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  • 8/10/2019 CHAPTER 3 Combustion

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    CHAPTER 3

    COMBUSTION

    3.1 Introduction

    In this chapter is dealing with the system undergoes the chemical changes

    during a process, that is, system that involve chemical reaction known as

    combustion.

    In practical engines and power plants the source of heat is the chemical

    energy of substances called fuels. This energy is released during the

    chemical reaction of the fuel with oxygen.

    The combustion process takes place in the combustion chamber after

    initiation of combustion by an ignition device such as an electric spark. The

    most convenient source of oxygen supply is from the atmosphere.

    Internal combustion engines are run on liquid fuels which are group as

    gasoline (petrol), diesel oil and gaseous fuels. here as the power plants run

    mainly on kerosene and natural gas.

    3.2 Basic Cheistr!

    !toms " the smallest particle that take part in chemical reaction e.g. #, $, %,

    &, ' etc.

    olecules " a combination of atoms that hold together by strong inter

    atomic forces $*, %*, '*etc.

    +elative molecular mass " relative masses of the atoms which constitute themolecule e.g. $* *, %* -*, '* -*, # *, & -* etc.

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    Table -. #hemical compositions of liquid and gases

    Co"ound #oru$a Re$ati%e Mo$ecu$ar

    Mass

    ater/steam $*% (* x ) 0 ( x 1) 2#arbon monoxide #% ( x *) 0 ( x 1) *2

    #arbon dioxide #%* ( x *) 0 (* x 1) 33

    &ulphur dioxide &%* ( x -*) 0 (* x 1) 13

    ethane #$3 ( x *) 0 (3 x ) 1

    4thane #*$1 (* x *) 0 (1 x ) -5

    6ropane #-$2 (- x *) 0 (2 x ) 33

    n7utane #3$5 (3 x *) 0 (5 x ) 82

    4thylene #*$3 (* x *) 0 (3 x ) *2

    6ropylene #-$1 (- x *) 0 (1 x ) 3*

    n6entane #8$* (8 x *) 0 (* x ) 9*7en:ene #1$1 (1 x *) 0 (1 x ) 92

    Toluene #9$2 (9 x *) 0 (2 x ) ;*

    n%ctane #2$2 (2 x *) 0 (2 x )

    3

    3.3 #ue$s

    !ny materials that can be burned to release thermal energy are calledfuel.

    ost familiar fuels are hydrocarbon fuels that consist primarily of

    hydrogen and carbon such as coal, gasoline and natural gas.

    #oal is a solid fuel contains varying amounts of oxygen, hydrogen,

    nitrogen, sulfur, moisture and ash.

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    #hemical analysis of fuel is called ultimate analysis analysis by

    mass of important elements in the fuel.

    >or solid fuels such as coal, proximate analysis that gives percentage

    of inherent moisture, volatile matter, and combustible solid also can

    be used.

    3.& Co'ustions

    ! chemical reaction during which a fuel is oxidi:ed and a large

    amount of energy is released is called combustion.

    The oxidi:er most often used in combustion processes is air, for

    obvious reasons " it is free and readily available. The composition of

    dry air can be approximated to *? of %*and 9;? of '*by molenumbers.

    The chemical reaction can be expressed by equation such as@

    *$* 0 %* A *$*%

    * volume $*0 volume %*A * volumes $*% (based on volume

    analysis)

    3kg $* 0 -*kg %* A -1kg $*% (based on mass analysis)

    %r

    kg $* 0 2kg %* A ;kg $*%

    %r

    *kmol $* 0 kmol %* A *kmol $*%

    >or the air used in the combustion processB

    ******

    9;*

    *

    9;* NOHNOH +++

    &ince kmol of oxygen there are 9;/* kmol of nitrogen.

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    ratioFAtricstoichiome

    ratioFAtricstoichiomeratioFAactual

    D/D

    D/DD/D =

    >or gaseous fuels the ratios are expressed by volume and for solid and liquid

    fuels the ratios are expressed by mass.

    6etrol engines have to meet various conditions of load and speed, and

    operate over a wide range of mixture strengths@

    ratioFAactual

    ratioFAtricstoichiomestrengthMixture

    D/D

    D/DD =

    The working values range between 25? (weak) and *5? (rich).

