Combustion Stoichiometry
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Transcript of Combustion Stoichiometry
1
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 STOICHIMETRY 2
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 STOICHIMETRY 3
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 STOICHIMETRY 4
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 STOICHIMETRY 5
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 STOICHIMETRY 6
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 STOICHIMETRY 7
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 STOICHIMETRY 9
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
COMBUSTION STOICHIMETRY 11
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
COMBUSTION STOICHIMETRY 12
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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COMBUSTION STOICHIMETRY 13
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
wCaHb + cO2+3.76cN2 mCO2+3.76cN2+nCO+oH2O + xCH4+yH2+zO2
COMBUSTION STOICHIMETRY 15
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
wCaHb + cO2+3.76cN2 mCO2+3.76cN2+nCO+oH2O + xCH4+yH2+zO2
o &w not part of analyses but can be calculatedby element balance etc
COMBUSTION STOICHIMETRY 16
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.
COMBUSTION STOICHIMETRY 17
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.
COMBUSTION STOICHIMETRY 18
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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.
COMBUSTION STOICHIMETRY 21
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.
COMBUSTION STOICHIMETRY 22
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.
COMBUSTION STOICHIMETRY 23
•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.
COMBUSTION STOICHIMETRY 24
• 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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COMBUSTION STOICHIMETRY 25
)/()/()(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 =
COMBUSTION STOICHIMETRY 26
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
)F,FuelofmassA,AirofmassAF =
C8H17 + aO2+ 3.76aN2 bCO2+ 3.76aN2+ cH2OTheoretical, everything should burn completely without excess air
nAir = 20.77gmolO2+78.1gmolN2= 98.9gmol Air
COMBUSTION STOICHIMETRY 27
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
COMBUSTION STOICHIMETRY 28
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 STOICHIMETRY 29
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 STOICHIMETRY 31
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
)F,FuelofmassA,AirofmassAF =
•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
COMBUSTION STOICHIMETRY 33
• Actual Air to fuel ratio: changing moles tomass
67.18254
4.4744, ==FuelggAirAFActual
airggmolAir
gAirgmolAir 4.47440.1296.163
=
Combustion Stoichiometry
• Actual and Theoretical Air to Fuel ratiosare related by dilution coefficiency (DC)
)/()/()(FAltheoretica
FAactualDCtCoefficienDilution =
Task 7: Use the data from the aboveexample to calculate DC.
COMBUSTION STOICHIMETRY 34
Combustion StoichiometryTask 7:Use the above data to calculate DC
)/()/()(FAltheoretica
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.
COMBUSTION STOICHIMETRY 35
%100)/(
)/()/(% XFAltheoretica
FAltheoreticaFAactualAirExcess −=
%26%10088.14
88.148.18% =−
= XAirExcess
Combustion Stoichiometry
• 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
COMBUSTION STOICHIMETRY 36
• Variousformulae to usedin calculating%excess Air
%100AirquiredReOAirExcessAirExcess% 2×=
%100Re
Re%2
22 ×−
=Oquired
quiredOEnteringOAirExcess
%100%22
2 ×−
=OExcessOEntering
OExcessAirExcess
Combustion Stoichiometry
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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COMBUSTION STOICHIMETRY 37
• C3H8 + 5O2 + 3.76N2 3CO2 + 4H2O + 3.76N2
– Theoretical Required O2
Combustion Stoichiometry
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
Combustion Stoichiometry
%100Re
Re%2
22 ×−
=Oquired
quiredOEnteringOAirExcess
%10027.2
27.290.2% ×−
=AirExcess
COMBUSTION STOICHIMETRY 39
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.
COMBUSTION STOICHIMETRY 40
• 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.
COMBUSTION STOICHIMETRY 41
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
COMBUSTION STOICHIMETRY 42
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 −++=
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COMBUSTION STOICHIMETRY 43
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
COMBUSTION STOICHIMETRY 44
2.1) I think you can try this question of thehomework again.
HOMEWORK 2: Stoichiometric Calculations
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