combustion stoichiometry

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COMBUSTION STOICHIMETRY 1

Combustion ProcessesBe to determine:

–Composition of air–Equivalent formulae/chemical reactions of fuel fromgiven composition of combustion fuels.–Composition of combustion products.–Equivalent formulas/chemical reaction of fuel fromgiven composition of products (fuels).–Actual combustion process–Stoichiometric combustion process.–Air to fuel Ratio, Dilution coefficient, %Excess Air.


Combustion Processes• What is combustion?

• Combustion is a process of releasing chemicalenergy by burning a combustible (able to burn)material in the present oxygen thus releasingenormous amounts of energy and gases.

• This energy is harnessed in various forms

Task 1:Classify different types of fuels as fossils and non-

fossil

C + O2 CO2 14. 093 Btu/lbH + 0.5O2 H2O 61.100 Btu/lbS + O2 SO2 61.100 Btu/lb


Combustion Fuels• A combustible material is called a fuel.• Combustible elements in fuels such as coal are

carbon, hydrogen and oxygen with other traceelements.

• A dry bituminous coal may contain 88%C, 6%H,4%O, 1%N and 1%S.

• The above composition exclude Ash andmoisture.

• Task 1: A “wee bit” of Chemistry 1:– Determine the equivalent formula for a fuel containing

85%C and 15%H (2.0 min)


Combustion FuelsDetermine the equivalent formula for a fuel

containing 85%C and 15%H1.0 Take A Basis of 100g samples such that 85% is

equal to 85g C and 15% is 15 g H2.0 Calculate number of moles, e.g. a and b in CaHb

Answer :C7.08H15

Task 1b :Determine the equivalent formula forbituminous coal may containing 88%C, 6%H,4%O, 1%N and 1%S. (2.0min)

CaHbScOdNe


Combustion Air• For combustion to take place, oxygen is required, oxygen is

not “freely” available, it is a component of air.• Task 2: Molecular mass of air is 28.967 g/mol, use data to

prove this (5 min)

1.0000100Total2.0160.00010.01H2

44.0030.00030.03CO2

39.9330.00940.94Ar28.0160.780378.03N2

32.000.209920.99O2

RelativeWeight

MolecularWeight

MoleFractions

%vol.Gas


Combustion Air

• Relative Weight = mole fraction X Molecular Wt

28.9671.0000100Total2.0160.00010.01H2

0.01344.0030.00030.03CO2

0.37639.9330.00940.94Ar21.86128.0160.780378.03N2

6.71732.000.209920.99O2

Reltve WeightMolecrWtMole Frac%vol.Gas

28.016 is pure nitrogen in the air, in combustionprocesses apparent nitrogen is used

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Combustion Air• Apparent nitrogen is the sum of pure nitrogen

plus other inert gases found in air i.e. 79.01%

2.01644.00339.94428.016Mol Wt

1.0000.000130.000370.01190.9876

Mole Fraction

79.010.010.030.9478.03Vol Relative wtGas

0.00026H2

28.161Tot

0.01628CO2

0.4753Ar27.6686N2

Mole Fraction =78.03/79.01 =0.9876

28.161 g/mol is the apparent nitrogen massCOMBUSTION STOICHIMETRY 8

Combustion Air

• In other words, dry combustion air supplies 3.76moles of air in every mole of oxygen.

2

2764.399.2001.79

OmolesNaparentmoless

=

2

2313.3717.6

25.22Ogram

Naparentgrams=

• Ratio of nitrogen to oxygen on volume basis

• Ratio of nitrogen to oxygen on mass basis

C + O2 + 3.76N2 CO2 + 3.76N2For hydrocarbons

CaHb + cO2+3.76cN2 mCO2+3.76cN2+nCO+xH2O+H2+zO2


COMBUSTION PRODUCTSCaHb + cO2+3.76cN2 mCO2+3.76cN2+nCO+xH2O+H2+zO2

Notice in the above reaction that O2, H2,CO, the sameamount of N2.in reactants and products and

O2 is the excess air that the required for completecombustion so that thee will no COs and H2

COs are H2 undesirable products resulting from the lack ofO2

NOx, will results if the combustion processes take place athigher temperatures

SOx are product sulphur containing fuels e.g. coal andsome fuel oils.

Typical complete combustion process yield no O2, H2,CO

CaHb + cO2+3.76cN2 mCO2+3.76cN2+xH2O COMBUSTION STOICHIMETRY 10

COMBUSTION STOICHIOMETRY

• Subscript a, b, subscript as in the ordinary molecularformula.

