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

## Presentation on theme: "Review of Chemical Thermodynamics Combustion MECH 6191 Department of Mechanical and Industrial Engineering Concordia University Lecture #1 Textbook: Introduction."— Presentation transcript:

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

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

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

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

Some hydrocarbon fuels at stoichiometric condition

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

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)….

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

Some hydrocarbon fuels combustion with air at stoichiometric condition

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 )

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)

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)

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:

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

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)

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)

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

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

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

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

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

By mass

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

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:

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

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

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

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

Similar presentations