Definition, Types of fuels, primary, secondary, solid, liquid and gaseous fuels advantages and disadvantages, Coal - Proximate analysis and Ultimate analysis, Petroleum - Fractional distillation, petrol, diesel, composition, Cracking, Moving bed catalytic cracking, Knocking, octane number, cetane nu...
Definition, Types of fuels, primary, secondary, solid, liquid and gaseous fuels advantages and disadvantages, Coal - Proximate analysis and Ultimate analysis, Petroleum - Fractional distillation, petrol, diesel, composition, Cracking, Moving bed catalytic cracking, Knocking, octane number, cetane number, LPG, CNG
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MATRUSRI ENGINEERING COLLEGE
DEPARTMENT OF SCIENCES AND HUMANITIES
SUBJECT NAME: CHEMISTRY
FACULTY NAME: VISHNU THUMMA
MATRUSRI
ENGINEERING COLLEGE
TOPIC: CHEMICAL FUELS
UNIT-IV CHEMICAL FUELS
4
COURSEOUTCOMES:Aftercompletionofcoursestudentswillbeableto
Identifyandanalyzedifferenttypesofchemicalfuelsfordomesticand
automobileapplications.
MATRUSRI
ENGINEERING COLLEGE
Definition:Achemicalfuelisasubstancewhichoncombustioninairor
oxygenproducessignificantamountsofheatwhichcanbeused
economicallyfordomesticandindustrialpurposes.
5
Example:
Wood,
Coal,
Crude Oil,
Natural Gas, Etc.
MATRUSRI
ENGINEERING COLLEGE
MODULE-1: INTRODUCTION TO CHEMICAL
FUELS
Classification of Fuels
Chemicalfuelsarebroadlyclassifiedassolid,liquidandgaseousbased
ontheirphysicalstateinwhichtheyareused.
Thesearefurtherclassifiedasprimaryandsecondaryfuels.
Afuelthatoccursinnatureiscalleda“primaryfuel”
Whileallothers,whicharederivedfromprimaryfuelsarecalled
“secondaryfuels”.
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MATRUSRI
ENGINEERING COLLEGE
Classification of Fuels
7
Physical
state
Primary fuels (Natural)
Secondary fuels
(Derived)
SolidWood, Coal, Lignite,
Bituminous, Anthracite, Peat
Charcoal,Coke
LiquidPetroleum(Crudeoil) Gasoline, Diesel,
Kerosene, Synthetic
petrol
GaseousNaturalgas Producer gas, Water
gas, Coal gas, LPG,
Biogas
MATRUSRI
ENGINEERING COLLEGE
High Calorific Value
Thecalorificvalueofafuelisthedirectmeasureofitsefficiencyasa
fuel.
Higherthecalorificvalue,moreefficientisthefuel.
Agoodfueltherefore,musthaveahighcalorificvalue.
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MATRUSRI
ENGINEERING COLLEGERequirements of a Good Fuel
Moderate Ignition Temperature
Ignitiontemperatureistheminimumtemperaturetowhichthefuelisto
beheatedtostartcombustion.
Thefuelshavingverylowignitiontemperaturecancausefirehazards
duringtheirhandling,applications,storageandtransportation.
Itisratherdifficulttoignitefuelifitsignitiontemperatureistoohigh.
Hence,moderateignitiontemperatureisthemostdesiredpropertyofa
fuel
11
MATRUSRI
ENGINEERING COLLEGERequirements of a Good Fuel
Low Ash Content
Formationofashduringcombustionisduetothepresenceof
inorganicmatterinthefuel.
Highashcontentlowersthecalorificvalueandposesproblemsinits
disposal.
Agoodfuelmusttherefore,befreefromashcontent.
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MATRUSRI
ENGINEERING COLLEGERequirements of a Good Fuel
Low Moisture Content
Thepresenceofhighpercentageofmoistureinfueldecreasesits
efficiency.
Italsoincreasestheignitiontemperatureandfuelcost.
Further,itlowersthecalorificvaluesincesomeoftheheatproducedis
utilizedtovaporizethemoisture.
Hence,fuelshouldhavelowmoisturecontent.
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MATRUSRI
ENGINEERING COLLEGERequirements of a Good Fuel
No Harmful Combustion Products
Agoodfuelmustnotproduceharmfulcombustionproductssuchas
CO,SO
2,NO,H
2S,smokeandclinkersduringcombustion.
Formationofsuchby-productscancauseharmfuleffectsonhealth.
Agoodfuelmustthereforeburnwithcleanflamewithoutproducing
undesiredby-products.
