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fuel
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Subject Contents
1. Fuel and Combustion
2. Steam Generator/ Boiler
3. Nozzles
4. Steam Turbines
5. Gas Turbines
6. Air Compressors
Recommended Books
i. Applied Thermodynamics for Engineering Technologists
By: T.D. Eastop & A. McConkey
ii. Basic Thermodynamics
By: Reynar Jeol
iii. Thermodynamics An Engineering Approach
By: Yunus A. Cengel & M.A. Boles
iv. Thermodynamics for Engineers
BY: Bhalchandra V. Karlekar
Reference Books
i Text book of Thermal Engineering
By: R.S. Khurmi
ii Thermal Engineering
By: Sarao
What is a Fuel ?
Any material that can be burned to release thermal energy is called a Fuel.
Most familiar fuels consist primarily of hydrogen and carbon.
They are called hydrocarbon fuels and are denoted by the general formula CnHm.
Type of Fuels
i. Liquid (gasoline)
ii. Solid (coal)
iii. Gas (natural gas)
Liquid Fuels
Furnace oil
Light diesel oil
Petrol / Gasoline
Kerosene
Ethanol
LSHS (low sulphur heavy stock)
Liquid Fuels and Calorific value
Fuel Oil Gross Calorific Value (kCal/kg)
Kerosene 11,100
Diesel Oil 10,800
L.D.O 10,700
Furnace Oil 10,500
LSHS 10,600
Contamination of Liquid Fuel
1. Sulphur content:
1. Depends on source of crude oil and less on the refining process
2. Furnace oil: 2-4 % sulphur
Effects: Sulphuric acid causes corrosion
2. Ash content
Inorganic material in fuel
Typically 0.03 - 0.07%
Effects: Corrosion of burner tips and damage to materials /
equipments at high temperature
3. Carbon residue
Tendency of oil to deposit a carbonaceous solid residue on a hot
surface
Residual oil: >1% carbon residue
4. Water content
Normally low in furnace oil supplied (
Solid Fuels
Coal classification
Anthracite: hard and geologically the oldest
Bituminous
Lignite: soft coal
Further classification:
semi- anthracite,
semi-bituminous, and
sub-bituminous
Physical properties
Heating or calorific value (GCV)
Moisture content
Volatile matter
Ash
Chemical properties
Chemical constituents: carbon, hydrogen, oxygen, sulphur
Solid Fuel Contents
1. Moisture content
0.5 to 10% of moisture in fuel
Effects:
Reduces heating value of fuel
Weight loss from heated and then cooled powdered raw coal
2. Volatile matter
Methane, hydrocarbons, hydrogen, CO, other
Typically 25-35%
Easy ignition with high volatile matter
Weight loss from heated then cooled crushed coal
3. Ash
Impurity that will not burn (5-40%)
Important for design of furnace
Ash = residue after combustion
4. Fixed carbon
Fixed carbon = 100 (moisture + volatile matter + ash)
Carbon + hydrogen, oxygen, sulphur, nitrogen residues
Heat generator during combustion
Analysis of Coal
1. Proximate analysis of coal
Determines only fixed carbon, volatile matter, moisture and ash
Useful to find out heating value (GCV)
Simple analysis equipment
2. Ultimate analysis of coal
Determines all coal component elements: carbon, hydrogen,
oxygen, sulphur, other
Useful for furnace design
(e.g flame temperature, flue duct design)
Laboratory analysis
Gaseous Fuels
Classification of gaseous fuels
1. Fuels naturally found in natureNatural gasMethane from coal mines
2. Fuel gases made from solid fuelGases derived from coalGases derived from waste and biomassFrom other industrial processes
3. Gases made from petroleumLiquefied Petroleum gas (LPG)Refinery gasesGases from oil gasification
4. Gases from some fermentation
Comparing of Fuels
Constituents Fuel Oil Coal Natural Gas
Carbon 84 41.11 74
Hydrogen 12 2.76 25
Sulphur 3 0.41 -
Oxygen 1 9.89 Trace
Nitrogen Trace 1.22 0.75
Ash Trace 38.63 -
Water Trace 5.98 -
Advantages of Gaseous Fuel
They are free from all solid and liquid impurities.
Simplest burners systems
Burner systems require least maintenance
Environmental benefits: lowest GHG and other emissions
They do not produce smoke and ash.
They undergo complete combustion with minimum air supply.
A good fuel should have:
Low ignition point
High calorific value
Freely burn with high efficiency, once it is ignited
Not produce harmful gases
Produce least quantity of smoke and gases
Economical , easy to store and convenient for transportation
Calorific Value of FuelsIt is amount of heat given out by complete combustion of one kg of
fuel, solid, liquid or gas.
