Chemical equilibrium at constant temperature and pressure#
As a first example, let’s compute the equilibrium composition of a stoichiometric CH\(_4\)-ideal air mixture at 3000 K and 1.01325 bar (1 atm). To obtain the equilibrium composition, we have to define the chemical species that can appear in the final state. The combustion of typical hydrocarbon fuels, such as methane, is a complex process that involves hundreds of chemical species. However, the number of species that have a significant impact on the properties of the final mixture is much smaller. Combustion Toolbox provides a list of predefined species that can be used to compute the equilibrium composition of a mixture. The list of predefined species can be found in the routine list_species.m. Alternatively, refer to the Using predefined chemical systems tutorial section.
For a preliminary analysis, we will consider a set of 23 species:
CO\(_2\), CO, H\(_2\)O, H\(_2\), O\(_2\), N\(_2\), Cgr,
C\(_2\), C\(_2\)H\(_4\), CH, CH\(_3\), CH\(_4\), CN, H,
HCN, HCO, N, NH, NH\(_2\), NH\(_3\), NO, O, OH,
which typically appear in the combustion of hydrocarbon fuels with air. This set of species is defined in the routine list_species.m as Soot formation, which also includes Ar.
Defining chemical system#
To initialize the Combustion Toolbox (CT) with this set of species, type the following at the MATLAB prompt:
self = App('Soot formation');
self = App({'CO2', 'CO', 'H2O', 'H2', 'O2', 'N2', 'Cbgrb', ...
'C2', 'C2H4', 'CH', 'CH3', 'CH4', 'CN', 'H', ...
'HCN', 'HCO', 'N', 'NH', 'NH2', 'NH3', 'NO', 'O', 'OH'});
Note
This section is under development. Stay tuned!
Setting the initial thermodynamic state#
To indicate the temperature [K] and pressure [bar] of the initial mixture, type:
self = set_prop(self, 'TR', 300, 'pR', 1 * 1.01325);
Note
This section is under development. Stay tuned!
Setting the initial chemical composition#
To indicate the species and chemical composition in the initial mixture:
For a stochiometric CH\(_4\)-ideal air mixture:
self.PD.S_Fuel = {'CH4'};
self.PD.N_Fuel = 1;
self.PD.S_Oxidizer = {'N2', 'O2'};
self.PD.N_Oxidizer = [7.52, 2];
This is the same as specifying a unit value for the equivalence ratio:
self.PD.S_Fuel = {'CH4'};
self.PD.S_Oxidizer = {'N2', 'O2'};
self.PD.ratio_oxidizers_O2 = [79, 21]/21;
self = set_prop(self, 'phi', 1);
The last two lines of code establish the proportion of the oxidizers species over O\(_2\) and the equivalence ratio, respectively. The number of moles are calculated considering that the number of moles of fuel is one by default.
For a lean-to-rich CH\(_4\)-ideal air mixtures:
self.PD.S_Fuel = {'CH4'};
self.PD.S_Oxidizer = {'N2', 'O2'};
self.PD.ratio_oxidizers_O2 = [79, 21]/21;
self = set_prop(self, 'phi', 0.5:0.01:4);
The last two lines of code establish the proportion of the oxidizers species over O\(_2\) and the equivalence ratio, respectively. The number of moles are calculated considering that the number of moles of fuel is one by default.
Note
This section is under development. Stay tuned!
Set chemical transformation#
Depending on the problem you want to solve, you may need to configure additional inputs. In this case, to compute the equilibrium composition at the defined temperature and pressure (TP), we have to set these values as
self = set_prop(self, 'pP', self.PD.pR.value, 'TP', 3000);
Note
Here, the pressure of the final state has been defined using the defined value of the initial state, self.PD.pR.value.
Note
This section is under development. Stay tuned!
Perform chemical calculations#
To solve the aforementioned problem, run
self = solve_problem(self, 'TP');
Note
This section is under development. Stay tuned!
Results#
The results are contained in self.PS. By default, solve_problem.m prints the results through the command window, which gives for the stoichiometric case (phi=1):
COMPUTING Nº MOLES FROM EQUIVALENCE RATIO (PHI).