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MATRUSRI
ENGINEERING COLLEGERequirements of a Good Fuel
Combustioncontrol:Onecanavoidalargewastageofvaluablefuelif
itscombustionratecanbeproperlyregulatedandburningcanbe
stoppedimmediatelyasandwhendesired.
Easeofavailability:Fuelmustbereadilyavailableinabundant
amountanditscostmustbeminimum.
Easeofstorageandtransport:Fuelsshouldnotdemandmorespace
andsophisticatedarrangementsfortheirstorageandtransportation.
Furthertheremustbeminimumrisksoffirehazards.
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MATRUSRI
ENGINEERING COLLEGERequirements of a Good Fuel
Solid Fuels -Advantages
➢Easy to transport.
➢Convenient to store without any risk of spontaneous explosion.
➢Cost of production is low.
➢Possess moderate ignition temperatures.
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MATRUSRI
ENGINEERING COLLEGE
Solid Fuels -Disadvantages
➢Ash content is high.
➢Heat is wasted during combustion (low thermal efficiency.
➢Clinker formation.
➢Combustion operations cannot be controlled easily.
➢Cost of handling is high.
➢Low calorific value.
➢Excess of air for complete combustion.
➢Cannot be used as internal combustion engine fuels.
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MATRUSRI
ENGINEERING COLLEGE
Liquid fuels -Advantages
➢Higher calorific value per unit mass than solid fuels.
➢Burn without forming dust, ash, clinkers, etc.
➢The firing is easier and also fire can be extinguished easily by stopping
the liquid fuel supply.
➢Easy to transport through pipes, and one man can easily regulate a
large number of furnaces simultaneously.
➢They can be stored indefinitely without any loss.
➢The flame produced by burning liquid fuels can easily be controlled by
adjusting the liquid fuel supply.
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MATRUSRI
ENGINEERING COLLEGE
Liquid fuels -Advantages
➢Clean in use and economic in labour, Loss of heat to chimney is very
low due to greater cleanliness.
➢Less excess of air for complete combustion, Less furnace space for
combustion.
➢No wear and tear of grate bars and cleaning of fires etc. unlike solid
fuels.
➢They can be used as internal combustion fuels
19
MATRUSRI
ENGINEERING COLLEGE
24
1. Which of the following is a primary fuel
a)Charcoal b) Coke
c) Wood d)None
2.The lowest Temperature at which the Fuel must be pre-heated so that it burns smoothly is
called
a)Critical Temperature b) Flame Temperature
c) Ignition Temperature d) Boiling point of fuel
3. One of the following isnot the quality of a good fuel
a) High calorific Value b) Moderate Ignition Temperature
c) Low moisture content d) Analysis of Flue gases
4.In which type of fuel given below, risk of fire hazard is greatest?
a) Solid fuel b) Liquid fuel
c) Gaseous fuel d) Natural gas
MATRUSRI
ENGINEERING COLLEGE
QUIZ
Itistheamountofheatreleasedwhenaunitmassorvolumeofafuelis
burntcompletelyinairoroxygen.
UnitsofHeat:
(1)'Calorie'istheamountofheatrequiredtoraisethetemperatureof
onegramofwaterthroughonedegreeCentigrade.
(1)"Kilocalorie"isequalto1,000calories.Itmaybedefinedas'the
quantityofheatrequiredtoraisethetemperatureofonekilogramof
waterthroughonedegreeCentigrade.
Thus:1kcal=1,000cal
(3)"BritishThermalunit"(B.T.U.)isdefinedas"thequantityofheat
requiredtoraisethetemperatureofonepoundofwaterthroughone
degreeFahrenheit.ThisistheEnglishsystemunit.
1B.T.U.=252cal=0.252kcal1kcal=3.968B.T.U.
25
MODULE-2: CALORIFIC VALUE
MATRUSRI
ENGINEERING COLLEGE
Higher Calorific Value (HCV) or Gross Calorific Value (GCV)
Itisdefinedastheamountofheatreleasedwhenunitquantityofafuel
isburntcompletelyinairandthecombustionproductsarecooledto
roomtemperature.
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MATRUSRI
ENGINEERING COLLEGE
Itisdefinedastheamountofheatreleasedwhenaunitquantityofafuel
isburntcompletelyinair,andtheproductsofcombustionareletoff
intotheatmosphere.
LCV=HCV–latentheatofsteam
=HCV– x587cal/g
Because 1 g hydrogen gives 9 g of water and the latent heat of steam is
587 cal/g.
