3. combustion thermodynamics

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    Prof. MohananDepartment of Mechanical Engg,

    NITK Surathkal

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    Thermodynamics : Deals with quantities of state at equilibrium

    Temperature

    Internal energy

    Enthalpy Entropy

    Availability and so on

    Chemical Thermodynamics :

    Deals with concentration of Reactants and chemicalinteractions.

    Rate of reaction -- > how fast the reaction takes place.

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    First Law Analysis for Reacting SystemConsider a constant pressure process in which nfmoles of fuel react withnamoles of air to produce npmoles of product:

    PnAnFn paf

    Applying First Law with state 1 being the reactants at P1, T1and state 2being products at P2, T2:

    Reactants Products

    Reactants Products

    ReactionState 1 State 2

    Q

    W

    )()( 121221 VVPUUQ

    WUQ

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    RRii

    PpiiRP ThnThnHH

    HH

    VPUVPU

    )()(

    )()(

    12

    111222

    HP< HR Q< 0 exothermic reaction

    HP> HR Q> 0 endothermic reaction

    )()( 1212 VVPUUQ

    First Law Analysis for Reacting System

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    Enthalpy of ReactionConsider the case where the final temperature of the products is the same asthe initial temperature of the reactants (e.g., calorimeter is used to measure Q).

    Reaction Q

    W

    P1=P2=PoT1=T2=To

    The heat released under this situation is referred to as the enthalpy ofreaction, HR ,

    fuelofkmolorkgperkJ:units)()(

    )()(

    Roii

    Poii

    RRii

    PpiiR

    ThnThn

    ThnThnH

    To

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    Enthalpy Scale for a Reacting System

    By international convention, the enthalpy of every element in its natural state(e.g., O2(g), N2(g), H2(g), C(s)) at STP has been set to zero

    0)298,1( ofhKatmh

    We need to take into account that for a reacting system the working fluidchanges molecularly from reactants to products while undergoing a process.

    Consider the following identity:

    )]298,1(),([)298,1(),( KatmhTPhKatmhTPh

    (note the notation convention)

    at STP

    i.e,

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    The enthalpy of all other substances at STP is simply the heat of formation

    of the substance, since it is formed from its elements, for example:

    )]298,1(),([),(,

    KatmhTPhhTPhii

    o

    ifi

    sensible enthalpychemical enthalpy T

    K ipdTc

    298 ,

    Therefore, the enthalpy of the ith component in a mixture is:

    Enthalpy Scale for a Reacting System

    olOHflOH

    o

    lOHfgHgOlOH

    hh

    hhhhQSTPrecall

    lOHgHgO

    )(,)(

    )(,)()()(

    222

    22

    2222

    2/1@

    )()()(2/1

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    It is the heat of reaction, in a reaction inwhich species alone is the product & thecomponent elements of the species are theonly reactants at standard state.(298.16k, 0.1

    Mpa)

    Heat of formation of the reactants are set tozero at standard state.

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    Heat of Formation for Different Fuels

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    Heat of formation = Heat of combustion (refers to species) = (refers to reaction)

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    For any reaction aiAi -- > aiAi Where ai& aiare stoichiometric coefficients and Ai

    refer to species.

    Heat of Combustion =

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    The maximum amount of energy is released from a fuel when reacted with astoichiometric amount of air and all the hydrogen and carbon contained in thefuel is converted to CO

    2and H

    2O

    This maximum energy is referred to as the heat of combustionor the heatingvalue and it is typically given per mass of fuel

    Heat of Combustion

    222224

    76.32

    )76.3(4

    NOHCONOHC

    HR(298K)

    alcohols

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    Fuel Energy

    density

    (MJ/L)

    Air-

    fuel

    ratio

    Specific

    energy

    (MJ/kg air)

