This lecture discusses the need to under understand and the basis if minimum air requirement for an combustion to happen

1. Stoichiometry of combustion
Problems on volumetric and
gravimetric analysis
By
Dr. S. PUGALENDHI, Dr. D. RAMESH, Er. T. AYISHA NAZIBA

2. Combustion reaction
• Stoichiometry of chemical reactions means that balanced
species reactions
• Fossil fuels are mainly compounds of carbon and hydrogen
(hydrocarbons - C m H n).
• The reaction of its oxidation can by written by the equation of
stoichiometry
• C m H n + (m + n/4)O 2 → mCO2 + n/2H2O
• 1 mol + (m + n/4)moles → (m +n/2)moles

3. Oxidizers
• In combustion processes the oxidizer could be:
1. Oxygen (O 2 )
2. Air (21%O 2 + 79%N 2 )
3. Air enriched with oxygen (O 2 > 21%)
4. Some compounds containing oxygen, like nitrogen
oxides: N 2 O

4. Combustion equations of solid fuels
• C + O2  CO2
• 1 mol + 1mol  1mol
• 12 kg + 32kg  44 kg
• 1 kg +8/3 kg 11/3 kg
• 2C + O2  2CO
• 2 mol + 1mol  2mol
• 24 kg + 32kg  56 kg
• 1 kg +4/3 kg 7/3 kg
• 2CO + O2  2CO2
• 2 mol + 1mol  2mol
• 56 kg + 32kg  88 kg
• 1 kg +4/7 kg 11/7 kg
• S + O2  SO2
• 1 mol + 1mol  1mol
• 32kg + 32kg  64kg
• 1 kg +1 kg 2 kg

5. Combustion equations of gaseous fuels
2H2 + O2  2H2O
2 vol + 1 vol  2 vol
4 kg + 32kg  36 kg
1 kg +8 kg 9 kg
CH4 + 2O2  CO2+2H2O
1 vol + 2 vol  1vol+2 vol
16 kg + 64kg 44kg+36kg
1 kg +4 kg 11/4kg+9/4kg
2CO + O2  2CO2
2 vol + 1 vol  2 vol
56 kg + 32kg  88 kg
1 kg +4/7 kg 11/7 kg
C2H4 + 3O2  2CO2+2H2O
1 vol + 3 vol  2vol+2 vol
28 kg + 96kg 88 kg+36kg
1 kg +24/7 kg 22/7kg+9/4kg

6. Theoretical air requirement
= 100/23 [(8/3C + 8H2 + S)-O2 ] kg
= 100/21 [(0.5CO+ 0.5H2 +2CH4 + 3C2H4)-O2 ]m3

7. Volumetric to Mass analysis

8. Mass to Volumetric analysis