27
LowerCalorific Value (LCV) or Net Calorific Value (NCV)
MATRUSRI
ENGINEERING COLLEGE
28
Theoretical Calculation of Calorific Value
TheHCVoffuelisthesumofcalorificvaluesofalltheconstituents
presentinfuel.
ConstituentHydrogenCarbonSulphur
HCV
(kcal/kg)
34500 8080 2240
Calorific values of fuel constituents
Dulong’sformulafor calorific value from the chemical composition of fuel
where, C,H,O and S are the percentage of carbon, hydrogen, oxygen and
sulphurin the fuel respectively.
In this formula, oxygen is assumed to be present in combination with
hydrogen as water
Kcal/kg
MATRUSRI
ENGINEERING COLLEGE
29
LCV = HCV –0.09H x 587 kcal/kg
Since,1partofHbymassgives9partsofH
2Oandlatentheatofsteam
is587kcal/kg.
MATRUSRI
ENGINEERING COLLEGE
30
Numerical Problem:
Calculate the gross and net calorific value of a coal sample having the following
composition; C=80%, H=7%, O=3%, S=3.5%, N=2.1% and ash=4%.
Kcal/kg
= 8828 kcal/kg
LCV = HCV –0.09H x 587
= 8828 –0.09 x 7 x 587 = 8458 kcal/kg
MATRUSRI
ENGINEERING COLLEGE
31
MATRUSRI
ENGINEERING COLLEGEQUIZ
1.Which of the following factor does not affect the gross calorific value?
a) CO
2
emissions from the fuel b) Latent heat produced in the fuel
c) Size of the fuel d) Moisture content of the fuel
2.HCVandLCVarerelatedby
a)HCV=LCV-0.09*587cal/g b)HCV=LCV-0.9*587Cal/g
c) HCV= LCV-98* 587 cal/g d) Both A and B
3. Amount of energy released when unit mass of the fuel is burnt, and the products of
combustion are let off into the atmosphere is called
a) High calorific value b) Low calorific value
c)Net calorific value d) Both B&C
4. Calorific value of the fuel can be calculated by
a) By using Seyler’sformulae b) By using Dulong’sformula
c) By using Davies formulae d) By using Nakamura’s formulae
MODULE-3: Solid fuel -Coal
Coalisahighlycarbonaceousmatterthathasbeenformedasaresultof
alterationofvegetablematter(e.g.,plants)undercertainfavourable
conditions.
ItischieflycomposedofC,H,N,andO,besidesnon-combustible
inorganicmatter.
Classification of Coal:
Coals are classified on the basis of their rank.
Rankisdefinedasthedegreeorextentofmaturationandisthereforea
qualitativemeasureofcarboncontents.
Peat,ligniteandsub-bituminouscoalsarereferredaslowrankcoals
whilebituminouscoalsandanthracitesareclassedashighrank.
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MATRUSRI
ENGINEERING COLLEGE
42
MATRUSRI
ENGINEERING COLLEGEQUIZ
1. Which of the following form of coal has the highest amount of carbon
a) Peat b) Bituminous
c) Lignite d) Anthracite
2. Proximate analysis of the coal checks which one of the following characteristics
a)moisture content b) Ash
c) volatile impurities d) All the above
3. Why is the high percentage of moisture undesirable for coal?
a) It increases the rate of combustion b) It increases the cost of the coal
c) It reduces the calorific value of coal d) It decreases its ignition temperature
4.Which of the following is a noncombustible substance in coal?
a) volatile matter b) ash content
c) fixed carbon d) all the above
44
Combustion: C + O
2→CO
2 ↑
H
2+ ½ O
2→H
2O ↑
Flue gases:2KOH + CO
2→K
2CO
3+ H
2O (in KOH tube)
CaCl
2+ 7 H
2O →CaCl
2.7H
2O (in CaCl
2tube)
MATRUSRI
ENGINEERING COLLEGE
ULTIMATE ANALYSIS
QUIZ
1.Which of the following constituents are determined in ultimate analysis of coal?
a) C,H b) N,S
c) ash d) all
2. The Kiejald’smethod is used to determine
a) S b) N
c) O d) C
3. Which of the following constituent decreases the available hydrogen coal?
a) S b) N
c) O d) C
4. Washings of bomb calorimeter is used to determine
a) S b) N
c) O d) C
53
MATRUSRI
ENGINEERING COLLEGE
Composition and uses of various fractions obtained from Crude Oil
62
S.No Name of fraction Boiling rangeComposition Applications
1Uncondensedgas <30
0
C C
1toC
4
Asdomesticorindustrialfuel
underthenameLPG.
