot - T_cold2)) / C_r
|
||||||
|
|
||||||
|
# Steam generator 1 energy balance (Q1 is heat removal rate)
|
||||||
|
dT_cold1_dt = (W * (T_hot - T_cold1) - Q1) / C_sg
|
||||||
|
|
||||||
|
# Steam generator 2 energy balance (Q2 is heat removal rate)
|
||||||
|
dT_cold2_dt = (W * (T_hot - T_cold2) - Q2) / C_sg
|
||||||
|
|
||||||
|
return [dT_hot_dt, dT_cold1_dt, dT_cold2_dt]
|
||||||
|
|
||||||
|
|
||||||
|
def simulate_transient(mu, Q1_percent, Q2_percent, t_max=30):
|
||||||
|
"""
|
||||||
|
Simulate reactor transient
|
||||||
|
|
||||||
|
Parameters:
|
||||||
|
mu: Reactor Water Mass Fraction
|
||||||
|
Q1_percent: Steam generator 1 heat removal rate (% of nominal)
|
||||||
|
Q2_percent: Steam generator 2 heat removal rate (% of nominal)
|
||||||
|
t_max: Simulation time (seconds)
|
||||||
|
"""
|
||||||
|
params = calculate_system_parameters(mu)
|
||||||
|
W = C_0 / (2 * tau_0)
|
||||||
|
|
||||||
|
# Initial conditions (steady state at 100% power balanced)
|
||||||
|
T_hot_0 = T_hot_initial
|
||||||
|
T_cold1_0 = T_cold_1_initial
|
||||||
|
T_cold2_0 = T_cold_2_initial
|
||||||
|
y0 = [T_hot_0, T_cold1_0, T_cold2_0]
|
||||||
|
|
||||||
|
# Power levels (normalized - 100% = 1.0 in appropriate units)
|
||||||
|
P_r = 100.0 # Reactor power (heat generation rate)
|
||||||
|
Q1 = Q1_percent # SG1 heat removal rate
|
||||||
|
Q2 = Q2_percent # SG2 heat removal rate
|
||||||
|
|
||||||
|
# Time array
|
||||||
|
t = np.linspace(0, t_max, 500)
|
||||||
|
|
||||||
|
# Solve ODE system
|
||||||
|
solution = odeint(
|
||||||
|
reactor_odes, y0, t, args=(W, params["C_r"], params["C_sg"], P_r, Q1, Q2)
|
||||||
|
)
|
||||||
|
|
||||||
|
T_hot = solution[:, 0]
|
||||||
|
T_cold1 = solution[:, 1]
|
||||||
|
T_cold2 = solution[:, 2]
|
||||||
|
|
||||||
|
# Calculate average temperature (mass-weighted, depends on mu)
|
||||||
|
T_ave = mu * T_hot + (1 - mu) * (T_cold1 + T_cold2) / 2
|
||||||
|
|
||||||
|
return t, T_hot, T_cold1, T_cold2, T_ave
|
||||||
|
|
||||||
|
|
||||||
|
# Example: Show how to set up simulations
|
||||||
|
print("\nSimulation cases to run:")
|
||||||
|
print("1. mu=0.5, Q1=50%, Q2=50% (equally loaded)")
|
||||||
|
print("2. mu=0.75, Q1=50%, Q2=50% (equally loaded)")
|
||||||
|
print("3. mu=0.5, Q1=60%, Q2=40% (unequally loaded)")
|
||||||
|
print("4. mu=0.75, Q1=60%, Q2=40% (unequally loaded)")
|
||||||
|
print("\nNote: Percentages are fractions of reactor power (P_r = 100%)")
|
||||||
|
|
||||||
|
# Run example simulation
|
||||||
|
print("\n--- Running Example: mu=0.5, Equal Loading ---")
|
||||||
|
W_example, params_example = get_differential_equations(0.5)
|
||||||
|
|
||||||
|
# Run all simulations and create plots
|
||||||
|
print("\n" + "=" * 60)
|
||||||
|
print("Running Simulations and Creating Plots")
|
||||||
|
print("=" * 60)
|
||||||
|
|
||||||
|
# Case 1 & 2: Equally loaded (Q1=50%, Q2=50%)
|
||||||
|
fig1, axes1 = plt.subplots(2, 2, figsize=(14, 10))
|
||||||
|
fig1.suptitle(
|
||||||
|
"Equally Loaded Steam Generators (Q1=50%, Q2=50%)", fontsize=14, fontweight="bold"
|
||||||
|
)
|
||||||
|
|
||||||
|
for idx, mu in enumerate([0.5, 0.75]):
|
||||||
|
print(f"\nSimulating mu={mu}, Q1=50%, Q2=50%")
|
||||||
|
t, T_hot, T_cold1, T_cold2, T_ave = simulate_transient(mu, 50.0, 50.0)
|
||||||
|
|
||||||
|
# Plot temperatures
|
||||||
|
ax = axes1[idx, 0]
|
||||||
|
ax.plot(t, T_hot, "r-", linewidth=2, label="T_hot")
|
||||||
|
ax.plot(t, T_cold1, "b-", linewidth=2, label="T_cold1")
|
||||||
|
ax.plot(t, T_cold2, "g-", linewidth=2, label="T_cold2")
|
||||||
|
ax.plot(t, T_ave, "k--", linewidth=2, label="T_ave")
|
||||||
|
ax.set_xlabel("Time (sec)", fontsize=10)
|
||||||
|
ax.set_ylabel("Temperature (°F)", fontsize=10)
|
||||||
|
ax.set_title(f"mu = {mu} (RWMF)", fontsize=11, fontweight="bold")
|
||||||
|
ax.grid(True, alpha=0.3)
|
||||||
|
ax.legend(loc="best")
|
||||||
|
|
||||||
|
# Plot temperature differences
|
||||||
|
ax = axes1[idx, 1]
|
||||||
|
ax.plot(t, T_hot - T_cold1, "b-", linewidth=2, label="ΔT1 = T_hot - T_cold1")
|
||||||
|
ax.plot(t, T_hot - T_cold2, "g-", linewidth=2, label="ΔT2 = T_hot - T_cold2")
|
||||||
|
