# ---
# jupyter:
#   jupytext:
#     text_representation:
#       extension: .py
#       format_name: light
#       format_version: '1.5'
#       jupytext_version: 1.14.4
#   kernelspec:
#     display_name: Python 3
#     language: python
#     name: python3
# ---

# +
import numpy as np
import scipy.constants as co
import matplotlib.pyplot as plt
from bolosKhai import parser, grid, solver2
np.seterr(divide='ignore', invalid='ignore')
# Create an energy grid for Boltzmann Solver
# This energy grid has unit in eV
gr = grid.QuadraticGrid(0, 20, 200)
bsolver = solver2.BoltzmannSolver(gr)

# Import data file, which contains the cross section data.
with open('Cross Section.dat') as fp:
    processes = parser.parse(fp)
processes = bsolver.load_collisions(processes)
bsolver.target['CH4'].density = 0.5
bsolver.target['Ar'].density = 0.5

##################################################
# INPUT
# We have electric field:
# E = E0 * exp (i * Omega * t)
bsolver.OmegaN = 0.10000E-11        # Omega / N
bsolver.kT = 400 * co.k / co.eV     # Gas - Temperature 400 K

# GUESS by Maxwell distribution function.
# Here we are starting with
# with an electron temperature of 6 eV
f0 = bsolver.maxwell(6.0)
mean_max = bsolver.mean_energy(f0)

def EEDF_AC(EN, f0):
    bsolver.grid = gr
    bsolver.EN = EN * solver2.TOWNSEND
    # After change any parameter we must initial the solver
    bsolver.init()
    f1 = bsolver.converge(f0, maxn=200, rtol=1e-4)
    mean1 = bsolver.mean_energy(f1)
    print('E/N = %.0f Td' % EN)
    print('Mean Energy 1 = %.4f  eV' % (mean1))

    # Get new grid
    newgrid = grid.QuadraticGrid(0, 10 * mean1, 200)
    bsolver.grid = newgrid
    bsolver.init()

    # Interpolate the previous EEDF over new grid
    f2 = bsolver.grid.interpolate(f1, gr)
    mean2 = bsolver.mean_energy(f2)

    # Find final EEDF
    f3 = bsolver.converge(f2, maxn=200, rtol=1e-5)
    mean3 = bsolver.mean_energy(f3)
    print('Mean Energy Inter-EEDF = %.4f eV' % mean2)
    print('Mean Energy Final-EEDF = %.4f eV \n'  % mean3)

    grid_EEDF = bsolver.cenergy
    return f3, grid_EEDF

# Range of Electric field / Number of electron - E0/N
# E = E0 * exp (i * Omega * t)
EN = np.linspace(1000,2000,11)
plt.figure()

for i in EN:
    EEDF, gr_EEDF = EEDF_AC(i, f0)
    plt.plot(gr_EEDF, EEDF, label='%.f Td' % i)

plt.xlabel('Mean Energy (eV)')
plt.ylabel('EEDF (eV$^\mathdefault{3/2}$)')
plt.xlim([0,3])
plt.ylim([0,0.85])
plt.legend()
plt.show()
# -



StarcoderdataJupyterScriptsDedupFiltered