Handling Mixtures With Cantera

1. Use the \"moles\" method/function of the A object and explain how it was calculated.

By using the combustion reaction of methane:

`CH_4+2(O_2+3.76N_2 )→CO_2+2H_2 O+7.52N_2`

We defined O2=0.21 and N2=0.71 that equal to 1.

A = mixture of O2 and N2, so A is contain 2 Moles

Knowing that:

`X_(species)=(N_(species)/(N_t))`

Where,

`X_(species)`= Mole Fraction of the species of interest.

`N_(species)`= No. of moles of the species of interest.

`N_t`= Total no. of moles from all species present in the mixture

`X_(O_2) = 2/(2+2*3.72) = 0.21`

`X_(N_2) = (2*3.72)/(2+2*3.72) = 0.79`

We define:

Moles=(Total number of moles)/(species Mole fraction)

`n_(O_2) = 2/0.21 = 9.52`                                 (1)

`n_(N_2) = (2*3.72)/0.79 = 9.52`                       (2)

Where n represent the no. of moles.

Noticably (1) = (2). Therefore, we can use each other.

2. Use the method mole_fraction_dict() of the A object and explain the result that you get. Here is how you need to use it. print(A.mass_fraction_dict())

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
print(A.mass_fraction_dict())


#The following step is used to apply the right molar balance
M=A+B
M.equilibrate(\'HP\')
print(M.T)

Output of the abobe code

These values above of N2 and O2 are the mass fraction on N2 and O2 respectively.

3. What is happenging in lines 15 and 16? Make sure you explain what 1546 means? Is that reasonable?

In the code line 15 gives the values of TPX(Temperature,Pressure,Mole fraction(s) values for

gas B), the number 1546 represents the number of moles of the object. This value can be

any value mayor than 0, because B.moles=1 indacates that only one mole of B is taken

in consideration.

4. What changes do you need to put in place to get the right value of AFT in line 22?. The right answer is 2224K.

To get the right value of AFT, we need to set A.moles to 9.52, we calculated this value

previously.