Using NASA-CEA
A. Miyoshi

Using NASA-CEA

The CEA is a computer program for chemical equilibrium calculation developed by Dr. S. Gordon, Dr. B. J. McBride, and co-workers in NASA Glenn Research Center.   It contains a thermodynamic data library, the PAC99 program which does regression and estimation of thermodynamic data, etc., as well as the main chemical-equilibrium code (CEA2).

How to Use CEA

Manual

Usave Overview

[Example-1] Methane-Oxigen-Nitrogen

An example of the equilibrium calculation for the methane-oxigen-nitrogen gas mixture is shown below.

Input

CH4_O2_N2_tp.inp (Right-click to download)
problem
  tp
  p(atm) = 1 3
  t(k)   = 300 350 900 2000
reactant
  name = CH4  moles = 0.11
  name = O2   moles = 0.21
  name = N2   moles = 0.79
end
  • Equilibrium calculation for CH4 / O2 / N2 gas mixture of molar ratio 0.11 : 0.21 : 0.79 at specified temperatures and pressures (tp).  Calculations are done for eight combinations of pressures (1 and 3 atm) and temperatures (300, 350, 900, and 2000 K).
  • The keywords prob[lem] and reac[tant] indicate the beginning of the input blocks for the conditions except for the reactant composition and for the reactants.  Since the calculations were done assuming the ideal gas, two state variables which are taken to be constants must be specified by their symbols.  In this example, the temperature (t) and the pressure (p) is held constant.  Other variables may also be specified, for example, hp (constant enthalpy and pressure = isobaric adiabatic), uv (constant internal energy and volume = isovolume adiabatic), and etc.
  • The relative amount of reactants can be specified by moles (moles) or weights (wt).  Unlike the mole fractions or weight percent, the sum needs not to be unity or one hundred.

Output

The major resuts can be found after the line containing "THERMODYNAMIC EQUILIBRIUM ... PROPERTIES AT ASSIGNED".  An extract is shown below.
               THERMODYNAMIC EQUILIBRIUM PROPERTIES AT ASSIGNED

                           TEMPERATURE AND PRESSURE

 CASE =                

             REACTANT                       MOLES         ENERGY      TEMP
                                                         KJ/KG-MOL      K
 NAME        CH4                          0.1100000         0.000      0.000
 NAME        O2                           0.2100000         0.000      0.000
 NAME        N2                           0.7900000         0.000      0.000

 O/F=    0.00000  %FUEL=  0.000000  R,EQ.RATIO= 1.047619  PHI,EQ.RATIO= 0.000000

 THERMODYNAMIC PROPERTIES

 P, BAR            1.0132   1.0132   1.0132   1.0132   3.0397   3.0397   3.0397   3.0397
 T, K              300.00   350.00   900.00  2000.00   300.00   350.00   900.00  2000.00
 RHO, KG/CU M    1.3484 0 9.6032-1 3.7013-1 1.6648-1 4.1432 0 3.0809 0 1.1104 0 4.9956-1
 H, KJ/KG        -3290.30 -2962.43 -2263.54  -683.63 -3321.66 -3061.66 -2263.55  -688.02
 U, KJ/KG        -3365.44 -3067.94 -2537.30 -1292.25 -3395.03 -3160.33 -2537.31 -1296.51
 G, KJ/KG        -5210.59 -5559.99 -10010.3 -20168.8 -5129.20 -5445.26 -9709.59 -19500.2
 S, KJ/(KG)(K)     6.4010   7.4216   8.6076   9.7426   6.0251   6.8103   8.2734   9.4061

 M, (1/n)          33.194   27.581   27.335   27.323   33.998   29.495   27.335   27.328
 MW, MOL WT        27.339   27.581   27.335   27.323   27.335   27.503   27.335   27.328
 (dLV/dLP)t      -1.03649 -1.00001 -1.00000 -1.00026 -1.01180 -1.17474 -1.00001 -1.00013
 (dLV/dLT)p        1.6437   1.0002   1.0000   1.0090   1.2079   3.5514   1.0001   1.0046
 Cp, KJ/(KG)(K)    4.2185   1.1279   1.3161   1.6012   2.2991  11.7487   1.3162   1.5554
 GAMMAs            1.1415   1.3649   1.3006   1.2395   1.1674   1.1466   1.3006   1.2458
 SON VEL,M/SEC      292.9    379.5    596.7    868.6    292.7    336.4    596.7    870.7

