—
Do we get the same equilibrium with and without inverse reactions?#
This example checks if the equilibrium of two mechanisms is the same:
one with inverse reactions,
one without inverse reactions.
Key formula used for reverse rate constant calculation:
\[k_b = \frac{k_f}{K_{eq}}\]
where:
\(k_b\): backward rate constant
\(k_f\): forward rate constant
\(K_{eq}\): equilibrium constant
The equilibrium constant is computed from Gibbs free energy:
\[K_{eq} = \exp\left(-\frac{\Delta G}{R T}\right)\]
where:
\(\Delta G\): Gibbs free energy change
\(R\): universal gas constant
\(T\): temperature
For two-temperature plasma reactions, in equilibrium, the electron temperature is equal to the gas temperature, and the rate constant is computed using the above formula.
# This is an option for the online documentation, so that each image is displayed separately.
# sphinx_gallery_multi_image = "single"
# sphinx_gallery_thumbnail_number = 9
Import the required libraries.#
from dataclasses import dataclass, field
import cantera as ct
import matplotlib.pyplot as plt
import numpy as np
from adjustText import adjust_text
from rizer.misc.ct_utils import load_goutier2025_mechanism_with_options
from rizer.misc.plt_utils import (
get_species_color,
get_species_in_latex,
get_text,
set_mpl_style,
)
from rizer.misc.utils import get_path_to_data
set_mpl_style(nb_columns=1)
Define mechanisms and parameters.#
gas_no_inverse = ct.Solution(
get_path_to_data(
"mechanisms",
"Goutier2025",
"builder",
"CH4_to_C2H2_forward_reactions.yaml",
),
name="gas",
)
gas_arrhenius_rates = load_goutier2025_mechanism_with_options(
thermo="gas",
electron_energy_distribution="maxwellian",
electronic_excited_states="all",
radiative_ionization_included=True,
)
gas_druyvesteyn_rates = load_goutier2025_mechanism_with_options(
thermo="gas",
electron_energy_distribution="Druyvesteyn",
electronic_excited_states="all",
radiative_ionization_included=True,
)
# Temperatures in Kelvin.
# For the mechanism with inverse reactions, we can go up to 50000 K,
# because the equilibrium is reached and the results are reliable.
temperatures_with_inverse = np.array(
[
1000,
2000,
5000,
10000,
15000,
20000,
30000,
40000,
50000,
]
)
# For the mechanism without inverse reactions, we cannot go higher than 15000 K,
# because the equilibrium is not reached and the results are not reliable.
temperatures_no_inverse = temperatures_with_inverse[temperatures_with_inverse < 15000]
# Pressure in Pascals.
P = ct.one_atm
# Initial mole fraction (here, we start with pure methane).
X_0 = "CH4:1"
# Species to plot (colors are assigned via get_species_color for consistency across examples).
species_to_plot = [
"CH4",
"CH4+",
"CH3",
# "CH3+",
# "CH2",
# "CH",
"H2",
"H",
"H(n=2)",
"H(n=3)",
"H(n=4)",
"H+",
"C",
"C(1D)",
"C(1S)",
"C(5So)",
"C+",
"C++",
# "C+++",
"e-",
"C2H2",
# "C2H",
"C2",
]
# Parameters for the reactor network.
simulation_time = 1e-2 # Simulation time in seconds
rtol = 1e-11 # Relative tolerance for the reactor network
max_steps = 100_000 # Maximum steps for the reactor network
max_time_step = 1e-5 # Maximum time step for the reactor network
Define the state groups for the different mechanisms.#
@dataclass
class StateGroup:
solution: ct.Solution
temperatures: np.ndarray
linestyle: str
linewidth: float
states: list[ct.SolutionArray] = field(default_factory=list)
no_inverse = StateGroup(
solution=gas_no_inverse,
temperatures=temperatures_no_inverse,
linestyle="--",
linewidth=5,
)
with_arrhenius_rates = StateGroup(
solution=gas_arrhenius_rates,
temperatures=temperatures_with_inverse,
linestyle="-",
linewidth=6,
)
with_druyvesteyn_rates = StateGroup(
solution=gas_druyvesteyn_rates,
temperatures=temperatures_with_inverse,
linestyle="-.",
linewidth=5,
)
all_states: list[StateGroup] = [
no_inverse,
with_arrhenius_rates,
with_druyvesteyn_rates,
]
Compute equilibrium and time evolution for each mechanism and each temperature.#
# Compute the equilibrium for the mechanism with inverse reactions.
states_eq = ct.SolutionArray(gas_arrhenius_rates, shape=temperatures_with_inverse.shape)
states_eq.TPX = temperatures_with_inverse, P, X_0
states_eq.equilibrate("TP")
# Compute the time evolution for each mechanism and each temperature.
for state_group in all_states:
print(f"Running equilibrium for {state_group.solution.name} mechanism")
for T in state_group.temperatures:
print(f"Running equilibrium for T={T} K")
# Initialize the reactor network for the current temperature.
state_group.solution.TPX = T, P, X_0
# `clone=False`: keep the reactor's integrated state on `state_group.solution`
# itself (the default `clone=True`, since Cantera 3.3, integrates an
# independent copy instead, leaving `state_group.solution` frozen at its
# initial state -- see the `states.append(state_group.solution.state, ...)`
# below).
r1 = ct.IdealGasConstPressureReactor(
state_group.solution, energy="off", clone=False
)
sim = ct.ReactorNet([r1])
sim.rtol = rtol # Set relative tolerance for the reactor network.
