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.

Octane C8H18, molar mass 114.2

In air, the ratio N2/O2 is 79/21 = 3.76

C8H18     + 12.5 O2 +   47 N2   = 8CO2   + 9 H2O +   47 N2      + C8H18(not burned)

114g …………….………………….352 g…….162 g…..1316 g…….114x2/100 = 2.28

100 mol ……………………………800 mol…900 mol….4700 mol…..2 mol

Total exhaust gases = 800 mol+900 mol+4700 mol+2 mol = 6402 mol

6.402 kmol x 30 kg/kmol = 192.06 kg exhaust total containin 2 mol (228 g) C8H18 unburned

C8H18 content in exhaust gases is

0.228 kg/ 192.06 kg = 1.2 x 10^-3 = 1200 x10^-6 kg/kg = 1200 ppm

Alternative calculation (simplified)

From 1 kg gasoline 2 % ( i.e. 0.020 Kg) is not burned.

The unburned hydrocarbon fraction in the exhaust is

0.020 Kg / 15 kg = 1300x 10⁶ kg/kg = 1300 ppm.

B. The stoichiometric air/fuel ratio is

C8H18               + 12.5 O2 +   47 N2   = ………………….

114 g C8H18…(12.5+47=)59.5 mol air or 59.5 mol x 22.4L/mol = 1333 L

Ratio 1333 L/114 g = 11.7 L air/g gasoline