It is expressed in terms of kJ/kg of fuel.
Types of calorific value
i. Gross/higher calorific value (H.C.V)
ii. Net/lower calorific value (L.C.V)
i. Gross/higher calorific value:
Total amount of heat produce by one kg of fuel
ii. Net/lower calorific value:
Net amount of heat used to produce steam per kg of heat
L.C.V = H.C.V Heat of steam formed during combustion
What is Combustion ?
A chemical reaction during which a fuel is oxidized and a large quantityof energy/ heat is released is called combustion .
Necessary Constituents of Combustion:
i. Fuel
ii. Supporter of combustion(Air/oxidizer)
iii. Ignition / kindling temperature
Ignition temperature for different fuels
Fuel Ignition Temperature (oC)
Gasoline / Petrol 260
Carbon 400
Hydrogen 580
Carbon monoxide 610
Methane 630
The oxidizer most often used in combustion processes is air, which is freeand readily available.
Pure oxygen O2 is used as an oxidizer only in some specialized applications, such as cutting and welding.
Oxygen is the key to combustion
Principle of Combustion
Composition of Air
Types of combustion Process
1. Complete combustion
2. Incomplete combustion
1. Complete Combustion Process
A combustion process is complete if all the carbon & hydrogen in
the fuel burns to form, CO2,& H2O.
All the sulfur (if any) burns to SO2.
It means, all the combustible components of a fuel are burned tocompletion during a complete combustion process.
2. Incomplete Combustion Process
The combustion process is incomplete if the combustion products
contain any unburned fuel or components such as, C, H2, CO, or OH.
Insufficient oxygen is an obvious reason for incomplete combustion,but it is not the only one.
Incomplete combustion occurs even when more oxygen is present inthe combustion chamber than is needed for complete combustion.
Another cause of incomplete combustion is dissociation, which becomes important at high temperatures.(During adiabatic combustion max. Temp. Lower than the calculatedone)
Chemical equations are balanced on the basis of the conservation ofmass principle (or the mass balance),
Stated as follows: The total mass of each element is conservedduring a chemical reaction.
That is, the total mass of each element on the right-hand side of thereaction equation (the products) must be equal to the total mass ofthat element on the left-hand side (the reactants).
Even though the elements exist in different chemical compounds in the reactants and products.
Example:
2H2 + O2 2H2OReactants Product
Theoretical / Stoichiometric Combustion Process
The ideal combustion process during which a fuel is burnedcompletely with theoretical air(stoichiometric air) is called thestoichiometric or theoretical combustion process.
For example,
the theoretical combustion of methane is notice that theproducts of the theoretical combustion contain no unburned
methane and no C, H2, CO, OH, or free O2.
Stoichiometric air:
Minimum mass / volume of air required for completedcombustion of 1kg of fuel is known as Stoichiometric air.
Actual combustion process
In actual combustion processes, it is common practice to use
more air than the stoichiometric amount to increase the chances
of complete combustion or to control the temperature of the
combustion chamber.
Excess Air
The amount of air in excess of the stoichiometric amount is called
excess air.
The amount of excess air is usually expressed in terms of thestoichiometric air as percent excess air or percent theoretical air.
For example,50 percent excess air is equivalent to 150 percent theoretical air,
200 percent excess air is equivalent to 300 percent theoretical air.
Deficiency of Air
Amounts of air less than the stoichiometric amount are called
deficiency of air and are often expressed as percent deficiency of air. For example, 90 percent theoretical air is equivalent to 10 percent deficiency of
air.
Air-Fuel RatioA frequently used quantity in the analysis of combustion processes
to quantify the amounts of fuel and air is the airfuel ratio (AF).
AF represents the amount of air used per unit mass of fuel during acombustion process
It is usually expressed on a mass basis and is defined as,
The mass m of a substance is related to the number of moles N through the relation m= NM, where M is the molar mass.
Example:
The airfuel ratio can also be expressed on a mole basis as the ratioof the mole numbers of air to the mole numbers of fuel. But we willuse the former definition.
The reciprocal of airfuel ratio is called the fuelair ratio.
Example:One k mol of octane (C8H18) is burned with air that contains 20 kmol of O2.Assuming the products contain only CO2, H2O, O2, and N2, determine themole number of each gas in the products and the airfuel ratio for thiscombustion process. Assume the molar mass of air is 29.0 kg/kmol.
The chemical equation for this combustion process,
X, y,z and w represent the unknown mole numbers of the product, theseunknowns are determined by applying the mass balance to each of theelements.
