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

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)")