1. Review of Chemical Thermodynamics Combustion MECH 6191 Department of Mechanical and Industrial Engineering Concordia University Lecture #1 Textbook: Introduction and Chapter 1

2. 1. Stoichiometry – First “approximation” of how much fuel reacts with how much oxidizer to give how much product 2. Energy balance – concerns with energetics, first law of thermodynamics Objective Basic question: How much energy is released? Fuel + oxidizer -> products of combustion + energy reactantsproducts

3. Combustion involves chemical changes: ex.H 2 + 1/2 O 2  H 2 O  In chemical reactions, the total mass of the system is conserved, but the mole of chemical species are not!! First basic task in combustion analysis:  Stoichiometry deals with the conservation of atomic species as the basic rule in figuring out the number of moles of the different product species for a given number of moles of reactants. - What does it mean when a mixture is said to be in stoichiometric condition? Combustion stoichiometry

4. Stoichiometric: just enough O 2 to consume the fuel for complete reaction, i.e. right amount of O 2 reacting with the fuel to form H 2 O and CO 2. Example: The overall chemical equation for the complete combustion of one mole of propane (C 3 H 8 ) with oxygen is: Elements cannot be created or destroyed, so C balance: b= 3 H balance: 2c= 8  c= 4 O balance: 2b + c = 2a  a= 5 Thus the above reaction is: # of moles species i.e. 16.7% (by mole) of propane burns with 83.3% oxygen for complete reaction Combustion stoichiometry  Carbon goes to CO 2  Hydrogen to H 2 O complete combustion

5. Some hydrocarbon fuels at stoichiometric condition

6. In many combustion technology, the fuel is usually burned with air For stiochiometric mixture with air instead of with pure oxygen:  For every molecule of O 2 we need from air, we bring in 3.76 moles of N 2 as well.  For stoichiometric analysis, N 2 is usually inert (no dissociation) and there is the same amount of mole of N 2 in the product as in the reactant. Fuel-air combustion

7. Oxidation by air Except for chemical rockets, most of the mobile power plants burn fuels with air (air-breathing engine device). Air is a mixture of gases including oxygen (O 2 ), nitrogen(N 2 ), argon (Ar), carbon dioxide (CO 2 ), water vapour (H 2 O)….

8. Composition of Standard Dry Air In combustion, since oxygen is the oxidizer in air, it is usually sufficiently accurate to consider dry air composed of 21% O 2 and 79% N 2 by volume (mole fraction). For every mole of O 2 there are 3.76 moles of N 2. The molecular weight of air is

9. Some hydrocarbon fuels combustion with air at stoichiometric condition

10. Combustion Stoichiometry The complete reaction of a general hydrocarbon C  H  with air is: The above equation defines the stoichiometric proportions of fuel and air. Example: For propane (C 3 H 8 )  = 3 and  = 8 C balance:  = b H balance:  = 2c  c =  / 2 O balance: 2a = 2b + c  a = b + c / 2  a =  +  /4 N balance: 2 ( 3.76)a = 2d  d = 3.76a/2  d = 3.76(  +  / 4 )

11. Combustion Stoichiometry The complete reaction of a general hydrocarbon C  H  O    with air is: C balance:  = b H balance:  = 2c  c =  /2 O balance: 2a+  = 2b + c  a =  +  /4 –  /2 N balance: 2(3.76)a+  = 2d  d =  /2 + 3.76(  +  /4 –  /2)

12. The mole fraction, x i, of any given species is defined as: The mixture internal energy U and enthalpy H (units: kJ) is: where are molar specific values (units: kJ/kmol) The mixture molar specific internal energy and enthalpy (units kJ/ kmol) is: The total number of moles and the mole concentration in the mixture is: Mixture rules (Basic definition)

13. The mixture internal energy U and enthalpy H (units: kJ) is: The mixture specific internal energy u and enthalpy h (units: kJ/kg) is: The mass fraction, y i, of any given species is defined as: The mass m of a mixture is equal to the sum of the mass of n components The mixture can be expressed either by its mass (m) or mole number (n). And the total density is:

14. Relationship: where M i = molar weight of species i where the mixture molecular weight, M, given by: Mixture rules (Basic definition)