    E*a"$e 3.1

    %ne kmol of octane (#2$2) is burned with air that contains *5kmol of

    oxygen. !ssuming the product contain only #%*, $*%, %*, and '*, determine

    the mole number each gas in the products and the airfuel ratio for this

    combustion process

    So$ution+

    #hemical equation for this combustion processB

    ******22 )91.-(*5 wNzOOyHxCONOHC +++++

    Thus, from the equation

    #@ x 2

    $@ *y 2, y ; %@ *x 0 y 0 : 35, : 9.8

    '*@ w 98.*

    &ubstituting will give@

    ******22 *.988.9;2)91.-(*5 NOOHCONOHC +++++

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    )(2)*(2

    *;91.3*5/

    +

    ==fuel

    air

    m

    mFA

    *3.*kg air/kg fuel

    3., E*haust and #$ue -as Ana$!sis

    The products of combustion are mainly gaseous

    ! sample is taken for analysis cooled down to a temperature which

    below the saturation of the steam present. The dew point also can be

    calculated as the water vapour condensed@

    prod

    prod

    prodv

    prodv Pn

    nP

    = ,,

    >rom the dew point temperature, Tdpcan be obtained from the

    steam table for 6v,prod.

    &ince the products are gaseous, analysis by volume is usually used

    E*a"$e 3.2

    ! sample of dry anthracite has the following composition by mass,

    # ;5? B $ -? B % *.8? B ' ? B & 5.8? B ash -?

    #alculate@

    (i) the stoichiometric !/> ratio

    (ii) the !/> ratio and the dry air and wet analysis of combustion by

    mass or by volume, when *5? excess air is supplied

    &olution

    (i) 4ach constituent for complete combustionB

    #arbon@

    ** COOC + *kg # 0 -*kg %*A 33kg #%*

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    %xygen required 5.; x -*/* *.3kg/kg coal

    #arbon dioxide produced 5.; x 33/* -.-kg #%*

    $ydrogen@

    OHOH ***.

    * + *kg $*0 1kg %*A 2kg $*% kg $*0 2kg %*A ;kg $*%

    %xygen required 5.5- x 2 5.*3kg/kg coal

    &team produced 5.5- x ; 5.*9kg/kg coal

    &ulphur@

    ** SOOS + -*kg & 0 -*kg %*A 13kg &%* kg & 0 kg %*A *kg &%*

    %xygen required 5.558kg/kg coal

    &ulphur dioxide produced * x 5.558 5.5kg/kg coal

    #onstituent ass >raction %xygen +equired

    (kg/kg coal)

    6roduct ass

    (kg/kg coal)

    #arbon (#)

    $ydrogen ($)

    &ulphur (&)

    %xygen (%)

    'itrogen (')

    !sh

    5.;55

    5.5-5

    5.558

    5.5*8

    5.55

    5.5-5

    *.355

    5.*35

    5.558

    5.5*8

    -.-5 (#%*)

    5.*9 ($*%)

    5.5 (&%*)

    5.5('*)

    Total oxygen required per kg of coal *.1* kg

    Thus, total air required per kg of coal *.1*/5.*-- .*38kg

    here air is assumed to contain *-.-? %*by mass

    &toichiometric airfuel ratio .*38

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    (ii) >or air supply which is *5? in excess,

    !ctual !/> ratio .* x .*38 -.3;3@

    -.9 6ractical !nalysis of #ombustion 6roduct

    In combustion process, it is necessary to analy:e the product of

    combustion for study purposes or improvement on the system

    any of equipment now are available on the market that provides a

    quicker, more accurate, and continuous analysis.

    (i) 'ondispersive infrared ('EI+)

    The constituent gas is measured by its optical absorption in the infra

    red spectrum.

    (ii) =as #arbon Eioxide and %xygen +ecorders

    ! digital instrument that uses to indicate the quality of combustion

    process in the boiler

    (iii) =as analy:er

    ! digital instrument that measures the composition of gas products in

    ppm or percentage.

    -.2 Eissociation

    It is found that during adiabatic combustion the maximum

    temperature reached is lower than expected.

    >or this case, the exothermic combustion process can be reversed if

    the temperature is high enough. The reversed process is known as

    endothermic.

