Question

A) Gasoline is burned in an engine with 98% efficiency. The remainder leaves in the exhaust. If the engine exhausts 15 kg of exhaust gases (average MW=30) for each kg of gasoline burned, determine the unburned hydrocarbon fraction in the exhaust in ppm. Characterize the gasoline by the octane molecule.

B) What would be the air/fuel ratio be in this mixture?

Answer 1

a)given the efficiency of engine = 98%

amount of unburnt hydrocarbon=(1-0.98)*15=0.3kg=3*10^-7ppm.

b) C8H18+25/2O2------->8CO2+9H2O

mol.wt of C8H18=114

mol.wt of O2=32

so oxygen fuel ratio=(25/2*32)/(1*114)=3.50

1 mole of air has 23.2 mass percent of O2,

so amount of air containing 3.5 mass percent of O2=3.5*100/23.2=15.08

So air-fuel ratio=15.08