• Other parameter e.g. c, m and x are coefficients in thereaction

CaHb + cO2+3.76cN2 mCO2+3.76cN2+xH2O

Task 3 : If a = 8, b = 18 :1. What is the name of the fuel?2. Balance the chemical reaction (atom balance).3. What are the values of c and x (5.0 min)

C8H18 + 12.5 O2+3.76 x 12.5 N2 8CO2+3.76 x12.5 N2+9H2O

Answers to Task 31. Octane?2. m = 8 and c = 12.5


Homework 2: Bituminous coal may containing 88%C, 6%H,4%O, 1%N and 1%S (on dry basis) and its completecombustion rxn is as follows:CaHbOcNdSe+fO2+3.76f N2 gCO2+hH2O + iSO2 +(3.76f+d)N2

1. Balance the reaction2. If 100kmol of bituminous coal and 105.06 kmol air is

supplied into the burner.2.1 Calculate kmol of each species as in the actual

chemical equation.2.2 Determine Dilution Coefficients (DC)2.2 Calculate %excess air.

COMBUSTION STOICHIOMETRY


The Actual Combustion Process

The five conditions for good combustion are:

• proper mixing of reactants,

• sufficient air,

• temperature above the ignition temperature,

• sufficient time for the reaction to occur, and

• a reactant density sufficient to propagate the

flame. (MATTρ)



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wCaHb + cO2+3.76cN2 mCO2+3.76cN2+nCO+oH2O + xCH4+yH2+zO2

Excess AirSince perfect mixing is never attained in practice, good

combustion can only be achieved by supplying excess air

for the process.Too much excess air however increases the losses in the

combustion process and increases NOx emissions.Excess air will only be revealed in the flue gas (fg) by

presence of O2 in an analyser

Task 4 : O2 is shown in the above reaction as z mols, if the flue gas

analyser shows, 8.7%CO2, 0.3%O2, 8.9% CO, 3.7%H2, 0.3%CH4and

78.1%N2. If a = 8 and b = 17, Answer the following (10.0min)COMBUSTION STOICHIMETRY 14

Task 41. Is the above reaction complete or incomplete And why,

(give any two observation from the data)?2. What is the % of excess oxygen?3. Is the analyses wet or dry and why4. Balance the equation (atoms or elements or molecules).5. Calculate %excess airAnswers to Task 4• Incomplete reaction and because of the present of CO,

CH4, H2 etc in the flue gas.• % of excess oxygen is equal to 0.3%O2• Dry, water is not included in the data.• Balancing the equation: Mole balance



Hydrogen balance :w x 17 =0.3(4)+3.7(2)+14.7(2), w =2.235

Task 4.3 :Mole balanceStep 1 A basis of 100gmol sample

0.30.3CH4

8.98.9CO

3.73.7H2

100%

78.1

8.7

0.3

%vol

Tot

N2

CO2

O2

Gas

FLUE GASANALYSER

78.1

8.7

100.

0.3

gmol

Step 2 substitute gmol data into equationcoefficients i.e. z = 0.3, y=3.7, m = 8.7, n = 8.9,

x = 0.3,3.76c = 78.1, therefore c= 78.1/3.76=20.77, a and bare given in the problem, therefore a = 7 and b = 17

2.235C8H17 + 20.77O2+3.76cN2 8.7CO2+78.1N2+8.9CO+oH2O + 0.3CH4+3.7H2+0.3O2

Oxygen(O2) balance :20.8= 8.7+8.9/2+o/2+0.3, o = 14.7


o &w not part of analyses but can be calculatedby element balance etc


Flue Gas Analysers• There various flue gas analyser in the market.• One of them is Orsat Flue gas analyser.• These analyser measure concentrations of the

flue gas• The gas chromatograph is a very sensitive

device that can be used to detect trace amountsof gases.

• This is however a sophisticated and expensiveapparatus.

• The Orsat apparatus is a relatively simple,compact and portable gas analyzer

• It is specifically designed to measure three of thecompounds found in the combustion products.


Orsat Apparatus• Since the gas is collected at room temperature

over water, it is usually assumed that any watervapour in the exhaust gas will have condensedand that any SO2 will have reacted with thewater vapour in the exhaust gas and in thecollecting bottle.

