# equivalenceRatio.py (Source)

```
"""
This example demonstrates how to set a mixture according to equivalence ratio
and mixture fraction.
Requires: cantera >= 2.6.0
Keywords: combustion, thermodynamics, mixture
"""
import cantera as ct
gas = ct.Solution('gri30.yaml')
# Define the oxidizer composition, here air with 21 mol-% O2 and 79 mol-% N2
air = "O2:0.21,N2:0.79"
# Set the mixture composition according to the stoichiometric mixture
# (equivalence ratio phi = 1). The fuel composition in this example
# is set to 100 mol-% CH4 and the oxidizer to 21 mol-% O2 and 79 mol-% N2.
# This function changes the composition of the gas object and keeps temperature
# and pressure constant
gas.set_equivalence_ratio(phi=1.0, fuel="CH4:1", oxidizer=air)
# If fuel or oxidizer consist of a single species, a short hand notation can be
# used, for example fuel="CH4" is equivalent to fuel="CH4:1".
# By default, the compositions of fuel and oxidizer are interpreted as mole
# fractions. If the compositions are given in mass fractions, an
# additional argument can be provided. Here, the fuel is 100 mass-% CH4
# and the oxidizer is 23.3 mass-% O2 and 76.7 mass-% N2
gas.set_equivalence_ratio(1.0, fuel="CH4:1", oxidizer="O2:0.233,N2:0.767", basis='mass')
# This function can be used to compute the equivalence ratio for any mixture.
# The first two arguments specify the compositions of the fuel and oxidizer.
# An optional third argument "basis" indicates if fuel and oxidizer compositions
# are provided in terms of mass or mole fractions. Default is mole fractions.
# Note that for all functions shown here, the compositions are normalized
# internally so the species fractions do not have to sum to unity
phi = gas.equivalence_ratio(fuel="CH4:1", oxidizer="O2:233,N2:767", basis='mass')
print(f"phi = {phi:1.3f}")
# If the compositions of fuel and oxidizer are unknown, the function can
# be called without arguments. This assumes that all C, H and S atoms come from
# the fuel and all O atoms from the oxidizer. In this example, the fuel was set
# to be pure CH4 and the oxidizer O2:0.233,N2:0.767 so that the assumption is true
# and the same equivalence ratio as above is computed
phi = gas.equivalence_ratio()
print(f"phi = {phi:1.3f}")
# Instead of working with equivalence ratio, mixture fraction can be used.
# The mixture fraction is always kg fuel / (kg fuel + kg oxidizer), independent
# of the basis argument. For example, the mixture fraction Z can be computed as
# follows. Again, the compositions by default are interpreted as mole fractions
Z = gas.mixture_fraction(fuel="CH4:1", oxidizer=air)
print(f"Z = {Z:1.3f}")
# By default, the mixture fraction is the Bilger mixture fraction. Instead,
# a mixture fraction based on a single element can be used. In this example,
# the following two ways of computing Z are the same:
Z = gas.mixture_fraction(fuel="CH4:1", oxidizer=air, element="Bilger")
print(f"Z(Bilger mixture fraction) = {Z:1.3f}")
Z = gas.mixture_fraction(fuel="CH4:1", oxidizer=air, element="C")
print(f"Z(mixture fraction based on C) = {Z:1.3f}")
# Since the fuel in this example is pure methane and the oxidizer is air,
# the mixture fraction is the same as the mass fraction of methane in the mixture
print(f"mass fraction of CH4 = {gas['CH4'].Y[0]:1.3f}")
# To set a mixture according to the mixture fraction, the following function
# can be used. In this example, the final fuel/oxidizer mixture
# contains 5.5 mass-% CH4:
gas.set_mixture_fraction(0.055, fuel="CH4:1", oxidizer=air)
print(f"mass fraction of CH4 = {gas['CH4'].Y[0]:1.3f}")