2Petroleumether 30
0
to70
0
C C
5toC
7
Asasolvent.
3GasolineorPetrolorMotorspirit.40
0
to120
0
CC
5toC
9
Asmotorfuel,solvent,indry
cleaning.
4NapthaofSolventspirit 120
0
to180
0
CC
9toC
10
Assolvent,drycleaning.
5Keroseneoil 180
0
to250
0
CC
10toC
16
Asjetenginefuel,toprepare
laboratorygas.
6Dieseloil 250
0
to320
0
CC
10toC
18
Asdieselenginefuel.
7Heavyoil 320
0
to400
0
CC
17toC
30
Forgettinggasolinebycracking.
8AsphaltorPetroleumcoke Above400
0
CC
30andaboveWaterproofing,roadmaking.
MATRUSRI
ENGINEERING COLLEGE
63
Gasoline or Petrol
➢Itisamixtureofhydrocarbonsfrompentanetooctane.
➢Itisobtainedbetween40to120
0
C.
➢Itscalorificvalueis11,250kcal/kg.
➢Itishighlyvolatileandinflammable.
➢Itisusedasfuelforinternalcombustionengine.
MATRUSRI
ENGINEERING COLLEGE
64
Diesel Oil
➢It is a mixture of higher hydrocarbons C
10to C
18.
➢It is obtained between 250 to 320
0
C.
➢Its calorific value is 11,000 kcal/kg.
➢It is used as a fuel for diesel engine.
MATRUSRI
ENGINEERING COLLEGE
65
Kerosene Oil
➢It is a mixture of hydrocarbons C
10to C
16.
➢It is obtained between 180 to 250
0
C.
➢Its calorific value is 11,100 kcal/kg.
➢It does not vaporize easily due to its high boiling point.
➢It is used as domestic fuel in stoves, as jet engine fuel and
for making oil gas.
MATRUSRI
ENGINEERING COLLEGE
66
MATRUSRI
ENGINEERING COLLEGE
1. Which type of process is used for the refining of petroleum?
a) Destructive distillation b) Zone distillation
c) Fractional distillation d) Steam distillation
2. Composition of gasoline or petrol is
a) C
10
-C
16
b) C
5
-C
9
c) C
10
-C
18
d) C
17
-C
32
3. Whichof the following fraction is used as jet fuel?
a) kerosene b) diesel
c) gasoline d) None
4.The uncondensed gases in fractional distillation method are used as
a) CNG b) LPG
b) Coal gas d) water gas
QUIZ
MODULE-6: CRACKING
Thedecompositionofbiggerhydrocarbonmoleculesintosimpler,low
boilinghydrocarbonsoflowermolecularweightiscalledcracking.
67
B.P.=174
0
C B.P.=36
0
C
C
10H
22→C
5H
12+ C
5H
10
Decane Pentane Pentene
MATRUSRI
ENGINEERING COLLEGE
OCTANE NUMBER (or) OCTANE RATING
Itisdefinedasthepercentageofiso-octaneinamixtureofiso-octane
andn-heptanethathasthesameknockingcharacteristicsofthefuel
underexamination.
For Ex: A fuel with octane number 60 will give the same knocking as
the mixture by volume of 60% iso-octane and 40% n-heptane.
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MATRUSRI
ENGINEERING COLLEGE
CETANE NUMBER (or) CETANE RATING
Itisdefinedasthepercentageofcetane(n-hexadecane)inamixtureof
cetaneand2-methylnaphthalenethatwillhavethesameignition
characteristicsasthedieselfuelunderexamination.
Cetanehas a very short ignition lag and hence its cetanenumber is taken
as 100.
Ontheotherhand2-methylnaphthalenehasalongignitionlagand
henceitscetanenumberistakenaszero.
81
MATRUSRI
ENGINEERING COLLEGE
CNG (Compressed Natural Gas)
Natural gas is a mixture of mainly methane (60-95%) with small amounts
of ethane, H
2, CO
2and He.
It is stored and supplied as compressed natural gas (CNG) at a pressure
of 16 to 25 bar in cylinders.
The calorific value of CNG is 12000-14000 kcal/m
3
.
85
MATRUSRI
ENGINEERING COLLEGE
MODULE-9: COMBUSTION
➢It is an exothermic chemical reaction.
➢It is accompanied by development of heat and light at a rapid rate,
so that temperature rises considerably.
Ex:C + O
2→CO
2+ 97 kcal
➢For proper combustion, the substance must be brought to its
ignition temperature.
MATRUSRI
ENGINEERING COLLEGE
Ignition Temperature
Itistheminimumtemperatureatwhichthe
substanceignitesandburnswithoutfurther
additionofheatfromoutside.