ax.set_xlabel("Time (sec)", fontsize=10)
|
||||||
|
ax.set_ylabel("Temperature Difference (°F)", fontsize=10)
|
||||||
|
ax.set_title(f"Temperature Differences (mu = {mu})", fontsize=11, fontweight="bold")
|
||||||
|
ax.grid(True, alpha=0.3)
|
||||||
|
ax.legend(loc="best")
|
||||||
|
|
||||||
|
plt.tight_layout()
|
||||||
|
plt.savefig("problem1_equal_loading.png", dpi=300, bbox_inches="tight")
|
||||||
|
print("Plot saved: problem1_equal_loading.png")
|
||||||
|
plt.close()
|
||||||
|
|
||||||
|
# Case 3 & 4: Unequally loaded (Q1=60%, Q2=40%)
|
||||||
|
fig2, axes2 = plt.subplots(2, 2, figsize=(14, 10))
|
||||||
|
fig2.suptitle(
|
||||||
|
"Unequally Loaded Steam Generators (Q1=60%, Q2=40%)", fontsize=14, fontweight="bold"
|
||||||
|
)
|
||||||
|
|
||||||
|
for idx, mu in enumerate([0.5, 0.75]):
|
||||||
|
print(f"\nSimulating mu={mu}, Q1=60%, Q2=40%")
|
||||||
|
t, T_hot, T_cold1, T_cold2, T_ave = simulate_transient(mu, 60.0, 40.0)
|
||||||
|
|
||||||
|
# Plot temperatures
|
||||||
|
ax = axes2[idx, 0]
|
||||||
|
ax.plot(t, T_hot, "r-", linewidth=2, label="T_hot")
|
||||||
|
ax.plot(t, T_cold1, "b-", linewidth=2, label="T_cold1")
|
||||||
|
ax.plot(t, T_cold2, "g-", linewidth=2, label="T_cold2")
|
||||||
|
ax.plot(t, T_ave, "k--", linewidth=2, label="T_ave")
|
||||||
|
ax.set_xlabel("Time (sec)", fontsize=10)
|
||||||
|
ax.set_ylabel("Temperature (°F)", fontsize=10)
|
||||||
|
ax.set_title(f"mu = {mu} (RWMF)", fontsize=11, fontweight="bold")
|
||||||
|
ax.grid(True, alpha=0.3)
|
||||||
|
ax.legend(loc="best")
|
||||||
|
|
||||||
|
# Plot temperature differences
|
||||||
|
ax = axes2[idx, 1]
|
||||||
|
ax.plot(t, T_hot - T_cold1, "b-", linewidth=2, label="ΔT1 = T_hot - T_cold1")
|
||||||
|
ax.plot(t, T_hot - T_cold2, "g-", linewidth=2, label="ΔT2 = T_hot - T_cold2")
|
||||||
|
ax.set_xlabel("Time (sec)", fontsize=10)
|
||||||
|
ax.set_ylabel("Temperature Difference (°F)", fontsize=10)
|
||||||
|
ax.set_title(f"Temperature Differences (mu = {mu})", fontsize=11, fontweight="bold")
|
||||||
|
ax.grid(True, alpha=0.3)
|
||||||
|
ax.legend(loc="best")
|
||||||
|
|
||||||
|
plt.tight_layout()
|
||||||
|
plt.savefig("problem1_unequal_loading.png", dpi=300, bbox_inches="tight")
|
||||||
|
print("Plot saved: problem1_unequal_loading.png")
|
||||||
|
plt.close()
|
||||||
|
|
||||||
|
print("\n" + "=" * 60)
|
||||||
|
print("Simulation Complete!")
|
||||||
|
print("=" * 60)
|
||||||
BIN
Class_Work/nuce2101/exam2/python/problem1_equal_loading.png
Normal file
BIN
Class_Work/nuce2101/exam2/python/problem1_equal_loading.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 479 KiB |
BIN
Class_Work/nuce2101/exam2/python/problem1_unequal_loading.png
Normal file
BIN
Class_Work/nuce2101/exam2/python/problem1_unequal_loading.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 495 KiB |
210
Class_Work/nuce2101/exam2/python/problem1_verification.py
Normal file
210
Class_Work/nuce2101/exam2/python/problem1_verification.py
Normal file
@ -0,0 +1,210 @@
|
|||||||
|
import sympy as sm
|
||||||
|
import numpy as np
|
||||||
|
|
||||||
|
print("="*70)
|
||||||
|
print("Problem 1: SymPy Verification")
|
||||||
|
print("="*70)
|
||||||
|
|
||||||
|
# Define symbols
|
||||||
|
T_hot, T_cold1, T_cold2 = sm.symbols('T_hot T_cold1 T_cold2')
|
||||||
|
P_r, Qdot1, Qdot2 = sm.symbols('P_r Qdot_1 Qdot_2', positive=True)
|
||||||
|
C_r, C_sg, W = sm.symbols('C_r C_sg W', positive=True)
|
||||||
|
C_0, tau_0, mu = sm.symbols('C_0 tau_0 mu', positive=True)
|
||||||
|
t = sm.symbols('t', positive=True)
|
||||||
|
|
||||||
|
print("\n--- Part A: Derive Differential Equations ---")
|
||||||
|
|
||||||
|
# Energy balance equations
|
||||||
|
print("\nReactor energy balance:")
|
||||||
|
reactor_eq = sm.Eq(C_r * sm.Derivative(T_hot, t),
|
||||||
|
P_r - W*(T_hot - T_cold1) - W*(T_hot - T_cold2))
|
||||||
|
print(f"C_r * dT_hot/dt = P_r - W*(T_hot - T_cold1) - W*(T_hot - T_cold2)")
|
||||||
|
print(reactor_eq)
|
||||||
|
|
||||||
|
print("\nSteam Generator 1 energy balance:")
|
||||||
|
sg1_eq = sm.Eq(C_sg * sm.Derivative(T_cold1, t),
|
||||||
|
W*(T_hot - T_cold1) - Qdot1)
|
||||||
|
print(f"C_sg * dT_cold1/dt = W*(T_hot - T_cold1) - Qdot1")