# Mixture fraction and equivalence ratio are invariant to the reaction progress.
# For example, they stay constant if the mixture composition changes to the burnt
# state or for any intermediate state. Fuel and oxidizer compositions for all functions
# shown in this example can be given as string, dictionary or numpy array
fuel = {"CH4":1} # provide the fuel composition as dictionary instead of string
gas.set_equivalence_ratio(1, fuel, air)
gas.equilibrate('HP')
phi_burnt = gas.equivalence_ratio(fuel, air)
Z_burnt = gas.mixture_fraction(fuel, air)
print(f"phi(burnt) = {phi_burnt:1.3f}")
print(f"Z(burnt) = {Z_burnt:1.3f}")
# If fuel and oxidizer compositions are specified consistently, then
# equivalence_ratio and set_equivalence_ratio are consistent as well, as
# shown in the following example with arbitrary fuel and oxidizer compositions:
gas.set_equivalence_ratio(2.5, fuel="CH4:1,O2:0.01,CO:0.05,N2:0.1",
oxidizer="O2:0.2,N2:0.8,CO2:0.05,CH4:0.01")
phi = gas.equivalence_ratio(fuel="CH4:1,O2:0.01,CO:0.05,N2:0.1",
oxidizer="O2:0.2,N2:0.8,CO2:0.05,CH4:0.01")
print(f"phi = {phi:1.3f}") # prints 2.5
# Without specifying the fuel and oxidizer compositions, it is assumed that
# all C, H and S atoms come from the fuel and all O atoms from the oxidizer,
# which is not true for this example. Therefore, the following call gives a
# different equivalence ratio based on that assumption
phi = gas.equivalence_ratio()
print(f"phi = {phi:1.3f}")
# After computing the mixture composition for a certain equivalence ratio given
# a fuel and mixture composition, the mixture can optionally be diluted. The
# following function will first create a mixture with equivalence ratio 2 from pure
# hydrogen and oxygen and then dilute it with H2O. In this example, the final mixture
# consists of 30 mol-% H2O and 70 mol-% of the H2/O2 mixture at phi=2
gas.set_equivalence_ratio(2.0, "H2:1", "O2:1", diluent="H2O", fraction={"diluent":0.3})
print(f"mole fraction of H2O = {gas['H2O'].X[0]:1.3f}") # mixture contains 30 mol-% H2O
print(f"ratio of H2/O2: {gas['H2'].X[0] / gas['O2'].X[0]:1.3f}") # according to phi=2
# Another option is to specify the fuel or oxidizer fraction in the final mixture.
# The following example creates a mixture with equivalence ratio 2 from pure
# hydrogen and oxygen (same as above) and then dilutes it with a mixture of 50 mass-%
# CO2 and 50 mass-% H2O so that the mass fraction of fuel in the final mixture is 0.1
gas.set_equivalence_ratio(2.0, "H2", "O2", diluent="CO2:0.5,H2O:0.5",
fraction={"fuel":0.1}, basis="mass")
print(f"mole fraction of H2 = {gas['H2'].Y[0]:1.3f}") # mixture contains 10 mass-% fuel
# To compute the equivalence ratio given a diluted mixture, a list of
# species names can be provided which will be considered for computing phi.
# In this example, the diluents H2O and CO2 are ignored and only H2 and O2 are
# considered to get the equivalence ratio
phi = gas.equivalence_ratio(fuel="H2", oxidizer="O2", include_species=["H2", "O2"])
print(f"phi = {phi:1.3f}") # prints 2
# If instead the diluent should be included in the computation of the equivalence ratio,
# the mixture can be set in the following way. Assume the fuel is diluted with
# 50 mol-% H2O:
gas.set_equivalence_ratio(2.0, fuel="H2:0.5,H2O:0.5", oxidizer=air)
# This creates a mixture with the specified equivalence ratio including the diluent:
phi = gas.equivalence_ratio(fuel="H2:0.5,H2O:0.5", oxidizer=air)
print(f"phi = {phi:1.3f}") # prints 2
```