 MOLE FRACTIONS

 CH4              0.00007  0.00450  0.00000  0.00000  0.00001  0.00307  0.00000  0.00000
 *CO              0.00000  0.00000  0.00330  0.01235  0.00000  0.00000  0.00330  0.01220
 *CO2             0.08937  0.09460  0.09491  0.08582  0.08929  0.09292  0.09491  0.08599
 *H               0.00000  0.00000  0.00000  0.00013  0.00000  0.00000  0.00000  0.00007
 *H2              0.00000  0.00003  0.01455  0.00597  0.00000  0.00001  0.01454  0.00589
 H2O              0.02872  0.18917  0.18188  0.19008  0.00935  0.12677  0.18188  0.19032
 NH3              0.00000  0.00001  0.00000  0.00000  0.00000  0.00000  0.00001  0.00000
 *NO              0.00000  0.00000  0.00000  0.00014  0.00000  0.00000  0.00000  0.00008
 *N2              0.70545  0.71170  0.70536  0.70497  0.70537  0.70969  0.70536  0.70515
 *O               0.00000  0.00000  0.00000  0.00001  0.00000  0.00000  0.00000  0.00000
 *OH              0.00000  0.00000  0.00000  0.00045  0.00000  0.00000  0.00000  0.00026
 *O2              0.00000  0.00000  0.00000  0.00008  0.00000  0.00000  0.00000  0.00003
 C(gr)            0.00879  0.00000  0.00000  0.00000  0.00891  0.00283  0.00000  0.00000
 H2O(L)           0.16760  0.00000  0.00000  0.00000  0.18707  0.06470  0.00000  0.00000

  * THERMODYNAMIC PROPERTIES FITTED TO 20000.K

    PRODUCTS WHICH WERE CONSIDERED BUT WHOSE MOLE FRACTIONS
    WERE LESS THAN 5.000000E-06 FOR ALL ASSIGNED CONDITIONS

 *C              *CH             CH2             CH3             CH2OH          
 CH3O            CH3OH           CH3OOH          *CN             CNN            
 ...
  • For example, "9.6032-1" found in the density (RHO, KG/CU M) output of the results for 1.0132 bar and 350 K is in the special number format used in the CEA for the output of the maximum digits in limited columns and is equivalent to "9.6032E-01" (9.6032 × 10−1).
  • Names of the chemical species in the equilibrium mole fractions without ( ) (parentheses) usually mean gas-phase speceis, H2O(L) means the liquid water, and C(gr) means graphite.  For more accurate information, see the file thermo.inp found in the folder where the CEA has been installed.
  • In the equilibrium calculations for constant T and p, only the composition of elements is essential.  All kind of the reactant inputs with the same composition of elements will give the same result.  For example, though it is not recommended, the reac dataset shown below will give the same result as the example input shown above.
      name = C(gr)  moles = 0.11
      name = H2     moles = 0.22
      name = NO     moles = 0.42
      name = N2     moles = 0.58

[Example-2] Mathane-Air

Although the example-1 shown above is intended to be the methane-air composition, you may alternatively or preferably specify it by using the name "Air" which is the mixture of standard dry air composition (N2 78.084%, O2 20.9476%, Ar .9365%, CO2 0.0319%).

Input-a

CH4air_tp_a.inp (Right-click to download)
problem  tp
  p(atm) = 1 3  t(k) = 300 350 900 2000
reactant
  fuel = CH4  moles = 0.11
  oxid = Air  moles = 1.00
end
  • In the example-1, all the reactant names were specified with name.  They may be specified by using fuel and oxid[ant] as in this example.  Compared to the case using name, no essential difference in the calculation is expected except for the outputs related to the ratio of the fuel to oxidant shown below.
     O/F=   16.41389  %FUEL=  5.742541  R,EQ.RATIO= 1.050163  PHI,EQ.RATIO= 1.050240
    (Compare it with the output of example-1, in which O/F, %FUEL, and PHI,EQ.RATIO are zero.)