OR
Thelowesttemperatureatwhichasubstance
catchesfireandundergoescombustionliberating
heatandlightiscalledignitiontemperature.
MATRUSRI
ENGINEERING COLLEGE
Calculation of Air Quantities: Principles
➢Substances always combine in definite proportionsand these
proportions are determined by the molecular masses of the substances
involved and the products formed.
➢For Ex: C + O
2→CO
2
Mass proportions: 12 32 44
➢The above equation indicates that mass proportions of carbon, oxygen,
and carbon dioxide formed are 12: 32: 44 respectively.
➢Similarly, Mass proportion of Hydrogen: 4:32:36
➢2H
2+ O
2→2H
2O
Mass proportions: 2x2=4 322x18=36
MATRUSRI
ENGINEERING COLLEGE
➢22.4L of any gas at STP has a mass equal to its 1 mole.
Ex: 22.4L of CO
2at STP will have a mass of 44g, because 44g is its
molecular mass.
➢Air contains 21% of oxygen by volume.
➢Mass percent of oxygen is 23 in air.
MATRUSRI
ENGINEERING COLLEGECalculation of Air Quantities: Principles
➢Available hydrogen = Mass of hydrogen –(Mass of oxygen/8)
2H
2+ O
2→2H
2O
Mass proportions: 4 32
➢Therefore, Theoretical amount of oxygen required for the complete
combustion of 1 kg solid or liquid fuel =
MATRUSRI
ENGINEERING COLLEGECalculation of Air Quantities: Principles
➢Since the percentage of oxygen in air by mass is 23, so amount of air
required theoretically for combustion of 1 kg of fuel =
MATRUSRI
ENGINEERING COLLEGECalculation of Air Quantities: Principles
Numerical Problems
✓Calculate the weight and volume of air required for the combustion of 3
kg of coal?
Solution:
Amount of air required for combustion of 3 kg C
MATRUSRI
ENGINEERING COLLEGE
Numerical Problems;
✓Agasusedinaninternalcombustionenginehadthefollowing
compositionbyvolume:H
2=45%,CH
4=36%,CO=15%,N
2=4%.Findthe
volumeofairrequiredforthecombustionof1m
3
ofthegas?
1 m
3
of the gas contains Combustion Reaction Volume of O
2required (m
3
)
H
2(45%) = 0.45 m
3
H
2+ ½ O
2→H
2O 0.45 x 0.5 = 0.225
CH
4(36%) = 0.36 m
3
CH
4 + 2 O
2→CO
2+ 2 H
2O 0.36 x 2 = 0.720
CO (15%) = 0.15 m
3
CO + ½ O
2→CO
2 0.15 x 0.5 = 0.075
N
2(4%) = 0.04 m
3
Total = 1.020
Therefore, volume of air required per m
3
of gas
MATRUSRI
ENGINEERING COLLEGE
✓Agaseousfuelhasthefollowingcompositionbyvolume.CH
4=5%,H
2=20%,
CO=25%,CO
2=6%andrestN
2.If20%excessofairisusedforcombustion,
thencalculatevolumeofairsuppliedperm
3
offuelandcompositionofdry
fluegases?
Substance Combustion Reaction Volume of O
2 Volume of dry
required(m
3
) product(m
3
)
CH
4(5%) = 0.05 m
3
CH
4 + 2 O
2→CO
2+ 2 H
2O 0.05 x 2 = 0.1CO
2= 0.05 x 1 = 0.05
H
2(20%) = 0.2 m
3
H
2+ ½ O
2→H
2O 0.20 x 0.5 = 0.1--
CO (25%)= 0.25 m
3
CO + ½ O
2→CO
2 0.25 x 0.5 = 0.125CO
2= 0.25 x 1 = 0.2525
CO
2(6%) = 0.06 m
3
from fuel CO
2= 0.06
N
2 (44%) = 0.44 m
3
from fuel N
2= 0.44
Total = 0.325 m
3
Total CO
2= 0.36 m
3
N
2= 0.44 m
3
Volume of air required using 20% excess per m
3
MATRUSRI
ENGINEERING COLLEGE
Calculation of dry products of gases:
•CO
2= 0.36 m
3
•O
2= 20% of 0.325 m
3
=
•N
2= 0.44 m
3
+ 1.857 m
3
x 79/100 of air = 1.907 m
3
Total volume of dry products = 0.36 + 0.065 + 1.907 = 2.332 m
3
MATRUSRI
ENGINEERING COLLEGE