Equivalence ratio

This example demonstrates how to set a mixture according to equivalence ratio and mixture fraction.

Requires: cantera >= 2.6.0

Tags: Python 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}")

phi = 1.000

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}")

phi = 1.000

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}")

Z = 0.055

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}")

Z(Bilger mixture fraction) = 0.055
Z(mixture fraction based on C) = 0.055

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}")

mass fraction of CH4 = 0.055

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}")

mass fraction of CH4 = 0.055

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}")

phi(burnt) = 1.000
Z(burnt) = 0.055

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

phi = 2.500

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}")

phi = 2.047

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

mole fraction of H2O = 0.300
ratio of H2/O2: 4.000

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

mole fraction of H2 = 0.100

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

phi = 2.000

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

phi = 2.000