sim.max_time_step = max_time_step
sim.reinitialize()
# Store the states in a SolutionArray.
states = ct.SolutionArray(state_group.solution, 1, extra={"t": [0.0]})
i = 0
while sim.time < simulation_time:
# Advance the reactor network by one step.
sim.step()
# Append the current state to the SolutionArray.
states.append(state_group.solution.state, t=sim.time) # type: ignore
i += 1
if i % 1000 == 0:
print(f"{sim.time:.2e} / {simulation_time:.2e}")
state_group.states.append(states)
Running equilibrium for gas mechanism
Running equilibrium for T=1000 K
9.89e-03 / 1.00e-02
Running equilibrium for T=2000 K
1.14e-04 / 1.00e-02
2.62e-03 / 1.00e-02
Running equilibrium for T=5000 K
5.89e-08 / 1.00e-02
2.57e-06 / 1.00e-02
1.57e-03 / 1.00e-02
Running equilibrium for T=10000 K
4.45e-09 / 1.00e-02
3.71e-07 / 1.00e-02
6.02e-06 / 1.00e-02
6.83e-03 / 1.00e-02
Running equilibrium for mechanism
Running equilibrium for T=1000 K
9.88e-03 / 1.00e-02
Running equilibrium for T=2000 K
9.49e-05 / 1.00e-02
2.36e-03 / 1.00e-02
Running equilibrium for T=5000 K
6.60e-08 / 1.00e-02
3.22e-06 / 1.00e-02
1.81e-03 / 1.00e-02
Running equilibrium for T=10000 K
4.71e-09 / 1.00e-02
3.75e-07 / 1.00e-02
3.14e-06 / 1.00e-02
3.73e-05 / 1.00e-02
7.57e-03 / 1.00e-02
Running equilibrium for T=15000 K
1.84e-09 / 1.00e-02
8.79e-08 / 1.00e-02
5.76e-07 / 1.00e-02
1.27e-05 / 1.00e-02
7.84e-03 / 1.00e-02
Running equilibrium for T=20000 K
9.95e-10 / 1.00e-02
3.57e-08 / 1.00e-02
1.90e-07 / 1.00e-02
3.91e-04 / 1.00e-02
Running equilibrium for T=30000 K
3.76e-10 / 1.00e-02
1.17e-08 / 1.00e-02
9.07e-08 / 1.00e-02
1.42e-04 / 1.00e-02
9.47e-03 / 1.00e-02
Running equilibrium for T=40000 K
2.22e-10 / 1.00e-02
3.27e-09 / 1.00e-02
2.94e-08 / 1.00e-02
5.46e-06 / 1.00e-02
5.66e-03 / 1.00e-02
Running equilibrium for T=50000 K
7.81e-11 / 1.00e-02
1.75e-09 / 1.00e-02
8.97e-09 / 1.00e-02
8.20e-07 / 1.00e-02
2.08e-03 / 1.00e-02
Running equilibrium for mechanism
Running equilibrium for T=1000 K
9.88e-03 / 1.00e-02
Running equilibrium for T=2000 K
1.18e-04 / 1.00e-02
2.61e-03 / 1.00e-02
Running equilibrium for T=5000 K
6.60e-08 / 1.00e-02
3.22e-06 / 1.00e-02
1.21e-03 / 1.00e-02
Running equilibrium for T=10000 K
4.71e-09 / 1.00e-02
3.76e-07 / 1.00e-02
3.54e-06 / 1.00e-02
5.49e-05 / 1.00e-02
7.97e-03 / 1.00e-02
Running equilibrium for T=15000 K
1.84e-09 / 1.00e-02
9.12e-08 / 1.00e-02
9.50e-07 / 1.00e-02
2.40e-05 / 1.00e-02
6.90e-03 / 1.00e-02
Running equilibrium for T=20000 K
9.95e-10 / 1.00e-02
4.17e-08 / 1.00e-02
5.58e-07 / 1.00e-02
3.42e-04 / 1.00e-02
Running equilibrium for T=30000 K
3.74e-10 / 1.00e-02
1.07e-08 / 1.00e-02
1.34e-07 / 1.00e-02
2.33e-03 / 1.00e-02
Running equilibrium for T=40000 K
1.87e-10 / 1.00e-02
2.99e-09 / 1.00e-02
2.96e-08 / 1.00e-02
1.10e-06 / 1.00e-02
7.16e-03 / 1.00e-02
Running equilibrium for T=50000 K
7.24e-11 / 1.00e-02
1.96e-09 / 1.00e-02
8.78e-09 / 1.00e-02
1.56e-07 / 1.00e-02
2.74e-03 / 1.00e-02
Plotting the results.#
for i, T in enumerate(temperatures_with_inverse):
print(f"Plotting results for T={T} K")
fig, ax = plt.subplots()