For Carbon(C): 8 = x x = 8For Hydrogen (H): 18 =2y y = 9For Oxygen (O) 20x2=2x+ y+2z z =7.5For Nitrogen (N2): (20)(3.76) =w w = 75.2
Now, substitute the values and yields
Note that the coefficient 20 in the balanced equation above represents the Number of moles of Oxygen, not the number of moles of air.
The latter is obtained by
20x3.76 =75.2 moles of nitrogen to the 20 moles of oxygen.
Therefore, total moles of air is 95.2
The air-fuel ratio (AF) is determine by
AF = 24.2 kg air / kg fuel
It mean 24.2 kg of air is required to burn each kilogram of fuel during thisCombustion process.
Combustion Equations of Solid Fuels
When carbon burns in sufficient quantity of oxygen and produced carbon dioxide along with large amount of heat.
C + O2 = CO2+Heat
1mol + 1mol =1mol
12kg + 32kg = 44kg
1kg + 8/3kg = 11/3
It means that 1kg of carbon requires 8/3 kg of oxygen for its complete combustion and produces 11/3 kg of carbon dioxide gas.
If sufficient oxygen is not available, then combustion process is not complete and produces carbon monoxide instead of carbon dioxide.
2C + O2 = 2CO
2mol + 1mol =2mol (by volume)
2x12kg + 2x16kg =2x28kg (by mass)
1kg + 4/3kg = 7/3
It means that 1kg of carbon requires 4/3 kg of oxygen and produces 7/3 kg of carbon monoxide.
If carbon monoxide burn further and produces carbon dioxide
2CO + O2 = 2CO2
2mol + 1mol =2mol (by volume)
2x28kg + 2x16kg =2x44kg (by mass)
1kg + 4/7kg = 11/7
It means that 1kg of carbon monoxide requires 4/7 kg of oxygen and
produces 11/7 kg of carbon dioxide.
When sulphur (if any) burns with oxygen and produced sulphur dioxide.
S + O2 = SO2
1mol + 1mol =1mol (by volume)
32kg + 2x16kg = 64kg (by mass)
1kg + 1kg = 2kg
It means that 1kg of sulphur requires 1 kg of oxygen for its complete combustion and produces 2 kg of sulphur dioxide gas.
Combustion Equations of Gaseous Fuels
Gaseous fuels are generally measured by volume (m3) than by mass
When carbon monoxide burns with oxygen and produced carbon dioxide.By Volume:
2CO + O2 = 2CO2
2volumes + 1vol =2vol
2m3 + 1m3 = 2m3
By mass:
2CO + O2 = 2CO2
2x28kg + 1x32kg= 2x44kg
1kg + 4/7 kg = 11/7kg
It means that 1kg of carbon monoxide requires 4/7 kg of oxygen and produces 11/7 kg of carbon dioxide gas.
When hydrogen burns with oxygen produces:By Volume:
2H2 + O2 = 2H2O
2volumes + 1vol =2vol
2m3 + 1m3 = 2m3
By mass:
2H2 + O2 = 2H2O
2x2kg + 1x32kg= 2x18kg
1kg + 8 kg = 9 kg
It means that 1kg hydrogen requires 8 kg of oxygen and produces 9 kg of water or steam.
For burning of CH4 with O2,
CH4 + 2O2 =CO2 + 2H2O
16kg + 2x32kg =44kg + 2x18kg
1kg + 4 kg = 11/4 kg +9/4kg
Now, consider 1kg of a Fuel, which contains C,H and S
Let, Mass of Carbon = C kg
Mass of Hydrogen = H kgMass of Sulphur = S kg
We know that for the complete combustion of fuel requires sufficient Oxygen for constituent element such as,
1kg Carbon (C) 8/3 kg of Oxygen
1kg Hydrogen (H) 8 kg of Oxygen
1kg Sulphur (S) 1 kg of Oxygen
Therefore, total oxygen required for complete combustion of 1 kg of fuel
= 8/3 C +8 H2 +1 S kg
If some quantity of oxygen is already present in the fuel, then
=[8/3 C +8 H2 + S kg]- O2 kg
We know that oxygen is obtained from the atmospheric air for the combustionWhich mainly consists of Nitrogen, Oxygen and negligible quantity ofCarbon dioxide and inert gases (Neon, Argon and Krypton).
For the calculations these negligible quantities are not considered.Therefore,
N2 = 77% and O2 = 23% By massN2 = 79% and O2 = 21% By volume
It is obvious that for obtaining 1kg of oxygen, amount of air req.
=100/23=4.35kg
=100/23[(8/3 C+8H2+S)-O2]
Example:A fuel has the following composition by mass,Carbon 86%, Hydrogen 11.75% and Oxygen 2.25%.Calculate the theoretical air supply per kg of fuel and the mass ofproducts of combustion per kg of fuel.
