Combustion analysis basic concepts and strategy

analysis of combustion.

1. Combustion Text Book : Thermodynamics by Yunus Cengel , 8 th Edition.

2. COMBUSTION Combustion is a chemical reaction in which certain element of the fuel combine with oxygen and releasing a large quantity of energy causing an increase in temperature of gases. Fuel + Oxygen = Products + Heat There are thousands of different hydrocarbon fuel components, which consist mainly of hydrogen and carbon but may also contain oxygen, nitrogen, and/or sulphur, etc. The oxygen necessary for combustion is obtained from air, which is oxygen diluted chiefly by nitrogen.

3. Composition of Air Table gives proportion of oxygen and nitrogen by volume as well as by mass of dry air. In combustion, oxygen is the reactive component of air. The properties of air vary geographically, with altitude and with time. It is usually sufficiently accurate to regard air as 21% oxygen and 79% inert gases taken as nitrogen. For every one mole of oxygen, there are 3.76 moles of nitrogen in air as per molar ratio. Gas Volume % Mass % Molar Mass Molar Fraction Molar Ratio O 2 N 2 A CO 2 20.95 78.09 0.93 0.03 23.16 75.55 1.25 0.04 32.00 28.01 38.95 44.01 0.21 0.79 1 3.76 Air 100.00 100.00 28.95 1.00 4.76

4. STOICHIOMETRY Most IC engines obtain their energy from the combustion of a hydrocarbon fuel with air, which converts chemical energy of the fuel to internal energy in the gases within the engine. The maximum amount of chemical energy that can be released (heat) from the fuel is when it reacts (combust) with a stoichiometric amount of oxygen. Stoichiometric oxygen (sometimes also called theoretical oxygen) is just enough to convert all carbon in the fuel to CO 2 and all hydrogen to H 2 O, with no oxygen left over.

5. Types of Combustion Reactions Stoichiometric Reaction . (Oxygen and fuel in exact amount required for combustion) Lean Mixture Reaction . (Oxygen in excess) Rich Mixture Reaction . (Fuel in excess)

6. Stoichiometric Reaction A stoichiometric reaction is defined such that the only products are carbon dioxide and water. The reactants are completely utilized in combustion and no reactant is in excess. The components on the left side of a chemical reaction equation which are present before the reaction are called reactants , while the components on the right side of the equation which are present after the reaction are called products. Chemical equations are balanced on a basis of the conservation of mass principle (or the mass balance), which can be stated as follows: The total mass of each element is conserved during a chemical reaction. For example, the stoichiometric reaction of propane would be C 3 H 8 + a O 2 = b CO 2 + d H 2 O

7. Stoichiometric Reaction (Balancing) For balancing, equate the quantity of each element on both sides (Number of atoms x moles) C 3 H 8 + a O 2 = b CO 2 + d H 2 O Carbon balance gives: b = 3; Hydrogen balance gives: 2 d = 8  d = 4; Oxygen balance gives: 2 a = 2 b + d  a = 5. Then, reaction equation becomes C 3 H 8 + 5 O 2 = 3 CO 2 + 4 H 2 O Task : Write stoichiometric reaction for combustion of Butane (C 4 H 10 ) with oxygen and balance it.

8. Using Air instead of Oxygen in Combustion Very small powerful engines could be built if fuel were burned with pure oxygen. However, the cost of using pure oxygen would be prohibitive, and thus is not done. Air is used as the source of oxygen to react with fuel. Nitrogen and argon are essentially chemically neutral and do not react in the combustion process. The complete (Stoichiometric) combustion of a general hydrocarbon fuel of average molecular composition C  H  with air would be C  H  +  (O 2 + 3.76 N 2 ) =  1 CO 2 +  2 H 2 O+  3 N 2 Where, -  is the total moles of air, - (O 2 + 3.76 N 2 ) is the composition of air. -  1 , 2 , 3 are the moles of products.

9. Equivalence Ratio Fuel/Air Equivalence Ratio: For actual combustion in an engine, the fuel/air equivalence ratio is a measure of the fuel-air mixture relative to stoichiometric conditions. It is defined as: where: F / A = m f / m a = fuel-air ratio; A/F = m a / m f = air-fuel ratio; m a = mass of air; m f = mass of fuel

10. Equivalence Ratio Relative Air/Fuel Ratio : The inverse of  , the relative air/fuel ratio  , is also sometimes used. For fuel-lean mixtures:  < 1 ,  > 1, oxygen in exhaust For stoichiometric mixtures:  =  = 1, maximum energy released from fuel. For fuel-rich mixtures:  > 1,  < 1, CO and fuel in exhaust.

11. Overall Combustion Reaction: At low temperatures, the overall combustion reaction can be written as C  H  O  N  + (  /  ) (O 2 + 3.76 N 2 ) =  1 CO 2 +  2 H 2 O +  3 N 2 +  4 O 2 +  5 CO +  6 H 2

12. Lean MIXTURE REACTIONS Fuel-air mixtures with more than or less than the stoichiometric air requirement can be burned. Combustion can occur, within limits, i.e., the proportions of the fuel and air must be in the proper range for combustion to begin. With more than stoichiometric air requirement, i.e., with lean mixture, there is extra oxygen left after combustion which shows up separately in the exhaust. The combustion reaction for lean mixtures thus can be written as; C  H  O  N  + (  /  ) (O 2 + 3.76 N 2 ) =  1 CO 2 +  2 H 2 O +  3 N 2 +  4 O 2

13. RICH MIXTURE REACTIONS With less than stoichiometric air requirement, i.e., with fuel-rich combustion, there is insufficient oxygen to oxidize fully the fuel C and H to CO 2 and H 2 O. The products are a mixture of CO 2 and H 2 O with carbon monoxide CO and hydrogen H 2 (as well as N 2 ). The combustion reaction for rich mixtures can be written as; C  H  O  N  + (  /  ) (O 2 + 3.76 N 2 ) =  1 CO 2 +  2 H 2 O +  3 N 2 +  5 CO +  6 H 2

14. COMBUSTION EFFICIENCY : Under lean operating conditions the amounts of incomplete combustion products are small. Under fuel-rich operating conditions these amounts become more substantial since there is insufficient oxygen to complete combustion. Combustion efficiency indicates the fraction of the fuel energy supplied which is released in the combustion process due to unburnt fuel. Figure shows how combustion efficiency varies with the fuel/air equivalence ratio for internal combustion engines.

15. Problem Write the balanced chemical reaction equation for one mole of iso-octane (C 8 H 18 ) burning with stoichiometric air. Calculate the air/fuel ratio for combustion of this fuel. Solution Hint: Write the equation for stoichiometric reaction and balance it. C 8 H 18 +  (O 2 + 3.76 N 2 ) =  1 CO 2 +  2 H 2 O+  3 N 2 Calculate air/fuel ratio by calculating the masses of air and fuel as per balanced equation.