Week 4.1- Handling Mixtures with Cantera

Aim: Calculation of properties of Air-Fuel mixture by varying the propertiess of air and fuel individually by using Python code

Theoritical Background:

Air–fuel ratio (AFR) - It is the mass ratio of air to a solid, liquid, or gaseous fuel present in a combustion process. The air-fuel ratio determines whether a mixture is combustible at all, how much energy is being released, and how much-unwanted pollutants are produced in the reaction. Typically a range of fuel to air ratios exists, outside of which ignition will not occur. These are known as the lower and upper explosive limits.

Equivalence Ratio : The equivalence ratio is defined as the ratio of the actual fuel/air ratio to the stoichiometric fuel/air ratio. If the equivalence ratio is equal to one, the combustion is stoichiometric. If it is < 1, the combustion is lean with excess air, and if it is >1, the combustion is rich with incomplete combustion.

Balanced Combustion Equation for Methane

Python Code:

import cantera as ct

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

Air = ct.Quantity(gas)

Air. TPX =  298.15, ct.one_atm, {'O2': 0.21, 'N2': 0.79}
Air.moles = 9.52	
print("The Mass fraction of Air is ", Air.mass_fraction_dict())
print("The Mass of Air is ", Air.mass, 'g')

Fuel = ct.Quantity(gas)
Fuel.TPX = 298.15, ct.one_atm,'CH4: 1546'
Fuel.moles = 1
print("nThe mass fraction of fuel is ", Fuel.mass_fraction_dict())
print("The mass of the fuel is ", Fuel.mass, 'g')

Mixture = Air + Fuel

Mixture.equilibrate('HP', solver = 'auto')
print("nThe AFT of the Mixuture is ", Mixture.T, 'K')
print("The Mass of the Mixuture is ", Mixture.mass, 'g')
print("nThe Mole fraction of the Mixuture is ", Mixture.mole_fraction_dict())
print("nThe Mass fraction of the Mixuture is ", Mixture.mass_fraction_dict())

Output:

1] "Air = ct.Quantity(gas)" ------> This statement mainly calls of Cantera to designate the species in mixutres

2] "Air. TPX = 298.15, ct.one_atm, {'O2': 0.21, 'N2': 0.79}" -----> This Statement signifies the Temperature, Pressure and Mole Fraction of air the Temperature and Pressure are 298 K and 1 bar respectively.

Mole fraction of O2 = 0.21, Mole Fraction of N2 = 0.79 . 1 Mole of air = 0.21 Moles of O2 + 0.79 Moles of N2

3] Fuel.TPX = 298.15, ct.one_atm,'CH4: 1546' ----> The Mole fraction of Methane is taken to be 1546. Only Methane is involved in the combustion Process

4] Mixture.equilibrate('HP', solver = 'auto')
print("nThe AFT of the Mixuture is ", Mixture.T, 'K')
print("The Mass of the Mixuture is ", Mixture.mass, 'g')

By using the Equilibrate method the mixture is solved to find out its properties. The Mass and Adiabatic Flame Temperature are printed as Output.

Results:

The Mass of Air is 274.65809279999996 g
The mass of the fuel is 16.043 g
The AFT of the Mixuture is 2224.4118993147486 K
The Mass of the Mixuture is 290.70109279999997 g