Solution:
C=0.86 kg, H2 = 0.1175kg, O2 = 0.0225kg
(i) Theoretical air supply per kgwe know that
Airtheo = 100/23[(8/3C+8 H2 +S)- O2] kg
Substitute the values Airtheo = 100/23[(8/3x0.86+8 x0.1175 +0)- 0.0225] kg
Airtheo = 13.96 14kg
(ii) Mass of products of combustion
We know that,
C+O2 =CO22H2 + O2 = 2H2O
We know that 1kg of carbon produces 11/3 kg of carbon dioxide and 1kg of hydrogen produces 9 kg of water.
Therefore, total mass of the products of combustion,
=11/3xC+9H2kg
= 11/3x0.86+9x0.1175= 4.21 kg
Example:Calculate the stoichiometric A/F ratio for the combustion of a sample of dry Anthracite of the following composition by massC90% ;H 3%; O2 2.5% ; N1% ; S 0.5% and ash 3%Determine the A/F ratio and the dry and wet analysis of products of combustionBy volume, when 20% excess air is supplied.
Constituent Mass/kg Combust:Equation
Oxygen (O2) req./ kg
Products / kg
C 0.9 C+O2 CO212kg+32 44 kg
0.9x 32/12= 2.4kg
0.9x 44/12 =3.3kg of CO2
H 0.03 2H2+O2 2H2O1kg + 8kg 9kg
0.03x8=0.24kg
0.03x9 =0.27 kg H2O
O 0.025 0.025kg -0.025kg
N 0.01 0.01kg N2
S 0.005 S+O2 SO232kg+32kg=64kg
0.005x32/32=0.005kg
0.005x64/32=0.01kg SO2
Ash 0.03
Total 1.0 2.62kg
From table O2 required per kg of coal =2.62 kg
Therefore,
(i) Air required per kg of coal = 2.62/0.233= 11.25kg
(Assume 23.3% O2 by mass)
(ii) N2 associated with this air =0.767x11.25 =8.63kg
Total N2 in product =8.63+0.01=8.64kg
The stoichiometric A/F ratio = 11.25/1
For an air supply which is 20% excess
We know that % excess air = A/F(Actual)- A/F(Stioch)/ A/F (Stioch)
A/F(Actual)=11.25+20/100x11.25=1.2x11.25
A/F(Actual)=13.5/1
Therefore,
N2 supplied =0.767x13.5=10.36kg
O2 supplied =0.233x13.5=3.144kg
In the products, we have
N2 =10.36+0.01= 10.37kg
and excess oxygen
O2 =3.144-2.62=0.524kg
Example:The volumetric analysis of a flue gas is CO214% ,CO 1%, O2 5% and N280%Calculate the flue gas composition by mass. Solution:By volumeCO2= 0.14m
3 ,CO=0.01m3 , O2 =0.05m3 and N2= 0.8m
3
The volumetric analysis may be converted into mass analysis by Completing following table
Constituent Volume of gas (m3)
(a)
Molecular mass
(b)
ProportionalMassc=axb
Mass in/kgd=c/c
%by massdx100
CO2 0.14 44 6.16 6.16/30.44=0.202
20.2%
CO 0.01 28 0.28 0.28/30.44=0.009
0.9%
O2 0.05 32 1.60 1.60/30.44=0.053
5.3%
N2 0.80 28 22.40 22.40/30.44=0.736
73.6%
Total 1.00 c=30.44 1.00 100
The flue gas composition by mass is given in the last column
CO2=20.2%CO = 0.9%O2 =5.3%
N2 = 73.6%
Example:A flue gas has the following percentage composition by massCO213.3% ,CO 0.95%, O2 8.35% and N277.4%
Calculate the flue gas composition by volume. Solution:By volumeCO2= 13.3% ,CO=0.95% , O2 = 8.35% and N2= 77.4%The mass analysis may be converted into volumetric analysis by Completingfollowing table
Constituent % mass analysis
(a)
Molecular mass
(b)
Proportionalvolumec=a/b
Volume in (m3)
d=c/c
%volumetricdx100
CO2 13.3 44 13.3/44=0.302
0.302/3.357=0.090
9.0
CO 0.95 28 0.95/28=0.034
0.034/3.357=0.010
1.0
O2 8.35 32 8.35/32=0.261
0.261/3.357=0.078
7.8
N2 77.4 28 77.4/28=2.76
2.76/3.357=0.822
82.2
Total 100.0 c=3.357 1.00 100
The flue gas composition by volume is given in the last column
CO2=9%CO = 1%O2 =7.8%
N2 = 82.2% Ans