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c49c95d666
Introduce a monkey patch to avoid executing specific examples that won't run under sphinx-gallery.
155 lines
7.1 KiB
Python
155 lines
7.1 KiB
Python
"""
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Equivalence ratio
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=================
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This example demonstrates how to set a mixture according to equivalence ratio
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and mixture fraction.
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Requires: cantera >= 2.6.0
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.. tags:: Python, combustion, thermodynamics, mixture
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"""
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import cantera as ct
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gas = ct.Solution('gri30.yaml')
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# %%
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# Define the oxidizer composition, here air with 21 mol-% O2 and 79 mol-% N2
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air = "O2:0.21,N2:0.79"
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# %%
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# Set the mixture composition according to the stoichiometric mixture
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# (equivalence ratio phi = 1). The fuel composition in this example
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# is set to 100 mol-% CH4 and the oxidizer to 21 mol-% O2 and 79 mol-% N2.
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# This function changes the composition of the gas object and keeps temperature
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# and pressure constant
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gas.set_equivalence_ratio(phi=1.0, fuel="CH4:1", oxidizer=air)
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# %%
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# If fuel or oxidizer consist of a single species, a short hand notation can be
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# used, for example ``fuel="CH4"`` is equivalent to ``fuel="CH4:1"``.
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# By default, the compositions of fuel and oxidizer are interpreted as mole
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# fractions. If the compositions are given in mass fractions, an
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# additional argument can be provided. Here, the fuel is 100 mass-% CH4
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# and the oxidizer is 23.3 mass-% O2 and 76.7 mass-% N2
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gas.set_equivalence_ratio(1.0, fuel="CH4:1", oxidizer="O2:0.233,N2:0.767", basis='mass')
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# %%
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# This function can be used to compute the equivalence ratio for any mixture.
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# The first two arguments specify the compositions of the fuel and oxidizer.
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# An optional third argument ``basis``` indicates if fuel and oxidizer compositions
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# are provided in terms of mass or mole fractions. Default is mole fractions.
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# Note that for all functions shown here, the compositions are normalized
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# internally so the species fractions do not have to sum to unity
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phi = gas.equivalence_ratio(fuel="CH4:1", oxidizer="O2:233,N2:767", basis='mass')
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print(f"phi = {phi:1.3f}")
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# %%
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# If the compositions of fuel and oxidizer are unknown, the function can
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# be called without arguments. This assumes that all C, H and S atoms come from
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# the fuel and all O atoms from the oxidizer. In this example, the fuel was set
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# to be pure CH4 and the oxidizer ``O2:0.233, N2:0.767`` so that the assumption is true
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# and the same equivalence ratio as above is computed
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phi = gas.equivalence_ratio()
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print(f"phi = {phi:1.3f}")
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# %%
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# Instead of working with equivalence ratio, mixture fraction can be used.
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# The mixture fraction is always kg fuel / (kg fuel + kg oxidizer), independent
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# of the basis argument. For example, the mixture fraction Z can be computed as
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# follows. Again, the compositions by default are interpreted as mole fractions
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Z = gas.mixture_fraction(fuel="CH4:1", oxidizer=air)
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print(f"Z = {Z:1.3f}")
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# %%
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# By default, the mixture fraction is the Bilger mixture fraction. Instead,
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# a mixture fraction based on a single element can be used. In this example,
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# the following two ways of computing Z are the same:
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Z = gas.mixture_fraction(fuel="CH4:1", oxidizer=air, element="Bilger")
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print(f"Z(Bilger mixture fraction) = {Z:1.3f}")
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Z = gas.mixture_fraction(fuel="CH4:1", oxidizer=air, element="C")
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print(f"Z(mixture fraction based on C) = {Z:1.3f}")
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# %%
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# Since the fuel in this example is pure methane and the oxidizer is air,
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# the mixture fraction is the same as the mass fraction of methane in the mixture
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print(f"mass fraction of CH4 = {gas['CH4'].Y[0]:1.3f}")
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# %%
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# To set a mixture according to the mixture fraction, the following function
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# can be used. In this example, the final fuel/oxidizer mixture
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# contains 5.5 mass-% CH4:
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gas.set_mixture_fraction(0.055, fuel="CH4:1", oxidizer=air)
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print(f"mass fraction of CH4 = {gas['CH4'].Y[0]:1.3f}")
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# %%
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# Mixture fraction and equivalence ratio are invariant to the reaction progress.
