Answer:
Explanation:
Benzene gas is burned with 95 percentage theoretical air during a steady â flow combustion process. The mole fraction of the CO in the products and the heat transfer from the combustion chamber are to be determined
Assumption
steady operating conditions exit .
Air and combustion gases are gases
Kinetic and potential energies are negligible
The fuel is burned with insufficient amount of air and thus the products will contain some CO as well as CO2, H2O and H2
Combustion equation is
C6H6 + ath (O2 + 3.76N2) â 6CO2 +3H2O +3.76ath N2
Where ath is the stoichiometric coefficient and is determined from the O2 balance
ath = 6+1.5 = 7.5
then the actual combustion equation can be written as
C6H6 + 0.95 * 7.5(O2 + 3.76N2) â xCO2 + (6-x) CO + 3H2O + 26.79N2
O2 balance: 0.95 * 7.5 = x +(6-x)/2 +1.5 â x=5.25
Thus C6H6 + 7.125(O2 +3.76 N2) â 5.25 CO2 + 0.75CO + 3H2O + 26.79N2
The mole fraction of CO in the products is
yCO = N CO/ N total = 0.75/5.25 + 0.75+3+26.79
=0.021 or 2.1% mole fraction of the CO in the products
