Week 4 - Handling Mixtures with Cantera

AIM: To work with the Quantity class of Cantera to create various mixtures.

Introduction:

Mixture is the predominant word that appears in training texts, operation manuals and maintenance manuals in the aviation world.

Air–fuel ratio is the ratio between the mass of air and the mass of fuel in the fuel–air mix at any given moment. The mass is the mass of all constituents that compose the fuel and air, whether combustible or not.

For example, a calculation of the mass of natural gas—which often contains carbon dioxide (CO2), nitrogen (N2), and various alkanes—includes the mass of the carbon dioxide, nitrogen and all alkanes in determining the value of mfuel.

For pure octane the stoichiometric mixture is approximately 15.1:1, or λ of 1.00 exactly.

In naturally aspirated engines powered by octane, maximum power is frequently reached at AFRs ranging from 12.5 to 13.3:1 or λ of 0.850 to 0.901.

Air-fuel ratio of 12:1 is considered as maximum output ratio, where as the air-fuel ratio of 16:1 is considered as maximum fuel economy ratio.

(1)

Here, A represents Air and by considering fuel(HydroCarbon).

CxHy + b(O2 + 3.76*N2) --> xCO2 +y/2H2O + 3.76b*N2

Mole_fraction of Air = 0.21,

Number of moles of O2 is 2 for stoichiometric condition, therefore 2/0.21 = 9.52 gives moles of air

CODE:

"""
Sample code to vary the Temperature and oxidizer individually
"""
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 = 9.52
print('Mole fraction in air=',A.mole_fraction_dict())
print('Mass fraction in air=',A.mass_fraction_dict())

B = ct.Quantity(gas)
B.TPX = 298.15,ct.one_atm,'CH4:1546'
B.moles = 1
print('Mole fraction of fuel',B.mass_fraction_dict())

# The following step is used to apply the right molar balance
M = A+B

# Calculate AFT
M.equilibrate('HP')
#print(M.report())
#print(gas.T)
print('Value of AFT=',M.T)

(2)

Mass fraction = No. of moles of compound * molecular weight.

No. of moles(O2) = 2, mass(O2) = 64gms

No. of moles(N2) = 7.52, mass(N2) = 210.56gms

Therefore,

Mole fraction(O2) = 64/(64+210.56) = 0.2331

Mole fraction(N2) = 210.56/(64+210.56) = 0.767

(3)

What 1546 means?

1546 is the Molar fraction of CH4.

Because of only 1 compound, its has no effect.

(4)

Here, we have the Mole fraction of O2 & N2,

No. of mole of A for O2 and N2 is 9.52,

we use the command A.moles = 9.52 to get the value of AFT.