We Assume a 100 kmole of dry mixture of products of combustion obtained from orsat analysis of products of combustion of a internal combustion engine.We will consider methane and octane n this process in order to make this example as close to actual process as possible. ANALYSIS OF PRODUCTS OF COMBUSTION OF OCTANE COMPOSITION IN PERCENTS 10% CO2 1% CO 4% H2 2% CH4 0.46% C8H18 1% SO2 2% NO2 79.54% N2 This dry mixture is composed of the above constituents and sum up to a 100% of mixture .You can see traces of CH4 and C8H18 in the products too.
in this step we try to balance the stichometric equation.We can equate moles of Carbon,Oxygen,Hydrogen etc in order to know moles in reactants that lead to this combustion. 2.085C8H18 + 1S + 25.691[O2 + 3.76N2] —-> 2CH4 + 0.46C8H18 + 4H2 + 10CO2 + 1CO + 8.88O2 + 1SO2 +2NO2 + 6.625H2O + 96.6N2 OXYGEN BALANCE ON BOTH SIDES OF EQUATION[MOLE BALANCE OF COMBUSTION OF OCTANE] : we can see that the mole balance is equal to be 25.69 on both sides which is equal to each other.This shows that moles of oxygen are balanced on both sides. 25.69 = 25.69
Similarly moles of hydrogen are balanced on both sides. 18.765 = 18.765 CARBON BALANCE ON BOTH SIDES OF EQUATION[MOLE BALANCE OF COMBUSTION OF OCTANE]: Also moles of carbon balance on both sides too. 16.68 = 16.68 COMBUSTION OF OCTANE EQUATION TAKES FORM GIVEN BELOW : ASSUMING 1KMOLES OF PRODUTS MIXTURE : now if we want to calaculate for 1 mole of fuel that undergoes combustion is given as under. 1C8H18 + 0.48S + 12.32182[O2 + 3.76N2] —-> 0.9592CH4 + 0.2206C8H18 + 1.92H2 + 4.7962CO2 + 0.4796CO + 4.259O2 + 0.4796SO2 +0.9592NO2 + 3.1775H2O + 46.3309N2 OXYGEN BALANCE ON BOTH SIDES OF EQUATION[MOLE BALANCE OF COMBUSTION OF OCTANE]: 12.32 = 12.32
Stichometric/Chemically Correct Equation is : 1C8H18 + 8.06355[O2 + 3.76N2] —-> 0.9592CH4 + 0.2206C8H18 + 1.92H2 + 4.7962CO2 + 0.4796CO + 0.4796SO2 + 0.9592NO2 + 3.1775H2O + 30.32N2 OXYGEN BALANCE ON BOTH SIDES OF EQUATION : 8.06355 = 8.06355 Excess Air : 53% AIR/FUEL RATIO : Air/Fuel Ratio : 14.89