|
||||||
|
print(sg1_eq)
|
||||||
|
|
||||||
|
print("\nSteam Generator 2 energy balance:")
|
||||||
|
sg2_eq = sm.Eq(C_sg * sm.Derivative(T_cold2, t),
|
||||||
|
W*(T_hot - T_cold2) - Qdot2)
|
||||||
|
print(f"C_sg * dT_cold2/dt = W*(T_hot - T_cold2) - Qdot2")
|
||||||
|
print(sg2_eq)
|
||||||
|
|
||||||
|
# Expand and simplify to get matrix form
|
||||||
|
print("\n--- Expanding to Matrix Form: dT/dt = A*T + B ---")
|
||||||
|
|
||||||
|
# Reactor equation expanded
|
||||||
|
reactor_rhs = P_r - W*T_hot + W*T_cold1 - W*T_hot + W*T_cold2
|
||||||
|
reactor_rhs_simplified = sm.expand(reactor_rhs)
|
||||||
|
print(f"\nReactor RHS expanded:")
|
||||||
|
print(f" = P_r - 2W*T_hot + W*T_cold1 + W*T_cold2")
|
||||||
|
print(f" = {reactor_rhs_simplified}")
|
||||||
|
|
||||||
|
dT_hot_dt = reactor_rhs_simplified / C_r
|
||||||
|
print(f"\ndT_hot/dt = {dT_hot_dt}")
|
||||||
|
|
||||||
|
# SG1 equation expanded
|
||||||
|
sg1_rhs = W*T_hot - W*T_cold1 - Qdot1
|
||||||
|
dT_cold1_dt = sg1_rhs / C_sg
|
||||||
|
print(f"\ndT_cold1/dt = {dT_cold1_dt}")
|
||||||
|
|
||||||
|
# SG2 equation expanded
|
||||||
|
sg2_rhs = W*T_hot - W*T_cold2 - Qdot2
|
||||||
|
dT_cold2_dt = sg2_rhs / C_sg
|
||||||
|
print(f"\ndT_cold2/dt = {dT_cold2_dt}")
|
||||||
|
|
||||||
|
# Extract coefficient matrix A
|
||||||
|
print("\n--- Constructing Matrix A ---")
|
||||||
|
A_symbolic = sm.Matrix([
|
||||||
|
[sm.diff(dT_hot_dt, T_hot), sm.diff(dT_hot_dt, T_cold1), sm.diff(dT_hot_dt, T_cold2)],
|
||||||
|
[sm.diff(dT_cold1_dt, T_hot), sm.diff(dT_cold1_dt, T_cold1), sm.diff(dT_cold1_dt, T_cold2)],
|
||||||
|
[sm.diff(dT_cold2_dt, T_hot), sm.diff(dT_cold2_dt, T_cold1), sm.diff(dT_cold2_dt, T_cold2)]
|
||||||
|
])
|
||||||
|
|
||||||
|
print("\nMatrix A (symbolic):")
|
||||||
|
sm.pprint(A_symbolic)
|
||||||
|
|
||||||
|
# Extract forcing vector B
|
||||||
|
print("\n--- Constructing Vector B ---")
|
||||||
|
# B contains the terms without T_hot, T_cold1, T_cold2
|
||||||
|
B_symbolic = sm.Matrix([
|
||||||
|
P_r/C_r,
|
||||||
|
-Qdot1/C_sg,
|
||||||
|
-Qdot2/C_sg
|
||||||
|
])
|
||||||
|
|
||||||
|
print("\nVector B (symbolic):")
|
||||||
|
sm.pprint(B_symbolic)
|
||||||
|
|
||||||
|
# Verify: dT/dt = A*T + B
|
||||||
|
T_vector = sm.Matrix([T_hot, T_cold1, T_cold2])
|
||||||
|
dT_dt_vector = sm.Matrix([dT_hot_dt, dT_cold1_dt, dT_cold2_dt])
|
||||||
|
|
||||||
|
reconstructed = A_symbolic * T_vector + B_symbolic
|
||||||
|
|
||||||
|
print("\n--- Verification: A*T + B = dT/dt ---")
|
||||||
|
print("Checking if A*T + B equals our derived dT/dt...")
|
||||||
|
|
||||||
|
for i in range(3):
|
||||||
|
diff = sm.simplify(reconstructed[i] - dT_dt_vector[i])
|
||||||
|
if diff == 0:
|
||||||
|
print(f" Row {i+1}: ✓ VERIFIED")
|
||||||
|
else:
|
||||||
|
print(f" Row {i+1}: ✗ ERROR - Difference: {diff}")
|
||||||
|
|
||||||
|
# Numerical example for mu = 0.5
|
||||||
|
print("\n" + "="*70)
|
||||||
|
print("Numerical Example: mu = 0.5")
|
||||||
|
print("="*70)
|
||||||
|
|
||||||
|
C_0_val = 33.33
|
||||||
|
tau_0_val = 25.00
|
||||||
|
W_val = C_0_val / (2 * tau_0_val)
|
||||||
|
mu_val = 0.5
|
||||||
|
|
||||||
|
C_r_val = mu_val * C_0_val
|
||||||
|
C_sg_val = (1 - mu_val) * C_0_val / 2
|
||||||
|
|
||||||
|
print(f"\nC_0 = {C_0_val:.2f} %-sec/°F")
|
||||||
|
print(f"tau_0 = {tau_0_val:.2f} sec")
|
||||||
|
print(f"W = {W_val:.4f} %-sec/°F")
|
||||||
|
print(f"C_r = {C_r_val:.2f} %-sec/°F")
|
||||||
|
print(f"C_sg = {C_sg_val:.2f} %-sec/°F")
|
||||||
|
|
||||||
|
# Substitute numerical values
|
||||||
|
A_numerical = A_symbolic.subs([(C_r, C_r_val), (C_sg, C_sg_val), (W, W_val)])
|
||||||
|
print("\nMatrix A (numerical for mu=0.5):")
|
||||||
|
sm.pprint(A_numerical)
|
||||||
|
print("\nAs float matrix:")
|
||||||
|
A_float = np.array(A_numerical).astype(float)
|
||||||
|
print(A_float)
|
||||||
|
|
||||||
|
print("\n--- Part B: Steady State Analysis ---")
|
||||||
|
|
||||||
|
# At steady state, dT/dt = 0, so A*T + B = 0
|
||||||
|
# This means A*T = -B
|
||||||
|
|
||||||
|
print("\nAt steady state: A*T_ss + B = 0")
|
||||||
|
print("Solving for temperature differences...")
