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Difference in Molar Enthalpies between EoS and DFT #68
Description
Contributions.IdealGas uses different methods for EoS and DFT (see below).
Joback gives very small values for the IdealGas molar enthalpy (as fat as I tested all in the range ~1e-13 * JOULE).
import numpy as np
from feos.si import *
from feos.pcsaft import *
from feos.eos import EquationOfState, Contributions
from feos.eos import State as EosState
from feos.eos import PhaseEquilibrium as EosPhaseEquilibrium
from feos.dft import HelmholtzEnergyFunctional, PhaseEquilibrium
from feos.dft import State as DftState
from feos.dft import PhaseEquilibrium as DftPhaseEquilibrium
substances = ['nitrogen', 'methane']
params = PcSaftParameters.from_json(substances=substances, pure_path='pure_parameters.json', binary_path='binary_parameters.json', search_option=IdentifierOption.Name)
eos = EquationOfState.pcsaft(params)
dft = HelmholtzEnergyFunctional.pcsaft(params)
vle_eos = EosPhaseEquilibrium.bubble_point(eos, temperature_or_pressure=110*KELVIN, liquid_molefracs=np.array([0.1, 0.9]))
vle_dft = DftPhaseEquilibrium.bubble_point(dft, temperature_or_pressure=110*KELVIN, liquid_molefracs=np.array([0.1, 0.9]))
print(vle_eos.liquid.helmholtz_energy_contributions()) # Gives QSPR as IdealGas contribution
print(vle_dft.liquid.helmholtz_energy_contributions()) # Gives Joback as IdealGas contribution
print(vle_eos.liquid.molar_enthalpy(contributions=Contributions.ResidualNvt) - vle_dft.liquid.molar_enthalpy(contributions=Contributions.ResidualNvt)) #Gives very good agreement in the ResidualNvt contribution
print(vle_eos.liquid.molar_enthalpy(contributions=Contributions.IdealGas) - vle_dft.liquid.molar_enthalpy(contributions=Contributions.IdealGas)) # Gives large discrepancy in IdealGas contribution
Activity
prehner commentedon Nov 8, 2022
Thanks for raising this issue. To be completely aware of what is happening, we would need the content of your
pure_parameters.jsonfile, but the fact that Joback gives ~0 and the other outputs suggest, that it does not contain Joback parameters. In this case the eos will fall back on the QSPR method whereas the functional will use Joback where all parameters are 0.It is indeed confusing behavior that we should attempt to make more transparent.
prehner commentedon Jan 12, 2023
Closed in favor of #109