15. H 2 + 0.5 (O 2 +3.76 N 2 ) = H 2 + 2.38 air Mole fraction of H 2 = n H2 /n total = 1/(2.38+1) = 0.295 Mole fraction of Air = n air /n total = 2.38/(2.38+1) = 0.705 C 8 H 18 + 12.5 (O 2 +3.76 N 2 ) = C 8 H 18 + 59.5 air Mole fraction of C 8 H 18 = n C8H18 /n total = 1/(59.5+1) = 0.01653 Mole fraction of Air = n air /n total = 59.5/(59.5+1) = 0.98347 For heavy hydrocarbons we need only about 1% fuel for stoichiometric combustion. Mixture rules (Basic definition)

16. H 2 + 0.5 (O 2 +3.76 N 2 ) = H 2 + 2.38 air weight percent of H 2 = mass H2 /mass total = n H2 *Mw H2 /(nH2*Mw H2 +n air *Mw air ) = 2/(2+2.38*29) = 2.816% Mixture rules (Basic definition)

17. Mixture rules For most of the combustion problems, the perfect gas equation of state holds accurately.  Each specie behaves as if the others are not present. Consider a system divided into different compartments with n A moles of specie A in a compartment of volume V A, n B moles in a Volume V B, etc. All the gases in the various compartments have the same pressure p and temperature T. Applying the perfect gas law: Partial pressure

18. Mixture rules The mole fraction of the various species is just the partial pressure ratios. Important relationship used to prepare a mixture in practice.  cannot measure mole fraction experimentally  instead monitor the pressure during preparation of a combustible mixture

19. Ideal Gas Model for Mixtures Let say I want to prepare a mixture with n A of species A and n B mole of species B with total pressure of 1 atm 1. Calculate the corresponding partial pressure for each species in the mixture 2. Evacuate your tank, fill up the tank with gas A until it reach the pressure p A 3. Continue to fill the tank with gas species B until the final desired total pressure p is reached. Method of partial pressure

20. Combustion Stoichiometry The fuel-air or air-fuel ratio refers to the ratio of the amount of fuel to the amount of air (by moles or mass) or vice versa for air-fuel ratio. The stoichiometric air/fuel ratio: Mass-based fuel/air ratio: Mole-based

21. Combustion Stoichiometry Substituting the respective molecular weights and dividing top and bottom by  one gets the following expression that only depends on the ratio of the number of hydrogen atoms to carbon atoms (  ) in the fuel. Note above equation only applies to stoichiometric mixture The stoichiometric mass based air/fuel ratio for C  H  fuel is: For typical petroleum based fuel: (A/F) stoich ~14.2-15

22. By mass

23. Off-stoichiometric Mixture Fuel-air mixtures with more than stoichiometric air (excess air) can burn. With excess air you have fuel lean combustion - excess air (i.e. O 2 ) in the product - major products: CO2, H2O, O2, N2 - minor products: HC, CO, H2, NO With less than stoichiometric air you have fuel rich combustion. - insufficient oxygen to oxidize all the C and H in the fuel to CO 2 and H 2 O. - Incomplete combustion where carbon monoxide (CO) and molecular hydrogen (H 2 ) also appear in the products. - major products: CO2, H2O, CO, H2, N2 - minor products: HC, O2, NO

24. The equivalence ratio, , is commonly used to indicate if a mixture is stoichiometric, fuel lean, or fuel rich. Off-Stoichiometric Mixtures stoichiometric  = 1 fuel lean  < 1 fuel rich  > 1 Stoichiometric mixture: Off-stoichiometric mixture:

25. Off-Stoichiometric Conditions Other terminology used to describe how much air is used in combustion: e.g. 150% theoretical air = 50% excess air = lean mixture 85% theoretical air = 15% deficient in air = rich mixture

26. The stoichiometric reaction is: 10% excess air is: 16 + 9 + 2 a = 1.1(12.5)(2)  a = 1.25, b = 1.1(12.5)(3.76) = 51.7 Consider a reaction of octane with 10% excess air, what is the equivalence ratio,  ? Example

27. Summary Combustion theory - Stoichiometry > reaction and element balance > definition: theoretical air, excess air, mass-based A/F or F/A, equivalence ratio Next class - Energy balance and chemical equilibrium Objective of this first review lecture - Be aware of different basic definition in thermochemistry