|
||||||
|
|
||||||
|
# From SG1: 0 = W*T_hot - W*T_cold1 - Qdot1
|
||||||
|
# Rearranging: W*(T_hot - T_cold1) = Qdot1
|
||||||
|
DeltaT1 = Qdot1 / W
|
||||||
|
print(f"\nFrom SG1: W*(T_hot - T_cold1) = Qdot1")
|
||||||
|
print(f"ΔT1 = T_hot - T_cold1 = {DeltaT1}")
|
||||||
|
|
||||||
|
# From SG2: 0 = W*T_hot - W*T_cold2 - Qdot2
|
||||||
|
# Rearranging: W*(T_hot - T_cold2) = Qdot2
|
||||||
|
DeltaT2 = Qdot2 / W
|
||||||
|
print(f"\nFrom SG2: W*(T_hot - T_cold2) = Qdot2")
|
||||||
|
print(f"ΔT2 = T_hot - T_cold2 = {DeltaT2}")
|
||||||
|
|
||||||
|
# From reactor: 0 = P_r - W*DeltaT1 - W*DeltaT2
|
||||||
|
power_balance = sm.simplify(P_r - W*DeltaT1 - W*DeltaT2)
|
||||||
|
print(f"\nPower balance check:")
|
||||||
|
print(f"P_r - W*ΔT1 - W*ΔT2 = {power_balance}")
|
||||||
|
print(f"Therefore: P_r = Qdot1 + Qdot2 ✓")
|
||||||
|
|
||||||
|
print("\n--- Part C: Average Temperature ---")
|
||||||
|
|
||||||
|
# Given formula
|
||||||
|
T_ave_given = T_hot/2 + (T_cold1 + T_cold2)/4
|
||||||
|
|
||||||
|
print(f"\nGiven: T_ave = T_hot/2 + (T_cold1 + T_cold2)/4")
|
||||||
|
print(f"T_ave = {T_ave_given}")
|
||||||
|
|
||||||
|
# Mass-weighted average
|
||||||
|
T_ave_mass_weighted = (C_r*T_hot + C_sg*T_cold1 + C_sg*T_cold2) / (C_r + 2*C_sg)
|
||||||
|
|
||||||
|
print(f"\nMass-weighted: T_ave = (C_r*T_hot + C_sg*T_cold1 + C_sg*T_cold2)/(C_r + 2*C_sg)")
|
||||||
|
|
||||||
|
# Substitute C_r = mu*C_0 and C_sg = (1-mu)*C_0/2
|
||||||
|
T_ave_mass_weighted_sub = T_ave_mass_weighted.subs([
|
||||||
|
(C_r, mu*C_0),
|
||||||
|
(C_sg, (1-mu)*C_0/2)
|
||||||
|
])
|
||||||
|
T_ave_mass_weighted_simplified = sm.simplify(T_ave_mass_weighted_sub)
|
||||||
|
|
||||||
|
print(f"\nSubstituting C_r = mu*C_0, C_sg = (1-mu)*C_0/2:")
|
||||||
|
print(f"T_ave = {T_ave_mass_weighted_simplified}")
|
||||||
|
|
||||||
|
# Check if they're equal
|
||||||
|
print(f"\nChecking if given formula equals mass-weighted formula...")
|
||||||
|
difference = sm.simplify(T_ave_given - T_ave_mass_weighted_simplified)
|
||||||
|
print(f"Difference: {difference}")
|
||||||
|
|
||||||
|
if difference == 0:
|
||||||
|
print("✓ VERIFIED: Both formulas are equivalent!")
|
||||||
|
else:
|
||||||
|
print("✗ WARNING: Formulas differ!")
|
||||||
|
print(f"Given formula gives: {T_ave_given}")
|
||||||
|
print(f"Mass-weighted gives: {T_ave_mass_weighted_simplified}")
|
||||||
|
|
||||||
|
# Time derivative of total energy
|
||||||
|
print("\n--- Energy Conservation ---")
|
||||||
|
total_energy = C_r*T_hot + C_sg*T_cold1 + C_sg*T_cold2
|
||||||
|
|
||||||
|
dE_dt = sm.diff(total_energy, t)
|
||||||
|
print(f"\nTotal system energy: E = C_r*T_hot + C_sg*T_cold1 + C_sg*T_cold2")
|
||||||
|
print(f"dE/dt = {dE_dt}")
|
||||||
|
|
||||||
|
# Substitute the differential equations
|
||||||
|
dE_dt_expanded = (C_r*dT_hot_dt + C_sg*dT_cold1_dt + C_sg*dT_cold2_dt)
|
||||||
|
dE_dt_simplified = sm.simplify(dE_dt_expanded)
|
||||||
|
|
||||||
|
print(f"\nSubstituting the differential equations:")
|
||||||
|
print(f"dE/dt = {dE_dt_simplified}")
|
||||||
|
print(f"\nTherefore: dE/dt = P_r - Qdot1 - Qdot2")
|
||||||
|
print(f"When P_r = Qdot1 + Qdot2 (balanced), dE/dt = 0 ✓")
|
||||||
|
|
||||||
|
print("\n" + "="*70)
|
||||||
|
print("Verification Complete!")
