FUEL AND COMBUSTION
ABE 111
Fuels
• three groups: gaseous, liquid and solid.
• gaseous fuels are mostly natural gas and producer gas
• liquid fuels are derived from petroleum
• most solid fuels are coal
• the main limitations on fuels are that it must be easily transportable
through the fuel system to the combustion chamber, and that the fuel
releases sufficient energy in the form of heat
• Petroleum fuel is derived from crude oil composed primarily of hydrogen
(~14%) and carbon (~86%)
• two basic combustible elements are carbon and hydrogen
Essential Characteristics of Engine Fuel
1. Must have a reasonably high energy value
2. Must vaporize at least partially at comparatively low temperatures
3. Fuel vapors must ignite and burn readily when mixed in proper
proportions with oxygen
4. Must be of such nature that it can be handled and transported with
comparative ease and safety
• general formula for hydrocarbons is CnHm
• main types of hydrocarbons are :
1. paraffins, CnH2n+2
2. olefins and naphthenes, CnH2n
3. diolefins, CnH2n-2
4. aromatics, CnH2n-6
• n varies from 1 to about 26 and m from 2 to 54
Paraffin Hydrocarbons
CnH2n+2
• Heptane, C7H16
H H H H H H H
H C C C C C C C H
H H H H H H H
Napthenes, CnH2n
• Octene, C8H16
H H H H H H H H
H C C C C C C C C H
H H H H H H
Diolefins, CnH2n-2
• Hexadien, C6H10
H H H H
C C C C C C
H H H H H H
Aromatics, CnH2n-6
• Benzene, C6H6
H
C
H C C H
H C C H
C
H
Combustion
• a chemical reaction in which certain elements of the fuel combine
with oxygen causing an increase in temperature
• a chemical reaction of carbon and hydrogen in the fuel with oxygen in
the air to form water and other exhaust products
• Liberates heat which raises the cylinder pressure and drives the
piston downward during the power stroke
• main combustible elements are carbon and hydrogen
• internal combustion engines produce air pollution emissions, due to
incomplete combustion of carbonaceous fuel
Combustion
1. Ideal/Theoretical/Stoichiometric Combustion
• All of the hydrogen and carbon in the fuel is converted into water and
carbon dioxide, respectively
2. Rich Combustion
• There is not enough oxygen to convert all of the carbon to CO2
• CO appears in the exhaust
3. Lean Combustion
• Excess oxygen is supplied for combustion
• O2 appears in the exhaust
Approximate Air Composition
By Weight:
23.2% oxygen
76.8% nitrogen and other trace elements
By Volume/Mole:
21% oxygen
79% nitrogen and other trace elements
Approximate Air Composition
Air-Fuel Ratio
• the unit weight or volume of air present in the cylinder per unit
weight or volume of fuel
• the air-fuel ratio necessary for complete combustion depends only
upon the composition of the fuel
Combustion Formula
• Complete combustion for carbon dioxide
C + O2 = CO2
- 1 molecule of carbon combined with 1 molecule of oxygen, forms 1
molecule of carbon dioxide
In the mol system,
1mol C + 1 mol O2 = 1mol CO2 (1)
Substituting for each molecule its molecular weight,
12 + 32 = 44
This shows that 12 units of weight of carbon combine with 32 units of
oxygen will form 44 units of carbon dioxide.
