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    Lecture 1: Basics of Combustion EngineeringCombustion Stoichiometry

    1.1 Introduction

    1.2 Definitions

    1.3 Combustion stoichiometry for gaseous fuels

    1.4 Combustion stoichiometry for liquid and solid fuels

    1.5 Combustibles burnout for solid fuels

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    Examples of industrial combustion systems

    Boiler fired with pulverised coal

    Pulverised coal flame

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    Examples of industrial combustion systems

    Modern reheating furnace

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    Examples of industrial combustion systems

    Flameless oxidation

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    1.2 Definitions

    Chemical Reactions, Atoms and Molecules in Combustion

    H

    1

    2 O H O2 2 2+

    The atoms are conserved (neither created nor destroyed)

    Molecules are not conserved

    Water is the productHydrogen and oxygen are

    reactants

    Atoms relevant in combustion are: C,H,O,N,S,Cl

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    1.1 Definitions

    Compounds of carbon and hydrogen are called hydrocarbons.

    HYDROCARBONS

    Aliphatic

    Alkanes CnH2n+2Alkenes CnH2nAlkynes C

    n

    H2n-2

    Alicyclic

    CH2

    CH2 CH2

    Aromatic

    Benzene

    Benzene derivatives

    Naphtalene

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    The first ten members of the unbranched-chain alkanes series are:

    CH4 methane C6H14 hexane

    C2H6 ethane C7H16 heptane

    C3H8 propane C8H18 octaneC4H10 butane C9H20 nonane

    C5H12 pentane C10H22 decane

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    Table 1.1 Names of aliphatic hydrocarbons

    CnH2n+1-CnH2n-2CnH2nCnH2n+2n

    C5H11- PentylC5H8 PentyneC5H10 PenteneC5H12 Pentane5

    C4H9-ButylC4H6 ButyneC4H8 ButeneC4H10 Butane4

    C3H7-PropylC3H6 ButyneC3H10PropeneC3H8 Propane3

    C2H5- EthylC2H2 EthyneC2H4 EtheneC2H6 Ethane2

    CH3- MethylCH4 Methane1

    Alkyl

    group

    AlkyneAlkeneAlkaneNo.of C

    atoms

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    Other molecules relevant in combustion are:

    Haloalkanes R-X CH3Cl (chloromethane)

    Alcohols R-OH C2H5OH (ethanol)

    Amines R-NH2 CH3NH2 (methylamine)

    Aldehyde R-COH CH3COH (ethanal)

    Ketons R-CO-R CH3COCH3 (propanone)

    Carboxylic Acid R-COOH CH3COOH (ethanoic acid)

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    Amount of substances, mole and mass fractions

    1 mole contains 6.023 x 1023 particles (atoms, molecules)

    For a mixture of species:

    n (total number of moles)=ni

    xn

    nii

    = M x Mmean i i=

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    Mole (volume) fractions and mass fractions:

    w i = =num ber of kg of species " i"

    total num ber of kg in the system

    = =

    n M

    n M

    x M

    x M

    i i

    k kk

    i i

    k kk

    xi = =number of moles of species "i" in 1kg of mixture

    total number of moles in 1kg of mixture

    = =

    w M

    M

    w M

    w M

    i i

    m e a n

    i i

    k kk

    /

    /

    /

    /1

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    Equation of state for gases and gas mixtures

    F p T c( , , ) = 0or

    F p T( , , ) =

    0The perfect gas equation:

    p V n R T= or cp

    R T=

    = =

    p M

    R T

    p

    R Tw

    M

    m e a n

    i

    i

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    The perfect gas law:

    p V n R T=

    Under constant pressure and temperature one mole (kmol)

    of any ideal gas occupies the same volume.

    At normal conditions:

    p = 760 Tr (1 Tr=133,322 N/m2)

    T = 273.15 K (0 C)

    1 kmol of gas = 22.418 mn3

    1 mol of gas = 22.418 dmn3

    R = 8,314 J/kmol/K

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    1.2 Combustion Stoichiometry for Gaseous Fuels

    Stoichiometric Combustion

    Combustion is said to be stoichiometric if fuel and oxidizer

    consume each other completely forming only CO2

    and H2O

    CH 2O 2H O CO4 2 2 2

    + + stoichiometric

    CH 3O 2H O CO O4 2 2 2 2+ + + lean

    CH O H O 0.5CO 0.5CH4 2 2 2 4+ + + rich

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    1.3 Combustion Stoichiometry for Gaseous Fuels

    Mole fraction of fuel in a stoichiometric mixture

    1 kmol fuel O products (CO H O)2 2+ + 2

    xfuel stoich in oxygen, _ _ =+

    =number of moles of fuel

    total number of moles (fuel oxygen)

    =+1

    1

    xfuel stoich in air, _ _ / . .=

    + =

    +1

    1 0 211

    1 4 762

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    For example:

    CO 0.5O CO2 2+ xfuel,stoich_in_oxygen = 1/1.5=2/3

    xfuel,stoich_in_air= 1/(1+0.5/0.21)=0.2958

    C H 5O 3CO 4H O3 8 2 2 2+ +

    Xfuel,stoich_in_oxygen = 1/(1+5)=1/6

    xfuel,stoich_in_air= 1/(1+5/0.21)=0.0403

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    Excess air ratio (air equivalence ratio)

    = =( / )

    ( / )

    ( / )

    ( / )

    x x

    x x

    w w

    w w

    air fuel

    air fuel stoich

    air fuel

    air fuel stoich

    Fuel equivalence ratio =1

    Rich combustion > 1< 1

    Stoichiometric combustion = 1

    Lean combustion > 1