• Consequently, it is assumed that the resultingdry gas sample is composed of CO2, O2, COand nitrogen.




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COMBUSTION STOICHIMETRY 19 COMBUSTION STOICHIMETRY 20

The Orsat Apparatus•A 100cm3 sample of exhaust gas is taken atroom temperature in the burette by using thelevelling water bottle to collect and transfer thegas sample.•Once the gas sample has been obtained, it isthen sequentially passed through three chemicalreactors in the device•A typical Orsat gas analyzer shown is used todetermine the molar fractions of carbon monoxide;oxygen and carbon dioxide in the dry exhaustgases.


The Orsat Apparatus

The first reactor contains an aqueous solution ofKOH, which preferentially removes any CO2 inthe gas sample.•The second reactor contains a solution ofpyrogallic acid in potassium hydroxide and water,and this solution preferentially removes anyoxygen in the sample.•The third reactor contains a solution of cuprouschloride in ammonia and this solution absorbsany CO present in the gas.


Determination of Air to Fuel Ratio• The actual air-fuel ratio for a given combustion process isnormally estimated from an experimental measurement of

the gaseous component of the exhaust gas.•There are several ways of experimentally determining theconcentration of the various gas compounds in a mixture ofgases.•These systems include

•the gas chromatography and•the orsat apparatus among others.


•It also includes some products of incomplete

combustion, including some unburnt fuel, carbon

monoxide, some hydroxyls and aldehydes along

with nitrogen, unused oxygen, ash particles, and

nitrogen oxides.

•All of these products except water, oxygen and

nitrogen are considered to be atmospheric

pollutants.


• By carefully measuring the decrease in samplevolume as the gas passes through eachchemical reactors, in series, and dividing eachdecrease by the original gas volume, the volumeor mole fractions of carbon dioxide, oxygen andcarbon monoxide in the dry exhaust gas areobtained.

• Any gas that remains after the sample has beenpassed through all three reactors (usuallyaround 80% )is assumed to be nitrogen.



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)/()/()(FAltheoretica

FAactualDCtCoefficienDilution =

%100)/(

)/()/( XFAltheoretica

FAltheoreticaFAactual −

The amount of air supplied to a combustion processcan either be expressed as the dilution coefficient orthe percent excess air.The dilution coefficient is defined as the ratio of theactual to the theoretical air-fuel ratio:

The percent excessair is defined as: =% Excess air

= 100 ( dilution coefficient-1)

air-fuel (A/F) ratio is the ratio of air to fuel on massbasis

)F,FuelofmassA,AirofmassAF =


nAir = 20.77gmolO2+78.1gmolN2= 98.9gmol Air

The Actual Air to Fuel RatioTask 4 is Used to calculate using air to fuel ratio.

Below is an actual processes chemical equation

3.11Fuelg9.252

gAir2868AF ==

2.24C8H17 + 20.77O2+78.1N2 8.7CO2+78.1N2+8.9CO+oH2O + 0.3CH4+3.7H2+0.3O2

Mass of Air = 98.9gmol Air x 29g/gmol =2868.1g air

Mass of Fuel = 2.24gmol C8H17 x 113g/gmol =253 g air

nFuel = 2.24gmol C8H17


C8H17 + aO2+ 3.76aN2 bCO2+ 3.76aN2+ cH2OTheoretical, everything should burn completely without excess air

nAir = 20.77gmolO2+78.1gmolN2= 98.9gmol Air


Theoretical Combustion Process

The five conditions for good combustion are:

• proper mixing of reactants,

• sufficient air,

• temperature above the ignition temperature,

• sufficient time for the reaction to occur, and

• a reactant density sufficient to propagate the flame.

(MATTρ)

•However, this is not always the case:

