Tutorial#
With this short tutorial you will familiarize with the basic functionalities of the Combustion Toolbox.
Getting Started#
To begin, start MATLAB and navigate to the folder where you have downloaded the Combustion Toolbox. To include files in PATH, run this command in the command window:
INSTALL()
First, using Combustion Toolbox, you have to initialize the tool (load databases, set default variables, etc.). To do that, type the following at the prompt:
self = App('Soot formation')
Here, Soot formation represents a predefined list of species to be considered as possible products (see the routine list_species.m).
If files contained in Combustion Toolbox are correctly defined, you should see something like this:
self =
struct with fields:
E: [1×1 struct]
S: [1×1 struct]
C: [1×1 struct]
Misc: [1×1 struct]
PD: [1×1 struct]
PS: [1×1 struct]
TN: [1×1 struct]
DB_master: [1×1 struct]
DB: [1×1 struct]
Setting the state#
Indicate the temperature [K] and pressure [bar] of the initial mixture, type:
self = set_prop(self, 'TR', 300, 'pR', 1 * 1.01325);
Indicate species and number of moles of each species in the initial mixture:
Individual study#
For example, for a stochiometric CH4-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.
Parametric study (several cases)#
The Combustion Toolbox also allows the computation of a range of values of different properties. For example, if we want to compute a range of values of the equivalence ratio, e.g., \(\phi \in [0.5, 5]\) with a 0.01 step size, do this:
self.PD.S_Fuel = {'CH4'};
self.PD.S_Oxidizer = {'N2', 'O2'};
self.PD.ratio_oxidizers_O2 = [79, 21]/21;
self = set_prop(self, 'phi', 0.5:0.01:5);
Chemical Equilibrium#
Depending on the problem you want to solve, you may need to configure additional inputs. For example, to compute the equilibrium composition at a defined temperature and pressure (TP), we have to set these values as
self = set_prop(self, 'pP', self.PD.pR.value, 'TP', 3000);
and to solve the aforementioned problem, run
self = solve_problem(self, 'TP');
The results are contained in self.PS. By default, this routine print the results through the command window (default: self.Misc.FLAG_RESULTS=true) 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
-----------------------------------------------------------
***********************************************************
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 CH4-ideal_air mixtures at 3000 [K] and 1.01325 [bar].
Figure 1: Example TP: variation of molar fraction for lean to rich CH4-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.
Congratulations!#
Congratulations you have finished the Combustion Toolbox MATLAB tutorial! You should now be ready to begin using Combustion Toolbox on your own (see the examples folder).