# Storage for the maximum mole fraction with the corresponding time
# of each species across all states.
# The third value is the value in the equilibrium.
max_fractions: dict[str, tuple[float, float, float]] = {
species: (
0.0, # Maximum mole fraction
0.0, # Corresponding time
0.0, # Value in the equilibrium
)
for species in species_to_plot
}
max_fractions["Other Species"] = (0.0, 0.0, 0.0)
for j, state_group in enumerate(all_states):
if i >= len(state_group.temperatures):
continue
states = state_group.states[i]
other_species = np.ones_like(states.t) * 100.0 # type: ignore
for species in species_to_plot:
if species not in state_group.solution.species_names:
continue
# Plot the time evolution of the mole fraction.
x = states.t # type: ignore
y = states(species).X * 100
other_species -= y[:, 0]
color = get_species_color(species)
ax.plot(
x,
y,
linestyle=state_group.linestyle,
linewidth=state_group.linewidth,
alpha=0.9,
color=color,
)
# Add the species name at the maximum point, with the same color as the line.
max_x = x[y.argmax()]
if max_x < 1e-14:
max_x = 1e-14
max_y = y.max()
# Update the maximum fraction and corresponding time for the species.
if j != 0:
# For the mechanism without inverse reactions, we cannot trust the results,
# so we skip the update of the maximum fraction.
if max_y > max_fractions[species][0]:
max_fractions[species] = (
max_y,
max_x,
states_eq[i](species).X[-1] * 100,
)
# Plot the remaining species as "Other Species"
ax.plot(
states.t, # type: ignore
other_species,
linestyle=state_group.linestyle,
linewidth=state_group.linewidth,
alpha=0.7,
color="black",
)
# Add the species name at the maximum point, with the same color as the line.
max_x = states.t[np.argmax(other_species)] # type: ignore
if max_x < 1e-14:
max_x = 1e-14
max_y = np.max(other_species)
# Update the maximum fraction and corresponding time for "Other Species".
if max_y > max_fractions["Other Species"][0]:
max_fractions["Other Species"] = (max_y, max_x, 0.0)
# Plot the equilibrium values as horizontal lines.
for species in species_to_plot:
if species not in states_eq.species_names:
continue
y_eq = states_eq[i](species).X * 100
color = get_species_color(species)
ax.hlines(
y=y_eq,
xmin=1e-14,
xmax=simulation_time,
linestyle=":",
color=color,
alpha=0.7,
linewidth=6,
)
# Add a text label for the species.
# For species present in the equilibrium (with a mole fraction above 1e-5),
# add it at the end of the simulation.
# For other species, add it at the maximum of all the states (unless their
# maximum mole fraction is below 1e-5, in which case we skip them).
texts = []
for species, (max_y, max_x, y_eq) in max_fractions.items():
if max_y < 1e-5 and species != "Other Species":
continue # Skip species with a maximum mole fraction below 1e-5.
if y_eq > 1e-5:
x = simulation_time
y = y_eq
else:
x = max_x
y = max_y
text = get_text(
x,
y,
(
get_species_in_latex(species)
if species != "Other Species"
else "Other Species"
),
ax=ax,
color=get_species_color(species) if species != "Other Species" else "black",
)
texts.append(text)
ax.set_title(f"$T_e=T_g={T}$ K, P={int(P / ct.one_atm):.1f} atm")
ax.set_xlabel("Time [s]")
ax.set_xlim(left=1e-14, right=simulation_time)
ax.set_xscale("log")
ax.set_ylabel("Mole fraction [%]")
ax.set_ylim(bottom=1e-5, top=100)
ax.set_yscale("log")
adjust_text(
texts,
avoid_self=False,
expand=(
1.2,
2,
), # expand text bounding boxes by 1.2 fold in x direction and 2 fold in y direction
arrowprops=dict(
arrowstyle="->", color="red"
), # ensure the labeling is clear by adding arrows
)
plt.show()
Plotting results for T=1000 K
Plotting results for T=2000 K
Plotting results for T=5000 K
Plotting results for T=10000 K
Plotting results for T=15000 K
Plotting results for T=20000 K
Plotting results for T=30000 K
Plotting results for T=40000 K
Plotting results for T=50000 K
Total running time of the script: (0 minutes 22.207 seconds)