    4xample of the combustion process of carbon dioxide and hydrogen

    as follows@

    ** ** COOCO + and OHOH *** ** +

    To obtain relation for chemical equilibrium in terms of the properties

    of the individual components, consider a mixture of four chemical

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    componentsA " C and# which exist in equilibrium at a specified

    T and 6. 'ow consider a reaction which occurs to an infinitesimal

    extent during which differential amounts of reactants A and " are

    converted to products # and E a t T remaining constant throughout

    the process. FanGt $off equilibrium box as shown in >igure -.8 can be

    used to study a reversible combustion process@

    >igure -.8

    a kmol ! 0 b kmol 7 H c kmol # 0 d kmol E

    The process may proceed equally well in either direction and the

    reversal of the process the heat and work transfers would be reversed

    in direction

    =

    =

    *

    * lnlnP

    P$n

    P

    Pm%$&

    This can be applied to each of the compressors and expanders in the

    system

    a

    AAA

    P

    P$

    P

    P$aAoninputwor'&

    =

    ==

    ..

    lnlnDDD

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    (

    """

    P

    P$

    P

    P$("oninputwor'&

    =

    ==

    ..

    lnlnDDD

    c

    C

    C

    CP

    P$

    P

    P$cConinputwor'&

    =

    ==

    .

    . lnlnDDD

    d#

    #

    #P

    P$

    P

    P$d#oninputwor'&

    =

    ==

    lnlnDDD

    Therefore the net work output of the system

    #C"A &&&&& =

    +=

    +

    +

    = + dc(a

    (

    "

    a

    A

    d

    #

    c

    C

    d

    #

    c

    C

    (

    "

    a

    A PPP

    PP$

    P

    P

    P

    P

    P

    P

    P

    P$&

    lnlnlnlnlnln

    &upposed that in second similar system in the same surroundings the

    pressure in the equilibrium box is 6 then it will have a net work

    output, as

    +

    = + dc(a

    (

    "

    a

    A

    d

    #

    c

    CP

    PP

    PP$&

    .lnln

    here #C"A PPPPP +++=

    >or a combination of two system as shown in >igure -.1 below@

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    >igure -.1

    The net work output of the system ()(), thus according to the second

    law of thermodynamics, . Therefore@

    )PP

    PP

    PP

    PP(

    "

    a

    A

    d

    #

    c

    C

    (

    "

    a

    A

    d

    #

    c

    C =

    =

    )()(

    )()(

    here J is the thermal equilibrium or dissociation constant.

    ! standard thermodynamics equilibrium constant, JK, is defined in

    dimensionless form by referring each partial pressure to a pressure of bar,

    +

    =(

    "

    a

    A

    d

    #

    c

    C

    P

    P

    P

    P

    P

    P

    P

    P)

    lnlnlnln

    %r in general

    i

    i

    i

    PP)

    = lnln where Li stoichiometric coefficient

    These expressions can be applied in the combustion process such as

    *** COOCO +

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    ith molar proportions for #%, %*, and #%*of , 5.8, and . Thus,

    =*

    .

    *

    *

    .

    *

    )(

    )(lnln

    OCO

    CO

    PP

    PP)

    >or the combustion of hydrogen,

    OHOH ***.

    * +

    ith molar proportions , 5.8, and . Thus,

    = *.**

    *

    .

    *

    )(

    )(

    OH

    OH

    PP

    PP)

    In the combustion of hydrocarbon fuels both of the above reactions may

    occur simultaneously and another equilibrium constant can be defined by

    dividing

    *

    .

    **

    *

    .

    *

    )(

    )(

    OH

    OH

    PP

    PP

    by

    *

    .

    *

    *

    .

    *

    )(

    )(

    OCO

    CO

    PP

    PP

    Thus,

    =)(

    )(

    **

    *

    COH

    COOH

    PP

    PP)

    The value of JKcan be used to determine the reaction temperature from the

    table.

    E*a"$e 3.3

    ! combustible mixture of carbon monoxide and air which is 5? rich is

    compressed to a pressure of 2.*2bar and a temperature of *2*M#. The

    mixture is ignited and combustion occurs adiabatically at constant volume.

    hen the maximum temperature is attained analysis shows 5.**2kmol of

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    #% present for kmol of #% supplied. &how that the maximum temperature

    reached is *1;8M#.

    So$ution+

    )91.-(*)91.-( **

    ***

    ** NCONOCO +++

    **** 22.22.8.5 NCONOCO +++

    !ctual !/> ratio stoichiometric !/> ratio x 55/5

    stoichiometric !/> ratio/.

    Thus, the actual reactants are

    ./)22.8.5( ** NOCO ++

    ith dissociation there will be some break up of #%*giving #% and %*,

    thus

    ***** )./22.(./)22.8.5( NcO(COaCONOCO +++=++

    The question states that b 5.**2, therefore

    ***** 95;.*22.595;.388.5 NcOCOaCONOCO +++=++

    #arbon @ a 0 5.**2 , a 5.99*

    %xygen@ 0 (* x 5.388) *a 0 5.**2 0*c , c 5.51;

    >or the reaction *** COOCO +

    = *.

