1. Thermo-chemistry of Engine Combustion
P M V Subbarao
Professor
Mechanical Engineering Department
A n Important Clue to Control Rate of Heat Release ….

2. Selection of Mixture Strength

3. Real Combustion & Model Testing

4. Results of Model Testing.
For a given fuel and required Power & Speed conditions.
Optimum composition of Exhaust Gas.
Optimum air flow rate.
Optimum fuel flow rate.
Molar Analysis of Dry Exhaust Products:
Mole fraction of CO2 : x1
Mole fraction of CO : x2
Mole fraction of O2 : x4
Mole fraction of N2 : x5

5. Stoichiometry of Actual Combustion
CXHY +e 4.76 (X+Y/4) AIR+ Moisture → P CO2 +Q H2O + T N2 + U O2 + V CO + WCH4
Conservation species:
Conservation of Carbon: X = P+V+W
Conservation of Hydrogen: Y = 2 Q+4W
Conservation of Oxygen : e (2X+Y/2) = 2P +Q +2U+V
Conservation of Nitrogen: e 3.76 (2X+Y/2) = T

6. For every 100 kg of fuel.
CXHY +e 4.76 (X+Y/4) AIR → P CO2 +Q H2O ++ T N2 + U O2 + V CO + W CH4

7. Dry Exhaust gases: P CO2 + T N2 + U O2 + V CO + W CH4 kmols.
Volume of gases is directly proportional to number of moles.
Volume fraction = mole fraction.
Volume fraction of CO2 : x1 = P * 100 /(P + T + U + V+W)
Volume fraction of N2 : x2= T* 100 /(P + T + U + V+W)
Volume fraction of O2 : x3= U * 100 /(P + T + U + V+W)
Volume fraction of CO : x4= V * 100 /(P + T + U + V+W)
Volume fraction of CH4 : x5= W * 100 /(P + T + U + V+W)
These are dry gas volume fractions.
Emission measurement devices indicate only Dry gas volume fractions.

8. x1 CO2 + x2 N2 + x3 O2 + x4 CO + x5 CH4 +x6H2O
Combustion equation for 100 kmoles of Dry Exhaust gas:
nCXHY +en 4.76 (X+Y/4) AIR →
x1 CO2 + x2 N2 + x3 O2 + x4 CO + x5 CH4 +x6H2O

9. Fuel Lean Mixtures : f <1 Fuel-rich Mixtures: f >1
Air-fuel Ratio:
Equivalence ratio:
Fuel Lean Mixtures : f <1
Fuel-rich Mixtures: f >1