Input-b

CH4air_tp_b.inp (Right-click to download)
prob  tp  phi(eq.ratio) = 1.05
  p(atm) = 1 3  t(k) = 300 350 900 2000
reac
  fuel = CH4
  oxid = Air
end
  • In the combusion problems, one often needs to set the "equivalence ratio" to a certain value.  When the reactants are properly specified with fuel and oxid, the ratio of the fuel(s) and the oxidant(s) can be specified with phi(eq.ratio).  The part of the output of this example related to the fuel-oxidant ratio is looks like:
     O/F=   16.41764  %FUEL=  5.741306  R,EQ.RATIO= 1.049924  PHI,EQ.RATIO= 1.050000

Input-c

CH4air_tp_c.inp (Right-click to download)
prob  tp  phi(eq.ratio) = 1.05  p(atm) = 1 3  t(k) = 300 350 900 2000
reac
  fuel = CH4  oxid = Air
output
  trace = 1e-10
end
  • This sample input shows an example for the ouput dataset beginning with the keyword outp[ut].  The option trace instructs the CEA to output species with mole fractions greater than or equal to the assigned value. (The default is 5E-06 = 5 × 10−6)   Since the value is 1E-10 in this example, all the species found with mole fractions greater than or equal to 1 × 10−10, in any set of the pressure and temperature, are printed.  A part of the output of this example is shown below.  It should be noted that the number format also changed to the special CEA format described in the output of example-1.
     MOLE FRACTIONS
    
     *Ar             8.3594-3 8.4370-3 8.3583-3 8.3546-3 8.3584-3 8.4097-3 8.3583-3 8.3563-3
     CH4             6.857 -5 4.709 -3 5.288 -8 8.665-16 7.266 -6 3.077 -3 4.752 -7 7.427-15
     *CO             1.022-11 1.496 -9 3.473 -3 1.289 -2 5.828-12 9.006-10 3.472 -3 1.275 -2
     *CO2            8.9170-2 9.4656-2 9.4964-2 8.5502-2 8.9098-2 9.2724-2 9.4965-2 8.5668-2
     COOH            0.000  0 2.070-32 1.089-14 3.007 -9 0.000  0 2.337-32 1.886-14 5.188 -9
     *H              0.000  0 5.730-33 1.434-11 1.279 -4 0.000  0 2.052-33 8.275-12 7.338 -5

Input-d

CH4air_tp_d.inp (Right-click to download)
prob  tp  phi(eq.ratio) = 1.05  p(atm) = 1 3  t(k) = 300 350 900 2000
reac  fuel = CH4  oxid = Air
outp  trace = 1e-10
  plot p t C(gr) H2O(L) H2O
end
  • This example uses a specification for the plotting output in the output dataset.  The variable listed after the option plot are printed in a text-format file with the extension plt (in this example, it is CH4air_tp_d.plt).  The variables p and t mean the pressure and the temperature, respectively.  By indicating the names of the species, their mole fractions are printed.  Below, the contents of the CH4air_tp_d.plt are shown.  The text-file in this format can be read into softwares like MS-Excel.  When using the modified version for command prompt, FCEA2m.exe, a file with an extension csv will be also created.  This file can be directly opened by MS-Excel. 
    #  p           t           C(gr)       H2O(L)      H2O         
       1.0132E+00  3.0000E+02  9.2118E-03  1.6748E-01  2.8712E-02
       1.0132E+00  3.5000E+02  0.0000E+00  0.0000E+00  1.8870E-01
       1.0132E+00  9.0000E+02  0.0000E+00  0.0000E+00  1.8108E-01
       1.0132E+00  2.0000E+03  0.0000E+00  0.0000E+00  1.8969E-01
       3.0397E+00  3.0000E+02  9.3334E-03  1.8695E-01  9.3430E-03
       3.0397E+00  3.5000E+02  3.2418E-03  6.4611E-02  1.2673E-01
       3.0397E+00  9.0000E+02  0.0000E+00  0.0000E+00  1.8108E-01
       3.0397E+00  2.0000E+03  0.0000E+00  0.0000E+00  1.8993E-01
    #  p           t           C(gr)       H2O(L)      H2O         

[Example-3] Calculation for the Adiabatic Flame

The following examples use hp (constant enthalpy and pressure = isobaric adiabatic) in the prob dataset.  For the reactant mixtures with fuels and air (oxygen), this is so-called adiabatic flame condition.

