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

Selection of Mixture Strength

Real Combustion & Model Testing

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

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

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

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.

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

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