    Heat of

    vaporizatio

    n

    Gasolineand

    biogasoline

    32 14.6 2.9 0.36 MJ/kg

    Butanol fuel 29.2 11.2 3.2 0.43 MJ/kg

    Ethanol fuel 19.6 9.0 3.0 0.92 MJ/kg

    Methanol 16 6.5 3.1 1.2 MJ/kg

    http://en.wikipedia.org/wiki/Energy_densityhttp://en.wikipedia.org/wiki/Air-fuel_ratiohttp://en.wikipedia.org/wiki/Air-fuel_ratiohttp://en.wikipedia.org/wiki/Air-fuel_ratiohttp://en.wikipedia.org/wiki/Specific_energyhttp://en.wikipedia.org/wiki/Heat_of_vaporizationhttp://en.wikipedia.org/wiki/Heat_of_vaporizationhttp://en.wikipedia.org/wiki/Heat_of_vaporizationhttp://en.wikipedia.org/wiki/Biogasolinehttp://en.wikipedia.org/wiki/Biogasolinehttp://en.wikipedia.org/wiki/Ethanol_fuelhttp://en.wikipedia.org/wiki/Methanolhttp://en.wikipedia.org/wiki/Methanolhttp://en.wikipedia.org/wiki/Ethanol_fuelhttp://en.wikipedia.org/wiki/Biogasolinehttp://en.wikipedia.org/wiki/Gasolinehttp://en.wikipedia.org/wiki/Heat_of_vaporizationhttp://en.wikipedia.org/wiki/Heat_of_vaporizationhttp://en.wikipedia.org/wiki/Heat_of_vaporizationhttp://en.wikipedia.org/wiki/Specific_energyhttp://en.wikipedia.org/wiki/Specific_energyhttp://en.wikipedia.org/wiki/Air-fuel_ratiohttp://en.wikipedia.org/wiki/Air-fuel_ratiohttp://en.wikipedia.org/wiki/Air-fuel_ratiohttp://en.wikipedia.org/wiki/Air-fuel_ratiohttp://en.wikipedia.org/wiki/Energy_densityhttp://en.wikipedia.org/wiki/Energy_densityhttp://en.wikipedia.org/wiki/Fuel
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    Heat of CombustionThere are two possible values for the heat of combustion depending onwhether the water in the products is taken to be saturated liquid or vapour.

    Tp

    T

    S

    hghfFrom steam tables:

    hfg= hghf > 0

    HR= HPHR < 0 (exothermic)

    The term higher heat of combustionis used when the water in the productsis taken to be in the liquid state (hH20= hf)

    The term lower heat of combustionis used when the water in the productsis taken to be in the vapour state (hH20= hg)

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    0

    h kJ/kg fuel)

    T

    Products with H2O (g)

    Products with H2O (l)

    Reactants

    hfg,H2Oper kgfuel

    hlo

    whhigh

    Heat of Combustion, graphical

    298K

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    Adiabatic Flame TemperatureConsider the following adiabatic constant pressure process:

    )()(

    0)()(

    1

    Rii

    Paii

    RRii

    Ppii

    ThnThn

    ThnThnQ

    For a constant pressure process, the final products temperature, Ta, is known

    as the adiabatic flame temperature (AFT).

    For a given reaction where the nis are known for both the reactants and theproducts, Tacan be calculated explicitly.

    FuelAir Productseactants

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    Adiabatic Flame Temperature

    P R

    o

    ifi

    o

    ifiR iiiP iaii

    Rii

    o

    ifiP

    iai

    o

    ifi

    Rii

    Paii

    hnhnKhThnKhThn

    KhThhnKhThhn

    ThnThn

    ,,1

    1,,

    1

    )298()()298()(

    )298()()298()(

    )()(

    o

    RHSensible heat of reactants(equal to 0 if T1= 298K)

    Sensible heat of products

    P R

    o

    ifi

    o

    ifiR

    T

    ipiP

    T

    ipi hnhndTcndTcn ia

    ,,298 ,298 ,OR

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    The flame temperatureis attained as in anadiabatic situation.

    Reactants are kept inan isolated cylinder

    with a piston tosimulate the pressure. They are sparked to

    initiate ignition. Reaction proceeds to

    completion. Temperature rises toadiabatic flametemperaure.

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    STEPS:

    Composition of the products is to be found out .

    Heat of combustion (HOC) is to be calculated. HOC is used to raise the temp from Toto Tf This rise in temp is in the form of sensible enthalpy

    rise.

    If all the HOC is used up to raise the temp to Tf, thenthe adiabatic flame temp can be obtained.