Input-a

CH4air_hp.inp (Right-click to download)
prob  hp
  phi(eq.ratio) = 1.05  p(atm) = 1 3
reac
  fuel = CH4  t(k)= 298.15
  oxid = Air  t(k)= 298.15
end
  • An adiabatic flame calculation for methane-air gas mixture.  For the type hp in the prob dataset, the temperature(s) of unburnt gas should be given in the reac dataset.  A part of the output is shown below.  This resut indicate the adiabatic flame temperatures to be 2231 K at 1 atm and 2249 K at 3 atm.
     THERMODYNAMIC PROPERTIES
    
     P, BAR            1.0132   3.0397
     T, K             2230.98  2249.09
     RHO, KG/CU M    1.4929-1 4.4487-1
     H, KJ/KG         -271.06  -271.06

Input-b

C3H6air_hp.inp (Right-click to download)
problem  hp  p(atm) = 1  phi(eq.ratio) = 1
reac
  fuel= C3H6,propylene  t(k)= 298.15
  oxid= Air             t(k)= 298.15
end
  • An adiabatic flame calculation for stoichiometric (equivalence ratio = 1) propylene-air mixture.  It should be noted that the name of the reactant must be registered in thermo.inp.  For C3H6, two isomers, propylene and cyclopropane, have been registered.  The name of the propylene must be C3H6,propylene as resistered in the thermo.inp.  The cyclopropane is registered with the name C3H6,cyclo-
  • A part of the output is shown below.  Though being an exactly stoichiometric mixture, the radicals such as OH or the species such as CO and NO present with a significant amount at temperatures of the adiabatic flame.
     THERMODYNAMIC PROPERTIES
    
     P, BAR            1.0132
     T, K             2332.68
     RHO, KG/CU M    1.4913-1
     H, KJ/KG          26.047
     U, KJ/KG         -653.41
     G, KJ/KG        -22336.7
     S, KJ/(KG)(K)     9.5867
    
     M, (1/n)          28.545
     (dLV/dLP)t      -1.00405
     (dLV/dLT)p        1.1160
     Cp, KJ/(KG)(K)    2.4353
     GAMMAs            1.1695
     SON VEL,M/SEC      891.4
    
     MOLE FRACTIONS
    
     *Ar              0.00864
     *CO              0.01776
     *CO2             0.11144
     *H               0.00067
     *H2              0.00344
     H2O              0.12291
     *NO              0.00310
     *N2              0.71922
     *O               0.00053
     *OH              0.00446
     *O2              0.00784

Input-c

C2H2_hp.inp (Right-click to download)
prob  hp  p(atm) = 1
reac
  name = C2H2,acetylene  t(k)= 298.15
end
  • It is well known that gaseous acetylene decomposes explosively.  The result of the isobaric adiabatic calculation starting from pure acetylene at ambient temperature and pressure is shown below.  The major products are solid carbon and hydrogen and temperature is raised as high as 2850 K.
     THERMODYNAMIC PROPERTIES
    
     P, BAR            1.0132
     T, K             2849.07
     RHO, KG/CU M    1.0633-1
      :
    
     MOLE FRACTIONS
      :
     *H               0.03179
     *H2              0.30807
     C(gr)            0.64748
    

[Example-4] Calculation for the Adiabatic Constant-Volume Equilibrium State

The following example uses uv (constant internal-energy and volume = constant-volume adiabatic) in the prob dataset.  For the reactant mixtures with fuels and air (oxygen), the result corresponds to the burnt equilibrium state in a constant-volume adiabatic vessel.  The temperature is higher than the (constant-pressure) adiabatic flame due to the lack of expansion.

Input

CH4air_uv.inp (Right-click to download)
prob  uv  phi(eq.ratio) = 1.05  rho(kg/m**3) = 1.1316
reac
  fuel = CH4  t(k) = 298.15
  oxid = Air  t(k) = 298.15
end
  • A constant-volume adiabatic burnt equilibrium state calculation for methane-air gas mixture.  For the type uv in the prob dataset, the density or the specific volume (volume per unit mass) should be given.  Here, the density is given by rho.  A part of the output is shown below.  The temperature is higher than the (constant-pressure) adiabatic flame.
     THERMODYNAMIC PROPERTIES
    
     P, BAR            9.0164
     T, K             2601.97
     RHO, KG/CU M    1.1316 0
     H, KJ/KG          436.17
     U, KJ/KG         -360.61
    The density required as an input can be, of courese, calculated by using the ideal gas law, but, it may become time consuming task for the complex mixtures.  Alternatively, the density can be obtained by, for example see below, a nearly meaningless calculation.