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# For example, they stay constant if the mixture composition changes to the burnt
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# state or for any intermediate state. Fuel and oxidizer compositions for all functions
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# shown in this example can be given as string, dictionary or numpy array
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fuel = {"CH4":1} # provide the fuel composition as dictionary instead of string
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gas.set_equivalence_ratio(1, fuel, air)
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gas.equilibrate('HP')
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phi_burnt = gas.equivalence_ratio(fuel, air)
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Z_burnt = gas.mixture_fraction(fuel, air)
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print(f"phi(burnt) = {phi_burnt:1.3f}")
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print(f"Z(burnt) = {Z_burnt:1.3f}")
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# %%
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# If fuel and oxidizer compositions are specified consistently, then
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# ``equivalence_ratio`` and ``set_equivalence_ratio`` are consistent as well, as
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# shown in the following example with arbitrary fuel and oxidizer compositions:
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gas.set_equivalence_ratio(2.5, fuel="CH4:1,O2:0.01,CO:0.05,N2:0.1",
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oxidizer="O2:0.2,N2:0.8,CO2:0.05,CH4:0.01")
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phi = gas.equivalence_ratio(fuel="CH4:1,O2:0.01,CO:0.05,N2:0.1",
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oxidizer="O2:0.2,N2:0.8,CO2:0.05,CH4:0.01")
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print(f"phi = {phi:1.3f}") # prints 2.5
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# %%
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# Without specifying the fuel and oxidizer compositions, it is assumed that
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# all C, H and S atoms come from the fuel and all O atoms from the oxidizer,
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# which is not true for this example. Therefore, the following call gives a
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# different equivalence ratio based on that assumption
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phi = gas.equivalence_ratio()
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print(f"phi = {phi:1.3f}")
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# %%
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# After computing the mixture composition for a certain equivalence ratio given
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# a fuel and mixture composition, the mixture can optionally be diluted. The
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# following function will first create a mixture with equivalence ratio 2 from pure
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# hydrogen and oxygen and then dilute it with H2O. In this example, the final mixture
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# consists of 30 mol-% H2O and 70 mol-% of the H2/O2 mixture at phi=2
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gas.set_equivalence_ratio(2.0, "H2:1", "O2:1", diluent="H2O", fraction={"diluent":0.3})
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print(f"mole fraction of H2O = {gas['H2O'].X[0]:1.3f}") # mixture contains 30 mol-% H2O
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print(f"ratio of H2/O2: {gas['H2'].X[0] / gas['O2'].X[0]:1.3f}") # according to phi=2
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# %%
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# Another option is to specify the fuel or oxidizer fraction in the final mixture.
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# The following example creates a mixture with equivalence ratio 2 from pure
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# hydrogen and oxygen (same as above) and then dilutes it with a mixture of 50 mass-%
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# CO2 and 50 mass-% H2O so that the mass fraction of fuel in the final mixture is 0.1
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gas.set_equivalence_ratio(2.0, "H2", "O2", diluent="CO2:0.5,H2O:0.5",
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fraction={"fuel":0.1}, basis="mass")
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print(f"mole fraction of H2 = {gas['H2'].Y[0]:1.3f}") # mixture contains 10 mass-% fuel
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# %%
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# To compute the equivalence ratio given a diluted mixture, a list of
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# species names can be provided which will be considered for computing phi.
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# In this example, the diluents H2O and CO2 are ignored and only H2 and O2 are
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# considered to get the equivalence ratio
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phi = gas.equivalence_ratio(fuel="H2", oxidizer="O2", include_species=["H2", "O2"])
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print(f"phi = {phi:1.3f}") # prints 2
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# %%
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# If instead the diluent should be included in the computation of the equivalence ratio,
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# the mixture can be set in the following way. Assume the fuel is diluted with
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# 50 mol-% H2O:
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gas.set_equivalence_ratio(2.0, fuel="H2:0.5,H2O:0.5", oxidizer=air)
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# %%
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# This creates a mixture with the specified equivalence ratio including the diluent:
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phi = gas.equivalence_ratio(fuel="H2:0.5,H2O:0.5", oxidizer=air)
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print(f"phi = {phi:1.3f}") # prints 2
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