Using NASA-CEA

CH4_O2_N2_tp.inp

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.

CH4air_tp_a.inp

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.)

CH4air_tp_b.inp

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

CH4air_tp_c.inp

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

CH4air_tp_d.inp

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         

CH4air_hp.inp

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 under 1 and 3 atom for methane-air gas mixture of equivalence ratio = 1.05.  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

C3H6air_hp.inp

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

C2H2_hp.inp

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
  

CH4air_uv.inp

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.

CH4air_tp_frozen.inp

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.

CH4air_sv_frozen.inp

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

CH4air_sp_frozen.inp

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

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

CH4steam-reforming_tp.inp

! 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.
  

CaCO3_tp.inp

! 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.

lastpart_of_thermo.inp

...
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.

furan-air.inp

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.