Input to obtain the density

CH4air_tp_frozen.inp (Right-click to download)
prob  tp  phi(eq.ratio) = 1.05  p(atm) = 1  t(k) = 298.15
reac  fuel = CH4  oxid = Air
only  CH4 N2 O2 Ar CO2
end
  • This is a calculation for specified temperature and pressure (tp), but the composition change was avoided by only keyword.  Here, the density in the following output (ρ = 1.1316×100 kg/m3) was used in the input for the constant-volume adiabatic calculation shown above.
     THERMODYNAMIC PROPERTIES
    
     P, BAR            1.0132
     T, K              298.15
     RHO, KG/CU M    1.1316 0
     H, KJ/KG         -271.07
     U, KJ/KG         -360.61
     G, KJ/KG        -2426.74
     S, KJ/(KG)(K)     7.2302
    This output will be used in the following calculation for adiabatic compression.

[Example-5] Calculation for Adiabatic Compression (Expansion)

Below, examples using sv and sp in the prob dataset will be shown.  Isentropic processes, such as adiabatic compression and expansion, can be calculated.  In many cases, the compression ratio (or the volume ratio) is known, and the sv will be used.  However, for example for the core-gas model calculation for the rapid compression machine experiments, the temperature is estimated from the measured pressure, and then the sp will be used.

Input-a

CH4air_sv_frozen.inp (Right-click to download)
prob  sv  phi(eq.ratio) = 1.05  s/r = 0.869588
  rho(kg/m**3) = 1.1316 3.3948 11.316
reac  fuel = CH4  oxid = Air
only  CH4 N2 O2 Ar CO2
end
  • When the sv is specified in the prob dataset, the specific entropy (entropy per unit mass) and the density (or the specific volume) should be given.  Here, the results from the "meaningless" calculation shown above are used.  The specific entropy is given by the value divided by the universal gas constant, 7.2302/8.31451 = 0.869588.  In this calculation, the states compressed at initial volume, 1/3 volume (three-times density), and 1/10 volume (ten-times density) are calculated for the methan-air mixture (which was assumed to be non-reactive).  The specific entropy was conserved as shown below.
     THERMODYNAMIC PROPERTIES
    
     P, BAR            1.0133   4.6099   23.364
     T, K              298.16   452.15   687.46
     RHO, KG/CU M    1.1316 0 3.3948 0 1.1316 1
     H, KJ/KG         -271.05  -102.54   172.01
     U, KJ/KG         -360.60  -238.34  -34.453
     G, KJ/KG        -2426.84 -3371.70 -4798.48
     S, KJ/(KG)(K)     7.2302   7.2302   7.2302

Input-b

CH4air_sp_frozen.inp (Right-click to download)
prob  sp  phi(eq.ratio) = 1.05  s/r = 0.869588
  p(bar) = 1.01325 10 23
reac  fuel = CH4  oxid = Air
only  CH4 N2 O2 Ar CO2
end
  • When the sp is given in the prob dataset, the specific entropy and the pressure must be given.  Similarly to above, the specific entropy input is in the value divied by the gas constant.  In this calculation, the states compressed to the initial pressure, 10 bar, and 23 bar are calculated for the non-reactive methan-air mixture.
     THERMODYNAMIC PROPERTIES
    
     P, BAR            1.0132   10.000   23.000
     T, K              298.16   554.67   684.81
     RHO, KG/CU M    1.1315 0 6.0030 0 1.1183 1
     H, KJ/KG         -271.06   14.197   168.78
     U, KJ/KG         -360.60  -152.39  -36.887
     G, KJ/KG        -2426.81 -3996.18 -4782.51
     S, KJ/(KG)(K)     7.2302   7.2302   7.2302

Caution

The calculations similar to above, but without only input, do NEVER corresponds to the combustion equilibrium states of the compressed gas mixtures.  For such a calculation, perform uv calculation with initial state equal to the compressed state.  For example, from the 1/10-volume compressed state of above calculation, the input will be as follows.
prob  uv  phi(eq.ratio) = 1.05  rho(kg/m**3) = 11.316
reac
  fuel = CH4  t(k) = 687.46
  oxid = Air  t(k) = 687.46
end

[Example-6] Equilibrium State at Given T and p

Below is an example calculation for the steam reforming of methane.

Input

CH4steam-reforming_tp.inp (Right-click to download)
! CH4 steam reforming
problem  tp
  p(atm) = 1
  t(k) = 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200
reac
  name = CH4  moles = 1
  name = H2O  moles = 1
omit C(gr)
outp
  plot p t CH4 H2O CO H2 CO2
end
  • The results will be shown below.