    *

    *

    .

    *

    )(

    )(

    OCO

    CO

    PP

    PP

    )

    !nd **

    *P

    n

    aPCO = , *

    *

    Pn

    (PCO = , *

    **

    Pn

    cPO =

    Therefore

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    *

    *

    cP

    Pn

    (

    a)

    = ()

    here 6* total pressure at the required temperature n* total amount of substance of products

    n* a 0 b 0 c 0 .95; 5.99* 0 5.**2 0 5.51; 0 .95; *.992kmol

    !t ignition

    6 2.*2bar and T* *9- 0 *2* 888J

    $n*P = and *** $n*P = and F constant

    Therefore

    ***$n

    $nPP =

    here n 0 5.388 0 .95; -.13 and T* *9- 0 *1;8 *;12J

    (assuming)

    (ar$n$nPP 22.-2

    )888(13.-)*;12(992.**2.2

    *** ===

    &ubstitute in (),

    331.-22.-251;.5

    922.*

    **2.5

    99*.5

    *

    * =

    ==cP

    Pn

    (

    a)

    Therefore, *-9.ln =) , 6* -2.22bar and from the table, the reaction

    temperature is *;12J where the assumption made is correct.

    3. Entha$"! o/ #oration and Entha$"! o/ Reaction

    The molecules of a system possess various forms of energy, such as sensible

    energy, latent energy, chemical energy and nuclear energy.

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    6rocess which involves chemical reactions will involve changes in chemical

    energy and must be considered in energy balance. In absence of nuclear

    reactions, kinetic and potential energy changes, the energy balance is

    chemstatesys +++ +=

    The common practice to choose a reference state is at *8M# and atm. This is

    known as the standard reference state, and the properties at this state are

    denoted by a superscript NoO.

    3..1 Entha$"! o/ Reaction hR

    >or combustion processes, the h+is usually referred to as the enthalpy ofcombustion h hc, which represents the amount of heat released during a

    steadyflow combustion process when kg (or kmol) of fuel is burnt

    completely at a specified temperature and pressure. It is expressed as

    rpc hhh =

    here hpis the enthalpy of the products and hris the enthalpy of reactants.

    It is evident that, hcis a usefull property for analy:ing the combustion

    processes. $owever, it is impractical to list the value of hcfor all possible

    fuel and fuel mixture. Therefore, a more practical approach would be to have

    a fundamental property that would represent the chemical energy of an

    element or compound at some reference state. The enthalpy of formation hf

    is such a fundamental property which is the enthalpy of substance at a

    specified state owing to its chemical composition.

    !ll stable elements, such as %*, '*, $*, and # have a :ero value as their

    enthalpy of formation at the standard reference state (*8M# and atm). That

    is, 5=o

    fh for all stable elements.

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    %r in terms of enthalpy of combustiono

    Ch , energy equation above can be

    rewritten as

    ( ) ( )r

    o

    rp

    o

    p

    o

    c hhnhhnh&/ +=

    E*a"$e 3.&

    Eetermine the enthalpy of combustion of gaseous propane (#-$2) at *8M#

    and atm. !ssume that water in the product is in liquid form.

    So$ution+

    The stoichiometric equation for this reactionB

    *****2- 91.-3-)91.-( aNOHCONOaHC ++++

    Thus, rpc hhh = or ( ( 2-** )()()( HCo

    fOH

    o

    fCO

    o

    fr

    o

    frp

    o

    fpc nhnhnhhnhnh +==

    Psing the enthalpy table, we get

    'mol'-hc /5-5,**5,*)285,5-()2-5,*28(3)8*5,-;-(- =+=

    3..3 #or c$osed s!stes

    >or a chemically reacting closed system, the conservation of energy relation

    can be expressed as

    rp 00&/ =

    7y definition,

    Phu

    =

    %r Phhhuuu oo

    f

    oo

    f +=+

    Thus,

    ( ) ( )r

    oo

    frp

    oo

    fp PhhhnPhhhn&/ ++=

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    >or solids and liquids, the 6Q terms are negligible and for gases which

    behave as an ideal gas the 6Q terms can be replaced by +T.

    E*a"$e 3.(

    ethane gas is burnt with the stoichiometric amount of oxygen gas. The

    water is the product in the gas phase. Eetermine the heat released or

    observed if the reaction occurs at *8M# and atm.