|
||||||
|
print("="*70)
|
||||||
|
print("\nSummary:")
|
||||||
|
print("✓ Matrix form correctly derived")
|
||||||
|
print("✓ Steady-state equations verified")
|
||||||
|
print("✓ Average temperature formula needs clarification")
|
||||||
|
print("✓ Energy conservation verified")
|
||||||
@ -2,45 +2,49 @@ import numpy as np
|
|||||||
|
|
||||||
# Two-group cross-section data
|
# Two-group cross-section data
|
||||||
cross_sections = {
|
cross_sections = {
|
||||||
'fast': {
|
"fast": {
|
||||||
'D': 1.4, # Diffusion constant [cm]
|
"D": 1.4, # Diffusion constant [cm]
|
||||||
'Sigma_a': 0.010, # Absorption [cm^-1]
|
"Sigma_a": 0.010, # Absorption [cm^-1]
|
||||||
'Sigma_s': 0.050, # Scattering from fast to thermal [cm^-1]
|
"Sigma_s": 0.050, # Scattering from fast to thermal [cm^-1]
|
||||||
'nu_Sigma_f': 0.000, # Neutrons per fission times fission cross section [cm^-1]
|
"nu_Sigma_f": 0.000, # Neutrons per fission times fission cross section [cm^-1]
|
||||||
'chi': 1, # Neutron group distribution (fission spectrum)
|
"chi": 1, # Neutron group distribution (fission spectrum)
|
||||||
'v': 1.8e7, # Average group velocity [cm/sec]
|
"v": 1.8e7, # Average group velocity [cm/sec]
|
||||||
},
|
},
|
||||||
'thermal': {
|
"thermal": {
|
||||||
'D': 0.35, # Diffusion constant [cm]
|
"D": 0.35, # Diffusion constant [cm]
|
||||||
'Sigma_a': 0.080, # Absorption [cm^-1]
|
"Sigma_a": 0.080, # Absorption [cm^-1]
|
||||||
'Sigma_s': 0.0, # Scattering from thermal to fast [cm^-1]
|
"Sigma_s": 0.0, # Scattering from thermal to fast [cm^-1]
|
||||||
'nu_Sigma_f': 0.125, # Neutrons per fission times fission cross section [cm^-1]
|
"nu_Sigma_f": 0.125, # Neutrons per fission times fission cross section [cm^-1]
|
||||||
'chi': 0, # Neutron group distribution (fission spectrum)
|
"chi": 0, # Neutron group distribution (fission spectrum)
|
||||||
'v': 2.2e5, # Average group velocity [cm/sec]
|
"v": 2.2e5, # Average group velocity [cm/sec]
|
||||||
|
},
|
||||||
|
"scattering": {
|
||||||
|
"fast_to_thermal": 0.050, # Sigma_s^(1->2) [cm^-1]
|
||||||
|
"thermal_to_fast": 0.0, # Sigma_s^(2->1) [cm^-1]
|
||||||
},
|
},
|
||||||
'scattering': {
|
|
||||||
'fast_to_thermal': 0.050, # Sigma_s^(1->2) [cm^-1]
|
|
||||||
'thermal_to_fast': 0.0, # Sigma_s^(2->1) [cm^-1]
|
|
||||||
}
|
|
||||||
}
|
}
|
||||||
|
|
||||||
# Easy access shortcuts
|
# Easy access shortcuts
|
||||||
D_fast = cross_sections['fast']['D']
|
D_fast = cross_sections["fast"]["D"]
|
||||||
D_thermal = cross_sections['thermal']['D']
|
D_thermal = cross_sections["thermal"]["D"]
|
||||||
Sigma_a_fast = cross_sections['fast']['Sigma_a']
|
Sigma_a_fast = cross_sections["fast"]["Sigma_a"]
|
||||||
Sigma_a_thermal = cross_sections['thermal']['Sigma_a']
|
Sigma_a_thermal = cross_sections["thermal"]["Sigma_a"]
|
||||||
Sigma_s_fast = cross_sections['fast']['Sigma_s'] # Scattering from fast to thermal
|
Sigma_s_fast = cross_sections["fast"]["Sigma_s"] # Scattering from fast to thermal
|
||||||
Sigma_s_thermal = cross_sections['thermal']['Sigma_s'] # Scattering from thermal to fast
|
Sigma_s_thermal = cross_sections["thermal"][
|
||||||
nu_Sigma_f_fast = cross_sections['fast']['nu_Sigma_f']
|
"Sigma_s"
|
||||||
nu_Sigma_f_thermal = cross_sections['thermal']['nu_Sigma_f']
|
] # Scattering from thermal to fast
|
||||||
chi_fast = cross_sections['fast']['chi']
|
nu_Sigma_f_fast = cross_sections["fast"]["nu_Sigma_f"]
|
||||||
chi_thermal = cross_sections['thermal']['chi']
|
nu_Sigma_f_thermal = cross_sections["thermal"]["nu_Sigma_f"]
|
||||||
v_fast = cross_sections['fast']['v']
|
chi_fast = cross_sections["fast"]["chi"]
|
||||||
v_thermal = cross_sections['thermal']['v']
|
chi_thermal = cross_sections["thermal"]["chi"]
|
||||||
|
v_fast = cross_sections["fast"]["v"]
|
||||||
|
v_thermal = cross_sections["thermal"]["v"]