    Tfis assumed first.

    If some heat is left in balance or heat absorbed islarger than available, keep computing the

    composition & checking the balance till heat balanceis satisfied to the required degree of accuracy.

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    Ta,

    Constant Pressure Adiabatic Flame Temperaturewith products at equilibrium

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    Constant Pressure Adiabatic Flame Temperaturewith products at equilibrium

    nitromethane

    octaneethanol

    hydrogen

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    Estimate the constant-pressure adiabatic

    flame temperature for the combustion of astoichiometric CH4-air mixture. The pressureis 1 atm and the initial reactant temperatureis 298 K.

    Use the following assumptions:1.Complete combustion(no dissociation) i.e.,the product mixture consists of onlyCO2,H2O,and N2.

    2. The product mixture enthalpy is estimatedusing constant specific heats evaluated at1200K(~0.5(Ti+Tad),where Tadis guessed tobe about 2100K)

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    Mixture composition:

    Properties(Appendices A and B)

    52.7,2,1

    52.721)76.3(2

    222

    222224

    NOHCO NNN

    NOHCONOCH

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    Species Enthalpy of Formation @298K

    h0f,i(kJ/kmol)

    Specific Heat@1200K

    cP,i(kJ/kmol-K)

    CH4 -74,831 -----

    CO2 -393,546 56.21

    H2O -241,845 43.87

    N2 0 33.71

    O2 0 ----

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    298)]-33.71((7.52)[0

    )]298(87.43845,241)[2(

    )298(21.56546,393)[1(

    )]298([

    831,74)0(52.7)0(2)831,74)(1(

    ,

    0

    ,

    ad

    ad

    ad

    adipifiprod

    react

    i

    prod

    iprod

    react

    iireact

    T

    T

    T

    TchNH

    kJH

    hNHhNH

    Equating Hreactto Hprodand solvingfor Tad yieldsTad=2318K

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    Constant Volume AFT

    Reaction Q

    W

    Consider the case where the piston is fixed and the cylinder is perfectly

    insulated so the process is adiabatic (Q= 0)

    )()(

    0)()(

    1

    Rii

    Paii

    RRii

    Ppii

    TunTun

    TunTunQ

    Note h = u + pv = u + RT, so

    T)R)(Th(nTRThnR

    iiP

    aii 1))((

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    p Riai

    iP

    iP R

    o

    ifi

    o

    ifiR

    iiiP

    aii

    TRnTRn

    KhnhnhnKhThnThn

    1

    ,,1

    )298()298()()(

    Extra term compared to constant pressure AFT ( term > 0)

    The AFT for a constant volume process is larger than for a constantpressure process.

    The AFT is lower for constant pressure process since there is Pdvworkdone

    Constant Volume AFT

    R

    iii

    o

    ifiP

    iiai

    o

    ifi TRKhThhnTRKhThhn )298()()298()( 1,,

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    Estimate the constant-volume adiabatic flametemperature for a stoichiometric CH4-airmixture using the same assumptions as inExample 2.5. Initial conditions are T i=298K,P=1 atm 101,325Pa).

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    The same composition and properties used inExample 3 apply here. We note, however, thatthe cp,ivalues should be evaluated at atemperature somewhat greater than 1200K,

    since the constant-volume Tadwill be higherthan the constant-pressure Tad. Nevertheless,we will use the same values as before.

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    First Law:

    0)(

    0)(

    adprodinitreacuprod

    ii

    reac

    ii

    adprodinitreacuprodreac

    TNTNRhNhN

    orTNTNRHH

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    kJT

    T

    T

    TH

    kJH

    ad

    ad

    ad

    adprod

    reac

    )298(5.397236,877

    )]29833.71((7.52)[0

    )]298(87.43845,241)[2(

    )]298(21.56546,393)[1(

    831,74)0(52.7)0(2)831,74)(1(

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    And

    where reac=prod=10.52kmol.Reassembling and solving for Tadyields Tad=2889K

    )298)(52.10(315.8)( adadprodinitreacu TTNTNR

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    Combustion by Irvin Glassman Fuels and combustion by Sharma and

    Chander Mohan

    Introduction to combustion byStephen.R.Turns