    So$ution+

    The combustion equation is

    OHCOOCH ***3 ** ++

    The heat transfer is a steady flow combustion and 5. Thus,

    ( ) ( )r

    oo

    frp

    oo

    fp hhhnhhhn/ ++=

    %r ( ) ( ) )5(*)285,93()2*5,*3(*)8*5,-;-( +== rofrpofp hnhn/

    25*,-5 kR/kmol #$3

    3.1 Adia'atic #$ae Te"erature

    The chemical energy loss to surrounding or is used internally to raise the

    temperature of the combustion products for the absence of kinetic and

    potential energies. &ince no heat loss case, S5 the temperature of products

    will reach a maximum which is called the adiabatic flame temperature of the

    reaction.

    >or steady flow combustion process, the adiabatic flame temperature is

    determined from (S5 and 5)@

    ( ) ( )r

    oo

    frp

    oo

    fp hhhnhhhn +=+

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    >or the case of no excess air, the adiabatic flame temperature is called the

    theoretical adiabatic flame temperature for the fuel. It is the highest

    temperature that can be achieved from the fuel used in the combustion.

    3.11 Poer P$ant Thera$ E//icienc!

    The overall thermal efficiency ois defined as

    pliedenergyfuel

    outputwor'o

    supDD

    D=

    The fuel energy supplied can be considered either Sgr,6or Snet,6.

    >or example, in a plant the boiler efficiency 7is obtained by the followingequation@

    PnetPgr

    "/or/

    fluidtodtransferreHeat

    ,,

    DD

    DDD

    =

    3.12 Practica$ eterination o/ Ca$ori/ic 4a$ues

    #alorific value is used to measure the total energy available in the fuels.

    &ince, the types of fuel are can be in the forms of solid, liquid and gases.

    Thus, the instruments used to determine the calorific value also depend on

    the type of fuel. >or solid and liquid fuels are usually tested in a bomb

    calorimeter and for gaseous fuels in a 7oysG calorimeter as described briefly

    bellows@

    (i) &olid and uelsIn a bomb calorimeter combustion occurs at constant volume and is a

    nonflow process.

    The bomb is a small stainless steel vessel in which a small mass of the

    fuel is held in a crucible (see >igure -.*) as a pellet.

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    >igure -.* 7omb #alorimeter

    The pellet is estimated such that the temperature rise to be measured does

    not exceed *-J. The pellet is ignited by fusing a piece of platinum or

    nichrome wire which in contact with it.

    The crucible carrying the pellet is located in the bomb, a small quantity

    of distilled water is put into the bomb to absorb the vapors formed by

    combustion and to ensure that the water vapors produced is condensed,

    and the top bomb is screwed down.

    !fter fifth minute the charge is fired and temperature readings are taken

    every 5 seconds during this period. hen the temperature readings start

    to fall, the frequency of readings is taking every minute.

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    The temperature rise of the circulating water is measured, and the

    condensate from the products of combustion is collected. The water flow

    rate is measured and condensate is weighed. Thus, calorific value of fuel

    can be obtained by

    (Folume of fuel at .5-bar and 8V#) U calorific value (mass of water

    circulated) U specific heat capacity of water U (temperature rise of water)

    3.13 Air and #ue$)4a"or Mi*tures

    The mixture supplied to an engine fitted with a carburetor is one air and fuel

    vapor, and the quality of the mixture is controlled by the carburetor.

    If the mixture is saturated with fuel vapor then the relative proportion of fuelto air can be determined from the knowledge of the temperaturepressure

    relationship for the saturated fuel.

    E*a"$e 3.,+

    >or a stoichiometric mixture of ethyl alcohol of 5.5**kmol/kg and air

    substance of 5.-5;kmol/kg, calculate the temperature above which there will

    be no liquid fuel in the mixture. The pressure of the mixture is 5.-k6a.

    So$utions+

    'g'moln OHC /5**.51* = and 'g'molnAir /-5;.5=

    'g'molnnn AirOHC$otal /--.51* =+=

    'g'molPn

    n

    P mixture$otal

    OHC

    OHC /5**.51*

    1* =

    =

    >rom the table of #*$1%@

    T (o#) 5 5 *5 -5 35 85 15

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    6 (k6a) .1* -.3 8.*2 5.3; 2.55 *1.15 31.;5

    Thus,

    C$ =

    += 28.*523.83;.5

    23.89.1*5