***********************************************************
-----------------------------------------------------------
Problem type: TP | phi = 1.0000
-----------------------------------------------------------
| REACTANTS | PRODUCTS
T [K] | 300.0000 | 3000.0000
p [bar] | 1.0132 | 1.0132
r [kg/m3] | 1.1225 | 0.1029
h [kJ/kg] | -254.5296 | 2574.2795
e [kJ/kg] | -344.7953 | 1589.5140
g [kJ/kg] | -2428.4002 | -30246.9221
s [kJ/(kg-K)] | 7.2462 | 10.9404
W [g/mol] | 27.6333 | 25.3293
(dlV/dlp)T [-] | | -1.0285
(dlV/dlT)p [-] | | 1.5830
cp [kJ/(kg-K)] | 1.0786 | 5.5609
gamma [-] | 1.3869 | 1.1680
gamma_s [-] | 1.3869 | 1.1357
sound vel [m/s]| 353.8198 | 1057.5349
-----------------------------------------------------------
REACTANTS Xi [-]
N2 7.1493e-01
O2 1.9005e-01
CH4 9.5023e-02
MINORS[+22] 0.0000e+00
TOTAL 1.0000e+00
-----------------------------------------------------------
PRODUCTS Xi [-]
N2 6.4771e-01
H2O 1.1162e-01
CO 5.8485e-02
OH 3.5936e-02
H2 3.0822e-02
CO2 2.8615e-02
H 2.7583e-02
O2 2.5914e-02
O 1.8096e-02
NO 1.5206e-02
N 1.1123e-05
NH 9.5805e-07
HCO 1.2464e-07
NH2 1.1860e-07
NH3 3.0265e-08
HCN 4.4103e-09
CN 4.0110e-10
CH 5.7109e-13
CH3 1.5595e-13
CH4 1.1358e-14
MINORS[+5] 0.0000e+00
TOTAL 1.0000e+00
-----------------------------------------------------------
***********************************************************
Note
The solve_problem.m function prints the results in the command window by default. To avoid this, write before:
self.Misc.FLAG_RESULTS = false
or modify its default value directly in miscellaneous.m.
Note
This section is under development. Stay tuned!
Postprocessed results (predefined plots for several cases)#
There are some predefined charts based on the selected problem, in case you have calculated multiple cases. Just calling the routine
post_results(self);
will reproduce Figure 1, which represents the variation of the molar fraction with the equivalence ratio for lean to rich CH\(_4\)-ideal air mixtures at 3000 [K] and 1.01325 [bar].
Figure 1: Example TP: variation of molar fraction for lean to rich CH$_4$-ideal air mixtures at 3000 [K] and 1.01325 [bar], a set of 26 species considered and a total of 451 case studies. The computational time was of 2.39 seconds using a Intel(R) Core(TM) i7-11800H CPU @ 2.30GHz.
Note
This section is under development. Stay tuned!
Summary#
% Initialize CT and define chemical system
self = App('Soot formation');
% Define initial chemical composition
self.PD.S_Fuel = {'CH4'};
self.PD.S_Oxidizer = {'N2', 'O2'};
self.PD.ratio_oxidizers_O2 = [79, 21]/21;
self = set_prop(self, 'phi', 1);
% Define initial thermochemical state
self = set_prop(self, 'TR', 300, 'pR', 1 * 1.01325);
% Set chemical transformation
self = set_prop(self, 'pP', self.PD.pR.value, 'TP', 3000);
% Perform chemical calculations
self = solve_problem(self, 'TP');
% Initialize CT and define chemical system
self = App('Soot formation');
% Define initial chemical composition
self.PD.S_Fuel = {'CH4'};
self.PD.S_Oxidizer = {'N2', 'O2'};
self.PD.ratio_oxidizers_O2 = [79, 21]/21;
self = set_prop(self, 'phi', 0.5:0.01:4);
% Define initial thermochemical state
self = set_prop(self, 'TR', 300, 'pR', 1 * 1.01325);
% Set chemical transformation
self = set_prop(self, 'pP', self.PD.pR.value, 'TP', 3000);
% Perform chemical calculations
self = solve_problem(self, 'TP');
% Postprocess results
post_results(self);