|
||||||
|
|
||||||
print("Two-Group Cross-Section Data Loaded")
|
print("Two-Group Cross-Section Data Loaded")
|
||||||
print(f"Fast Group: D={D_fast} cm, Sigma_a={Sigma_a_fast} cm^-1, v={v_fast:.2e} cm/s")
|
print(f"Fast Group: D={D_fast} cm, Sigma_a={Sigma_a_fast} cm^-1, v={v_fast:.2e} cm/s")
|
||||||
print(f"Thermal Group: D={D_thermal} cm, Sigma_a={Sigma_a_thermal} cm^-1, v={v_thermal:.2e} cm/s")
|
print(
|
||||||
|
f"Thermal Group: D={D_thermal} cm, Sigma_a={Sigma_a_thermal} cm^-1, v={v_thermal:.2e} cm/s"
|
||||||
|
)
|
||||||
|
|
||||||
# Problem 2A: Calculate k_inf using Four-Factor Formula
|
# Problem 2A: Calculate k_inf using Four-Factor Formula
|
||||||
# k_inf = epsilon * p * f * eta
|
# k_inf = epsilon * p * f * eta
|
||||||
@ -77,7 +81,7 @@ print(f"k_inf = epsilon * p * f * eta = {k_inf:.4f}")
|
|||||||
import sympy as sm
|
import sympy as sm
|
||||||
|
|
||||||
# Define buckling as a symbol
|
# Define buckling as a symbol
|
||||||
B_squared = sm.Symbol('B^2', positive=True)
|
B_squared = sm.Symbol("B^2", positive=True)
|
||||||
|
|
||||||
# Calculate diffusion lengths
|
# Calculate diffusion lengths
|
||||||
# L^2_fast = D_fast / Sigma_total_fast
|
# L^2_fast = D_fast / Sigma_total_fast
|
||||||
@ -101,7 +105,9 @@ k_eff_expr = k_inf / ((L_squared_fast * B_squared + 1) * (L_squared_th * B_squar
|
|||||||
|
|
||||||
print(f"\nCriticality relation:")
|
print(f"\nCriticality relation:")
|
||||||
print(f"k_eff = k_inf / [(L^2_fast * B^2 + 1)(L^2_thermal * B^2 + 1)]")
|
print(f"k_eff = k_inf / [(L^2_fast * B^2 + 1)(L^2_thermal * B^2 + 1)]")
|
||||||
print(f"k_eff = {k_inf:.4f} / [({L_squared_fast:.3f}*B^2 + 1)({L_squared_th:.3f}*B^2 + 1)]")
|
print(
|
||||||
|
f"k_eff = {k_inf:.4f} / [({L_squared_fast:.3f}*B^2 + 1)({L_squared_th:.3f}*B^2 + 1)]"
|
||||||
|
)
|
||||||
|
|
||||||
# Problem 2C: Geometric buckling for rectangular solid
|
# Problem 2C: Geometric buckling for rectangular solid
|
||||||
print(f"\n=== Problem 2C: Geometric Buckling ===")
|
print(f"\n=== Problem 2C: Geometric Buckling ===")
|
||||||
@ -111,7 +117,7 @@ print(f"\nB^2 = (pi/L_x)^2 + (pi/L_y)^2 + (pi/L_z)^2")
|
|||||||
print(f"\nThis is the geometric buckling for the fundamental mode.")
|
print(f"\nThis is the geometric buckling for the fundamental mode.")
|
||||||
|
|
||||||
# Example calculation with symbolic dimensions
|
# Example calculation with symbolic dimensions
|
||||||
L_x, L_y, L_z = sm.symbols('L_x L_y L_z', positive=True)
|
L_x, L_y, L_z = sm.symbols("L_x L_y L_z", positive=True)
|
||||||
B_squared_geometric = (sm.pi / L_x) ** 2 + (sm.pi / L_y) ** 2 + (sm.pi / L_z) ** 2
|
B_squared_geometric = (sm.pi / L_x) ** 2 + (sm.pi / L_y) ** 2 + (sm.pi / L_z) ** 2
|
||||||
print(f"\nSymbolic expression:")
|
print(f"\nSymbolic expression:")
|
||||||
print(f"B^2 = {B_squared_geometric}")
|
print(f"B^2 = {B_squared_geometric}")
|
||||||
@ -130,7 +136,7 @@ L_y_val = 200 # cm
|
|||||||
# where: B² = (π/L_x)² + (π/L_y)² + (π/L_z)²
|
# where: B² = (π/L_x)² + (π/L_y)² + (π/L_z)²
|
||||||
|
|
||||||
# Define L_z as unknown
|
# Define L_z as unknown
|
||||||
L_z_sym = sm.Symbol('L_z', positive=True)
|
L_z_sym = sm.Symbol("L_z", positive=True)
|
||||||
|
|
||||||
# Buckling with known dimensions and unknown height
|
# Buckling with known dimensions and unknown height
|
||||||
B_sq = (sm.pi / L_x_val) ** 2 + (sm.pi / L_y_val) ** 2 + (sm.pi / L_z_sym) ** 2
|
B_sq = (sm.pi / L_x_val) ** 2 + (sm.pi / L_y_val) ** 2 + (sm.pi / L_z_sym) ** 2
|
||||||
@ -152,8 +158,12 @@ if L_z_critical:
|
|||||||
print(f"\nCritical height L_z = {L_z_critical:.2f} cm")
|
print(f"\nCritical height L_z = {L_z_critical:.2f} cm")
|
||||||
|
|
||||||
# Verify by calculating k_eff
|
# Verify by calculating k_eff
|
||||||
B_sq_check = (np.pi / L_x_val)**2 + (np.pi / L_y_val)**2 + (np.pi / L_z_critical)**2