therefore,
1 lb C + 2.67 lb O2 = 3.67 lb CO2
for air, each pound of O2 is accompanied by 3.31 lb of N2
• 1lb C + 2.67lb O2 +8.84 lb N2 = 3.67 lb CO2 + 8.84 lb N2
this means: 1 lb of carbon requires 2.67 + 8.84 = 11.51 lb of air
In air, for every 1 mol of O2 there are 3.76 mol of N2
eq. (1)
1 mol C + 1 mol O2 + 3.76 mol N2 = 1 mol CO2 + 3.76 mol N2
dividing the equation by the molecular weight of carbon, and
for gases be expressed in ft3
1 mol of gas = 379 ft3
1 lb C + 31.6 ft3 O2 + 118.8 ft3 N2 = 31.6 ft3CO2 + 118.8 ft3 N2
1 lb carbon requires 31.6 + 118.8 = 150.4 ft3 of air at
standard conditions
if combustion is conducted with excess air,
1 lb C + (1 + e) 150.4 ft3 of air = 31.6 ft3CO2 + (e) 31.6ft3 CO2 + (1 + e)
118.8 ft3 N2
Example: Carbon Monoxide
C + O2 = CO
C + (1/2)O2 = CO
1 mol C + ½ mol O2 = 1 mol CO
therefore,
1 mol C + ½ mol O2 + 1.88 mol N2 = 1 mol CO + 1.88 mol N2
1 lb C + 15.8 ft3 O2 + 59.4 ft3 N2 = 31.6 ft3 CO + 59.4 ft3 N2
which means that
1 lb carbon when forming carbon monoxide uses 75.2
ft3 of air
Ex. Combustion of CO to CO2
1 mol CO+ ½ mol O2+ 1.88 mol N2 =1 mol CO2 +1.88 mol N2
1 ft3 CO + 0.5 ft3 O2+ 1.88 ft3 N2 = 1 ft3 CO2 +1.88 ft3 N2
1 ft3 CO requires 2.38 ft3 of air
Ex. Combustion of H2 to produce H2O
1 mol H2+ ½ mol O2 +1.88 mol N2=1 mol H2O +1.88 mol N2
1 ft3 H2 +0.5 ft3 O2+1.88 ft3 N2 = 0.0475 lb H2O +1.88 ft3 N2
[2(1) + 16] ÷379
= 0.0475
Ex. Combustion of Hydrogen Sulfide, H2S
H2S + O2 +3.76N2 = H2O + SO2 + 3.76N2
1 mol H2S + 1.5 molO2 +(1.5)(3.76) mol N2 = 1 mol H2O
+ 1 mol SO2 +(1.5)(3.76) mol N2
1 ft3 H2S +7.14 ft3 of air = 0.0475 lb H2O +1 ft3SO2 +5.64 ft3
N2
Formulas for Hydrocarbons
• All hydrocarbons burns in the presence of a sufficient amount of
oxygen
• Hydrocarbons burn to carbon dioxide and water
• The combustion equations may be written and all computations
made as though carbon and hydrogen were separate.
• Methane, CH4
CH4 + O2 +3.76N2 = CO2 + H2O + 3.76N2
1 mol CH4 + 2 mol O2 +(2)3.76 mol N2 = 1 mol CO2 + 2 mol
H2O + (2)(3.76) mol N2
1 ft3 CH4 + 2 ft3 O2 +7.52 ft3 N2 = 1 ft3 CO2 + 0.095 lb H2O +
7.52 ft3 N2
1 ft3 CH4 + 9.52 ft3 of air = 1 ft3 CO2 + 0.095 lb H2O + 7.52 ft3 N2
• Benzene, C6H6
C6H6 + O2 +3.76N2 = CO2 + H2O + 3.76N2
1 mol C6H6 + 7.5 mol O2 +(7.5)(3.76) N2 = 6 mol CO2
+ 3 mol H2O + (7.5)(3.76) mol N2
1 lb C6H6 + 36.4 ft3 O2 +137 ft3 N2 = 29.2 ft3 CO2 + 0.69 lb
H2O + 137 ft3 N2
1 lb C6H6 + 173.4 ft3 of air = 29.2 ft3 CO2 + 0.69 lb H2O +
137 ft3 N2
Write the combustion formula of the following hydrocarbons
• Ethylene, C2H4
• Octane, C8H16
Alcohols
Ethyl Alcohol, C2H6O
C2H6O + O2 + 3.76N2 = CO2 + H2O + 3.76N2
1 mol C2H6O + 1 mol O2 + 3.76 mol N2 = 1 mol CO2 +
1 mol H2O + 3.76 mol N2
1 mol C2H6O + 3 mol O2 + (3.76)(3) mol N2 = 2 mol CO2 +
3 mol H2O + (3.76)(3) mol N2
1 lb C2H6O + 24.7 ft3 O2 + 92.9 ft3 N2 = 14.6 ft3 CO2 + 1.17 lb H2O +
92.9 ft3 N2
1 lb C2H6O + 117 ft3 of air = 14.6 ft3 CO2 + 1.17 lb H2O + 92.9 ft3 N2
Methyl Alcohol, CH4O
CH4O + O2 + 3.76N2 = CO2 + H2O + 3.76N2
1 mol CH4O + 1.5 mol O2 + (1.5)(3.76) mol N2 = 1 mol CO2 +
2 mol H2O + (1.5)(3.76) mol N2
1 lb CH4O + 17.7 ft3 O2 + 66.7 ft3 N2 = 11.8 ft3 CO2 + 1.13 lb
H2O + 66.7 ft3 N2
1 lb CH4O + 83.7 ft3 of air = 11.8 ft3 CO2 + 1.13 lbvH2O + 66.7
ft3 N2
Thank you!