CxHy + aO2+ 3.76aN2 bCO2+ 3.76aN2+ cH2O


Theoretical Amounts

1. 18 C = 18C,2. 38 H = (38/2)H = 19H2

3. Always balance O2 last:2(17.9) O2 in CO2 +19O2 in H2O=27.4(54.8/2)O2 =27.4 O2

C18H38 +27.4O2+ 3.76 x 27.4N2 18CO2+ 3.76x27.4N2+19H2O

2.235C8H17 + 20.8O2+78.1N2 8.7CO2+8.9CO+14.7H2O + 0.3CH4+3.7H2+ 78.1N2+ 0.3O2

Theoretical combustion for1.0 gmol of fuel mixture

CxHy + 20.8O2+78.1N2 8.7CO2+78.1N2+8.9CO+14.7H2O + 0.3CH4+3.7H2+0.3O2

1. C: 2.235x8 = 18 = x2. H: 2.235x17 = 38 = y

C18H38 +27.4O2+ 103N2 18CO2+ 103N2+19H2O

Theoretical combustion for 1.0 gmol of C18H38

C18H38 + 130.9 Air 18CO2+ 103N2+19H2O

C18H38 + 27.4 (O2+3.76N2) 18CO2+ 103N2+19H2O

CxHy + a (O2+3.76N2) bCO2+ cN2+dH2O

Theoretical combustion for hydrocarbons


Combustion Stoichiometry• Actual combustion process for 1.0 mol C18H38

• Let us use the data from the previous examplesC17.9H38+ 20.8O2+78.1N2 8.7CO2+8.9CO+0.3CH2+14.7H2O+0.3O2+78.1N2

C18H38 +27.4O2+ 103.5N2 18CO2+ 103N2+19H2O

Consistence, round 17.9 in C17.9 to 18 in C18 and acomplete combustion becomes

Observe that there is:27.4O2+103.5N2 = 130.9 Theoretical Ai r

In the theoretical combustion equationAlso observe that there is:

20.8O2+78.1N2 = 98.9 Actual AirIn the actual combustion equation COMBUSTION STOICHIMETRY 30

Combustion Stoichiometry

Theoretical Air is the air that is theoretical required tocomplete the reaction.

Theoretical Ai r is sometimes referred to as stoichiometricair.

If the combustion is taking place with air less thanstoichiometric air it is referred to as sub-stoichiometriccombustion.

The actual combustion processes took place under sub-stoichiometric combustion.

Combustion processes are normally carried out in excessair environment.

Although, there is 0.3molO2 in products of the actualcombustion equation, that oxygen might be dueinefficiency of the process I.e. improper mixing

Let us suppose, now the above reaction takes with excessair of about 163.7 gmol air.



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Combustion Stoichiometry• This is 34.25 gmolO2 and 129.37gmolN2

• Now the actual combustion equation isC18H38 + 34.3O2+ 129.4N2 18CO2+129.4N2+6.9O2 +19H2O

•Theoretical and Actual quantities are relatedby Air to Fuel (AF) Ratio

Actual air = 34.25 gmolO2+129.37gmolN2 =163.7gmolAir

Theoretical air = 27.4 gmolO2+103.5gmolN2 =130.9gmolAir

C18H38 +27.4O2+ 103.5N2 18CO2+ 103N2+19H2O


•There is a Theoretical and Actual Air to Fuel(AF) Ratio COMBUSTION STOICHIMETRY 32

• Theoretical A/F: Changing gmoles to mass

38183818

38183818 2540.12540.1 HgC

HgmolCHCgHgmolC

=

Theoretical neededOxygen to complete

the reaction27.4O2+ 103N2=130.4 gmol Air

airggmolAir

gAirgmolAir 6.37810.1299.130

=

•Task 6: Use the previous example to calculateTheoretical and Actual Air to Fuel ratios

8.14254

6.3781, ==g

gAFTheo


• Actual Air to fuel ratio: changing moles tomass

67.18254

4.4744, ==FuelggAirAFActual

airggmolAir

gAirgmolAir 4.47440.1296.163

=


• Actual and Theoretical Air to Fuel ratiosare related by dilution coefficiency (DC)


FAactualDCtCoefficienDilution =

Task 7: Use the data from the aboveexample to calculate DC.


Combustion StoichiometryTask 7:Use the above data to calculate DC


FAactualDCtCoefficienDilution = 26.188.1476.18 ==

100)1(% XtCoefficienDilutionAirExcess −=

Dilution coefficient can be used to calculate%air.

Task 8:Use the above data to calculate %Excess Air

%26%100)126.1(% =−= XAirExcess100)1(% XtCoefficienDilutionAirExcess −=

There are various other ways to calculate%excess air.