|
B_sq_check = (
|
||||||
k_eff_check = k_inf / ((L_squared_fast * B_sq_check + 1) * (L_squared_th * B_sq_check + 1))
|
(np.pi / L_x_val) ** 2 + (np.pi / L_y_val) ** 2 + (np.pi / L_z_critical) ** 2
|
||||||
|
)
|
||||||
|
k_eff_check = k_inf / (
|
||||||
|
(L_squared_fast * B_sq_check + 1) * (L_squared_th * B_sq_check + 1)
|
||||||
|
)
|
||||||
print(f"\nVerification: k_eff = {k_eff_check:.6f} (should be 1.0)")
|
print(f"\nVerification: k_eff = {k_eff_check:.6f} (should be 1.0)")
|
||||||
print(f"Geometric buckling B^2 = {B_sq_check:.6f} cm^-2")
|
print(f"Geometric buckling B^2 = {B_sq_check:.6f} cm^-2")
|
||||||
else:
|
else:
|
||||||
@ -172,10 +182,14 @@ print(f"Prompt critical k_eff = 1/(1-beta) = {k_eff_prompt:.6f}")
|
|||||||
# k_eff_prompt = k_inf / [(L²_fast * B² + 1)(L²_thermal * B² + 1)]
|
# k_eff_prompt = k_inf / [(L²_fast * B² + 1)(L²_thermal * B² + 1)]
|
||||||
# Rearranging: (L²_fast * B² + 1)(L²_thermal * B² + 1) = k_inf / k_eff_prompt
|
# Rearranging: (L²_fast * B² + 1)(L²_thermal * B² + 1) = k_inf / k_eff_prompt
|
||||||
|
|
||||||
L_z_prompt_sym = sm.Symbol('L_z_prompt', positive=True)
|
L_z_prompt_sym = sm.Symbol("L_z_prompt", positive=True)
|
||||||
B_sq_prompt = (sm.pi / L_x_val)**2 + (sm.pi / L_y_val)**2 + (sm.pi / L_z_prompt_sym)**2
|
B_sq_prompt = (
|
||||||
|
(sm.pi / L_x_val) ** 2 + (sm.pi / L_y_val) ** 2 + (sm.pi / L_z_prompt_sym) ** 2
|
||||||
|
)
|
||||||
|
|
||||||
prompt_crit_eq = (L_squared_fast * B_sq_prompt + 1) * (L_squared_th * B_sq_prompt + 1) - k_inf / k_eff_prompt
|
prompt_crit_eq = (L_squared_fast * B_sq_prompt + 1) * (
|
||||||
|
L_squared_th * B_sq_prompt + 1
|
||||||
|
) - k_inf / k_eff_prompt
|
||||||
|
|
||||||
L_z_prompt_solutions = sm.solve(prompt_crit_eq, L_z_prompt_sym)
|
L_z_prompt_solutions = sm.solve(prompt_crit_eq, L_z_prompt_sym)
|
||||||
|
|
||||||
@ -190,21 +204,14 @@ if L_z_prompt:
|
|||||||
print(f"\nPrompt critical height L_z = {L_z_prompt:.2f} cm")
|
print(f"\nPrompt critical height L_z = {L_z_prompt:.2f} cm")
|
||||||
|
|
||||||
# Verify
|
# Verify
|
||||||
B_sq_prompt_check = (np.pi / L_x_val)**2 + (np.pi / L_y_val)**2 + (np.pi / L_z_prompt)**2
|
B_sq_prompt_check = (
|
||||||
k_eff_prompt_check = k_inf / ((L_squared_fast * B_sq_prompt_check + 1) * (L_squared_th * B_sq_prompt_check + 1))
|
(np.pi / L_x_val) ** 2 + (np.pi / L_y_val) ** 2 + (np.pi / L_z_prompt) ** 2
|
||||||
|
)
|
||||||
|
k_eff_prompt_check = k_inf / (
|
||||||
|
(L_squared_fast * B_sq_prompt_check + 1)
|
||||||
|
* (L_squared_th * B_sq_prompt_check + 1)
|
||||||
|
)
|
||||||
print(f"\nVerification: k_eff = {k_eff_prompt_check:.6f}")
|
print(f"\nVerification: k_eff = {k_eff_prompt_check:.6f}")
|
||||||
print(f"Difference from delayed critical: {L_z_prompt - L_z_critical:.2f} cm")
|
print(f"Difference from delayed critical: {L_z_prompt - L_z_critical:.2f} cm")
|
||||||
else:
|
else:
|
||||||
print("No positive real solution found!")
|
print("No positive real solution found!")
|
||||||
|
|
||||||
# Problem 2F: Effect of people near trough
|
|
||||||
print(f"\n=== Problem 2F: Effect of People Near Trough ===")
|
|
||||||
print(f"If flux is not zero at edges and people are nearby:")
|
|
||||||
print(f" - People act as neutron reflectors/scatterers (water in human body)")
|
|
||||||
print(f" - This reduces neutron leakage from the trough")
|
|
||||||
print(f" - Reduced leakage → system becomes MORE reactive")
|
|
||||||
print(f" - To achieve criticality, LESS fissile material is needed")
|
|
||||||
print(f"\nConclusion: Critical height would DECREASE")
|
|
||||||
print(f" The actual critical height with people nearby < {L_z_critical:.2f} cm (bare reactor)")
|
|
||||||
print(f" This is a SAFETY CONCERN - human reflection can inadvertently make")
|
|
||||||
print(f" systems critical at lower liquid levels than calculated!")
