Handling Mixtures with Cantera

A. In the given code, mole fractions of O2 & N2 are given as 0.21 & 0.79

we know that mole fraction = (number of moles of that species)/ (Total number of moles)

from the reaction of methane (CH4)

CH4+2O2+2(3.76)N2=CO2+2H2O+2(3.76)N2\">CH4 + 2O2 + 2(3.76) *N2 = CO2 + 2H2O + 2(3.76) *N2CH4 + 2O2 + 2(3.76) *N2 = CO2 + 2H2O + 2(3.76) *N2

let\'s calculate the mole fraction of O2(A gas quantity) as example 

O2 mole fraction (for A gas quantity) = (number of moles of O2)/ (Total number of moles)

 O2 mole fraction (for A gas quantity) = 2/(2+(2*3.76))

O2 mole fraction (for A gas quantity) = 2/9.52 = 0.21

N2 mole fraction (for A gas quantity) = (2*3.76)/(2+(2*3.76)) = 7.52/9.52 = 0.79

Now, for moles of A = (no. of moles of O2)/ (O2 mole fraction)

Now, for moles of A = (no. of moles of N2)/ (N2 mole fraction).

for moles of A (with O2) = 2/0.21 = 9.52 (Approx.)                                                                                                                                                                                   equation-1

for moles of A (with N2) = (2*3.76)/0.79 = 9.52 (Approx.)                                                                                                                                                                         equation-2

for calculating moles of A, we can use either equation-1 or equation-2 which is same. So, A.moles = (2/0.21) or (7.52/0.79) which is shown in my code of line 10.

B. By using A.mass_function_dict() we are converting the mole fraction of O2 & N2 into the mass fractions of O2 & N2 and print(A.mass_function_dict()) will print those values as dictionaries.

 C. In the code line 15 & 16, the B gas quantity has given values of TPX (Temperature, Pressure & Mole fractions(X) of all species in B. Here, 1546 refers to the number of mole fractions of CH4 & moles of B is already mentioned as one (1).

 D. In order to get the Adiabatic Flame Temperature, moles of A should be set. I have already calculated for moles of A in point (2) which is (2/0.21) or ((2*3.76)/0.79) = (7.52/0.79). The results were calculated with constant enthalpy & constant pressure process which is shown above with AFT = 2224.461 (K). 

Program

import matplotlib.pyplot as plt
import numpy as np
import cantera as ct

gas = ct.Solution(\'gri30.cti\')

A = ct.Quantity(gas)
A.TPX = 298.15,ct.one_atm,{\'O2\':0.21,\'N2\':0.79}
A.moles = (2*3.76)/0.79
print(A.mass_fraction_dict())

B = ct.Quantity(gas)
B.TPX = 298.15,ct.one_atm,\'CH4:1546\'
B.moles = 1

M = A + B
M.equilibrate(\'HP\')
print(C.T)

Result

{\'N2\': 0.7670829987741036, \'O2\': 0.23291700122589645}
2224.461438930544