[Example-7] Dissociation Equilibrium of Calcium Carbonate

It is well known that the calcium carbonate (CaCO3) reversibly dissociates to carbon dioxide (CO2) and calcium oxide (CaO), or the calcium oxide reversibly adsorbs carbon dioxide.
CaCO3(s) CaO(s) + CO2
This example calculates the temperature variation of the equilibrium gas-phase partial pressure of CO2.  Here, a small amount of Ar is added as an initial component, which is a tip to apply problem tp to this type of problem.  Without this, the calculation will fail at low temperatures because there is no solution to satisfy the specified pressure.  Also it is essential to set the total pressure to 10 atm.  With 1 atm, no solution could be found at high temperatures where the equilibrium partial pressure exceeds 1 atm.

Input

CaCO3_tp.inp (Right-click to download)
! decomposition of CaCO3(cr)
problem  tp
 p(atm) = 10
 t(k) = 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300
reac
  name = CaCO3(cr)  moles = .9
  name = Ar         moles = .1
outp
  plot t p Ar CO2 CaCO3(cr) CaO(cr)
end
  • Small manipulation of the results will give the equilibrium partial pressure pCO2 as shown below.
    Since a small amount of CO + O2 may be formed at high temperature, the mole fraction of total gas was calcualted as 1 − [sum of mole fractions of CaCO3(cr) and CaO(cr)] and the pCO2 was derived from the total pressure and gas-phase mole fraction.  Above around 1160 K, the equilibrium partial pressure exceeds 1 atm.  The partial pressure at 800 K is found to be 36.9 Pa which is smaller than the atmospheric partial pressure of CO2 (1 atm × 400 ppm = 40.53 Pa), implying that at low temperatures, CaO(s) can chemically adsorb atmospheric CO2.

[Example-8] Adiabatic Flame Temperature of Furan – Addition of Thermodynamic Data

This example calculates the adiabatic flame temperature of furan.  Although the thermodynamic database included in CEA covers most of the compounds existing in typical equilibrium states, the unusual compounds like furan, for example the bio-fuels which may be used in the future, are not included.  However, the calculation of adiabatic flame temperature only requires the chemical composition and standard enthalpy of formation of the starting material (or fuel), which can be easily added to the thermodynamic database, thermo.inp.

Editing thermo.inp

lastpart_of_thermo.inp (Right-click to download: This is only an example input at around the bottom part, and cannot be prcessed directly with CEA!)
Edit the bottom part of the thermo.inp before END REACTANTS and add furan as a reactant.
...
RP-1              Mehta et.al. AIAA 95-2962 1995. Hcomb(high) = 19923.BTU/#     
 0 gll/00 C   1.00H   1.95    0.00    0.00    0.00 1   13.9761830     -24717.700
    298.150      0.0000  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0            0.000
C4H4O,furan       Furan                                                         
 0 am1209 C   4.00H   4.00O   1.00    0.00    0.00 0   68.0739600     -34700.000
    298.150      0.0000  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0            0.000
END REACTANTS                                                                   
  • The part added is shown by red.  Although the details should follow the manual of CEA, briefly:
    The first line shows the name of the compound and comments.  Although the name was given here in the form [chemical formula, name], according to the standard of CEA, other names may be used provided that the name consists of letters allowed for the CEA names.  The am1209 in the second line is an optional ID code and may be left blank.  The chemicla formula follows by repeating [element symbol, number].  The part 68.0739600 is the molecular weight.  The last value in the second line is the standard enthalpy of formation of gaseous furan (−34.7 kJ mol−1) taken from the NIST Chemistry WebBook.  The units should be J mol−1.  Leave the third line as this example, usually.  The columns for inputs are read as formated input of FORTRAN, and the column positions are important.
  • Run fcea2.exe or fcea2m.exe, and type thermo (enter).  This name is a special name for thermodynamic database, and this process creates the new thermo.lib in the current directory.  The procedure below should be done with this new thermo.lib residing in the current directory.

Input

furan-air.inp (Right-click to download)
problem case=furan-air hp p(atm)=1 phi,eq.ratio=1
reac
  fuel= C4H4O,furan  wt%=100  t(k)= 298.15
  oxid= Air          wt%=100  t(k)= 298.15
end
  • Once the reactant (fuel) has been successfully registered in thermo.lib, the calculation procedure is exactly the same as pre-registered compounds.