|
|
||||||
|
|||||||
@ -25,7 +25,7 @@ print(f"The Start Up Rate is: {sur:.3f}")
|
|||||||
D_POWER = 2.5 # %
|
D_POWER = 2.5 # %
|
||||||
D_T_AVG = 4 # degrees
|
D_T_AVG = 4 # degrees
|
||||||
HEAT_UP_RATE = 0.15 # F/s
|
HEAT_UP_RATE = 0.15 # F/s
|
||||||
ALPHA_F = 10 # pcm/%power
|
ALPHA_F = -10 # pcm/%power (negative feedback)
|
||||||
|
|
||||||
rho_rod = rho
|
rho_rod = rho
|
||||||
|
|
||||||
|
|||||||
99
Class_Work/nuce2101/exam2/python/problem4.py
Normal file
99
Class_Work/nuce2101/exam2/python/problem4.py
Normal file
@ -0,0 +1,99 @@
|
|||||||
|
import numpy as np
|
||||||
|
import matplotlib.pyplot as plt
|
||||||
|
|
||||||
|
# Problem 4: PWR Power Increase Transient (25% → 50% Steam Load)
|
||||||
|
|
||||||
|
print("=" * 70)
|
||||||
|
print("Problem 4: PWR Power Increase Transient")
|
||||||
|
print("=" * 70)
|
||||||
|
|
||||||
|
# Given Initial Conditions (25% power steady state)
|
||||||
|
T_ave_initial = 500 # °F - Average temperature
|
||||||
|
T_h_initial = 510 # °F - Hot leg temperature
|
||||||
|
T_c_initial = 490 # °F - Cold leg temperature
|
||||||
|
P_initial = 25 # % - Initial power level
|
||||||
|
P_final = 50 # % - Final power level
|
||||||
|
|
||||||
|
print("\n--- Initial Conditions (25% Power) ---")
|
||||||
|
print(f"T_ave = {T_ave_initial} °F")
|
||||||
|
print(f"T_h = {T_h_initial} °F")
|
||||||
|
print(f"T_c = {T_c_initial} °F")
|
||||||
|
print(f"Power = {P_initial}%")
|
||||||
|
|
||||||
|
# Calculate initial ΔT across core
|
||||||
|
DeltaT_initial = T_h_initial - T_c_initial
|
||||||
|
print(f"ΔT across core = {DeltaT_initial} °F")
|
||||||
|
|
||||||
|
print("\n" + "=" * 70)
|
||||||
|
print("Part A: Chronological Physical Impacts")
|
||||||
|
print("=" * 70)
|
||||||
|
|
||||||
|
print("\n" + "=" * 70)
|
||||||
|
print("Part B: Final State Comparison")
|
||||||
|
print("=" * 70)
|
||||||
|
|
||||||
|
# For a PWR at steady state:
|
||||||
|
# Power ∝ mass flow rate × Cp × ΔT
|
||||||
|
# If flow rate is constant (no pump speed change):
|
||||||
|
# P_new / P_old = ΔT_new / ΔT_old
|
||||||
|
|
||||||
|
# Calculate new ΔT
|
||||||
|
DeltaT_final = DeltaT_initial * (P_final / P_initial)
|
||||||
|
print(f"\nΔT calculation:")
|
||||||
|
print(f"ΔT_initial = {DeltaT_initial} °F at {P_initial}% power")
|
||||||
|
print(f"ΔT_final = ΔT_initial × (P_final/P_initial)")
|
||||||
|
print(f"ΔT_final = {DeltaT_initial} × ({P_final}/{P_initial}) = {DeltaT_final} °F")
|
||||||
|
|
||||||
|
# Estimate final temperatures
|
||||||
|
# Assume T_ave returns to approximately initial value (no automatic control,
|
||||||
|
# but moderator feedback will drive it back toward equilibrium)
|
||||||
|
# This is a simplification - in reality, T_ave might be slightly different
|
||||||
|
|
||||||
|
# Method 1: Assume T_ave = constant (strong moderator feedback)
|
||||||
|
T_ave_final_method1 = T_ave_initial
|
||||||
|
T_h_final_method1 = T_ave_final_method1 + DeltaT_final / 2
|
||||||
|
T_c_final_method1 = T_ave_final_method1 - DeltaT_final / 2
|
||||||
|
|
||||||
|
print(f"\n--- Method 1: Assume T_ave constant (strong moderator feedback) ---")
|
||||||
|
print(f"T_ave_final = {T_ave_final_method1} °F")
|
||||||
|
print(
|
||||||
|
f"T_h_final = T_ave + ΔT/2 = {T_ave_final_method1} + {DeltaT_final/2} = {T_h_final_method1} °F"
|
||||||
|
)
|
||||||
|
print(
|
||||||
|
f"T_c_final = T_ave - ΔT/2 = {T_ave_final_method1} - {DeltaT_final/2} = {T_c_final_method1} °F"
|
||||||
|
)
|
||||||
|
|
||||||
|
# Method 2: More realistic - Tc drops somewhat, Th rises more
|
||||||
|
# The SG cooling will lower Tc, but not back to initial Tc
|
||||||
|
# Let's estimate Tc drops by ~5-10°F, then Th rises to accommodate ΔT
|
||||||
|
T_c_drop = 5 # °F - estimated drop in Tc due to increased SG cooling
|
||||||
|
T_c_final_method2 = T_c_initial - T_c_drop
|
||||||
|
T_h_final_method2 = T_c_final_method2 + DeltaT_final
|
||||||
|
T_ave_final_method2 = (T_h_final_method2 + T_c_final_method2) / 2
|
||||||
|
|
||||||
|
print(f"\n--- Method 2: Estimate Tc drop due to SG cooling ---")
|
||||||
|
print(f"Assume Tc drops by ~{T_c_drop}°F due to increased steam flow")
|
||||||
|
print(f"T_c_final = {T_c_final_method2} °F")
|
||||||
|
print(
|
||||||
|
f"T_h_final = T_c + ΔT = {T_c_final_method2} + {DeltaT_final} = {T_h_final_method2} °F"
|
||||||
|
)
|
||||||
|
print(f"T_ave_final = {T_ave_final_method2} °F")
|
||||||
|
|
||||||
|
print(f"\n--- Comparison Table ---")
|
||||||
|
print(
|
||||||
|
f"{'Parameter':<20} {'Initial (25%)':<20} {'Final (50%) M1':<20} {'Final (50%) M2':<20}"
|
||||||
|
)
|
||||||
|
print(f"{'-'*80}")
|
||||||
|
print(f"{'Power':<20} {P_initial:<20} {P_final:<20} {P_final:<20}")
|
||||||
|
print(
|
||||||
|
f"{'T_h (°F)':<20} {T_h_initial:<20} {T_h_final_method1:<20.1f} {T_h_final_method2:<20.1f}"
|
||||||
|
)
|
||||||
|
print(
|
||||||
|
f"{'T_c (°F)':<20} {T_c_initial:<20} {T_c_final_method1:<20.1f} {T_c_final_method2:<20.1f}"
|
||||||
|
)
|
||||||
|
print(
|
||||||
|
f"{'T_ave (°F)':<20} {T_ave_initial:<20} {T_ave_final_method1:<20.1f} {T_ave_final_method2:<20.1f}"
|
||||||
|
)
|
||||||
|
print(
|
||||||
|
f"{'ΔT (°F)':<20} {DeltaT_initial:<20} {DeltaT_final:<20.1f} {DeltaT_final:<20.1f}"
|
||||||
|
)
|
||||||
Loading…
x
Reference in New Issue
Block a user