%100)/(

)/()/(% XFAltheoretica

FAltheoreticaFAactualAirExcess −=

%26%10088.14

88.148.18% =−

= XAirExcess


• Task 9: Use Task 6 solution to calculate %excessair using the above formula

100)(

76.3)(

)(%

22

2 ×

−=

prodprod

prod

nOnN

nOAirExcess

100.)(

76.3)(

)(%

22

2 ×

−=

prodprod

prod

xOxN

xOAirExcess

• Other ways of calculating %Excess Air

• Where:• ‘prod’ denotes products• x mole fractions• n is moles


• Variousformulae to usedin calculating%excess Air

%100AirquiredReOAirExcessAirExcess% 2×=

%100Re

Re%2

22 ×−

=Oquired

quiredOEnteringOAirExcess

%100%22

2 ×−

=OExcessOEntering

OExcessAirExcess


Exercise 1:A boiler use 20 kg propane based fueland a 400kg air to make steam. Calculate %excess air by various methods.



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• C3H8 + 5O2 + 3.76N2 3CO2 + 4H2O + 3.76N2

– Theoretical Required O2


22

2

83

8383 27.20.10.5

440.120 kgmolO

kgmolOkgmolO

HkgCHkgmolCHkgmolC

=

22 90.2

2921

290.1400 kgmolO

AirkgmolkgmolO

kgAirAirkgmolkgmolAir =

–Actual entering O2

Excess O2 = Entering O2 – Required O2

Excess O2 = 2.90 kgmol O2 – 2.27kgmol O2

Excess O2 = ……….kgmol O2COMBUSTION STOICHIMETRY 38

• Using one of theformulae.

• Try all otherformulae toprove if you canget the sameresults


%100Re

Re%2

22 ×−

=Oquired

quiredOEnteringOAirExcess

%10027.2

27.290.2% ×−

=AirExcess


The air Circuit:• A fan draws in air and forces it through aheater where it passes over plates heated onthe other side by exhaust furnace gases ontheir way to the chimney.•The hot air is ducted partly as primary air tothe underside of the moving chain grate andpartly as secondary air above the firebed.•This air provides the oxygen necessary for thecomplete combustion of the coal supplied fromthe automatic stokers on to the fire-grate.


• The hot products of combustion rise and circulateround the water tubes, usually being directed bybaffles so that they pass over the tubes threetimes.

• On leaving the boiler, the gases pass over thetubes of the economizer and the plates of the airheater to be discharged at the chimney.

• A large supply of cooling water is required toextract the latent heat from the steam in order tocondense it.

• When the power station is sited near a river,cooling water for the condensers is taken fromthe river and is pumped through the condensertubes and finally discharged downstream.


HOMEWORK 2• 1.0 Take Basis of 100g such

that % composition becomemasses.

• Calculate no of moles.• 1.1 Balance the rxn:

edcba NOSHC

141

164

321

16

1288 NOSHC

071.025.003.0633.7 NOSHC

CaHbOcNdSe + fO2+3.76f N2 gCO2+hH2O+iSO2 +(3.76f+d)N2

CaHbOcNdSe + fO2+ 3.76f N2 gCO2+ hH2O + iSO2+(3.76f+d)N2

C7.33H6O0.25N0.071S0.03 + fO2+ 3.76f N2 gCO2+ hH2O + iSO2+(3.76f+d)N2

Carbon balance: 7.33 C, LHS = 7.33CO3 , RHSHydrogen Balance: 6 H, LHS = 3H2O, RHS


HOMEWORK 2: Balancing the rxn continues

Recapping and now

Sulfphur Balance: 0.03 S, LHS = 0.0312SO2,RHS

C7.33H6O0.25N0.071S0.03 + fO2+ 3.76f N2

7.33CO2+ 3H2O + 0.0312SO2+ (3.76f+0.036)N2

d moles of N in fuel = 0.0710.071N, LHS = 2dN, RHS

036.02071.0rhs,N ==

Balance O2 last because is a stand alone substanceÓO2, LHS = ÓO2, RHS

036.02071.0d ==

RHS,O864.833.7230312.0 2=++=+ LHS,O

225.0

2

225.033.7

230312.0LHS,O 22 −++=



8


f = 8.7474.8125.0864.8LHS,O2 =−=

Recuperating and nowC7.33H6O0.25N0.071S0.03 + 8.74O2+ 3.76 x 8.74 N2

7.33CO2+ 3H2O + 0.0312SO2+ (3.76 x8.74+0.036)N2

C7.33H6O0.25N0.071S0.03 + 8.74O2+ 32.85 N2

7.33CO2+ 3H2O + 0.0312SO2+ 32.88N2

HOMEWORK 2: Balancing the rxn continues


2.1) I think you can try this question of thehomework again.

HOMEWORK 2: Stoichiometric Calculations



combustion stoichiometry

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