it covers all the fuel cycles in thermodynamics as well as complete analysis of the fuels.

1. FUEL AND COMBUSTION
BY SONENDRA

2. Fuel- Introduction
• A fuel is a combustible substance that generates heat when burnt in
presence of oxygen.
• Carbon and Hydrogen are main constituents of a fuel.
• Fuel also contains sulphur, oxygen and nitrogen. A solid fuel contains, 2
to 30% ash.
• In a fuel carbon, hydrogen and sulphur are combustible elements
whereas nitrogen and ash are incombustible elements.

3. CLASSIFICATION OF FUELS
Solid fuels- Natural solid fuels
1. Wood- It is used as domestic fuel.
- It has calorific value of about 10500 KJ/Kg
-It consists of about 48.5% carbon, 6% H2, 1.5 %ash ,43.5% O2 and 0.5% N2.
2. Peat- It is partially carbonised and decomposed material formed mainly due to
transformation of buried vegetation.
- It is spongy substance with high moisture(80%).
-calorific value= 14650 KJ/kg.
-C=58%, H2=6.3%, O2= 30.8%, N2= 0.9% and ash= 4%
3. Lignite(brown coal)-It is low grade coal used domestic use.
-It is intermidiate stage between bituminous and peat.
- Calorific Value=12 500KJ/Kg
-C=66%, H =5%, O =20%, N =1, S=1% and ash=3.5

4. CONTINUED
4. Bituminous Coal- It is commonly used form of coal. It is
available in caking and non-caking form.
-Calorific Value=32000KJ/Kg.
-C=81%, H2=5%, O2=8%, N2=1.5%, S=1%, Ash=3.5%
5. Anthracite Coal= Highest quality of fuel
-Calorific Value=35000 KJ/Kg.
-C=91%, H2%=3, O2=2.5%, N2=0.5% , S=0.5%, Ash= 2.5%
.

5. Artificial solid fuels
1.Coke-
• It is prepared by heating bituminous coal at 5000-70000C for 15-18 hours in a
closed chamber in absence of air.
•During this process, volatile matter is removed.
•Coke is hard, porous and smokelesss.
•It is used in producing gas producing plants and blast furnaces. It contains 85-
95% carbon.
2. Pulverised Coal-Dry coal is pulverised in pulverised in pulverizing mill.
•It burns most efficiently when mixed with air and hence it is used in most of
thermal power generating plants.
3. Wood charcoal-It is prepared by strong heating of wood in a limited supply of
air below 2800oC. It is used for domestic purpose.

6. ADVANTAGES OF SOLID FUEL-
•They can be easily transported.
•They have low ignition temperature.
•They involve low production cost.
DISADVANTAGES OF SOLID FUEL-
•They have large percentage of ash content.
•They produce large amount of smoke.
•The necessity of various fuel handling equipments (conveyor, crushers, hoppers,
bunkers), increases cost of fuel handling.
•They have lower calorific value when compared to liquid fuels.

7. Liquid Fuels
• Liquid fuels are availabe in
the form of crude petroleum.
• Petroleum is mixture of
hydrocarbons that can be
grouped into the following 4
categories.
- paraffins(CnH2n+2)
-oleffins(CnH2n)
-Naphthalenes(CnH3n)
-Aromatics(CnH4n-2)
• The impurities present in the
crude oil are eliminated by the
process of distillation.
1. Gasoline-It is lightest liquid petroleum fraction.
•It is highly volatile in nature.
•Specific gravity=0.7.-0.78
•It is obtained between 65 and 2000C
•It is light yellow in colour.
•It is used as fuel in automobiles and aeroplanes.
•calorific value=47000
2. Diesel-It is obtained between 200 and 3700C.
•Calorific value=45500
•specific gravity=0.86-0.95
•It is used as fuel in automobiles and diesel
engines.

8. 3. Kerosene-It is obtained between 150 to 3000C.
•Specific gravity=0.78-0.85
•It is used in lamps,stoves, jet fuel and aviation gas
turbines.
Adavantages of liquid fuels
•They have high calorific value as compared to solid
fuels.
•higher combustion efficiency due to uniform mixing
of fuel and air.
•they burn without forming ash, smoke thud
eliminating ash disposal problem.
•They can be stored and handled easily.
Disadvantages of liquid fuels-
•Leakage loss during transportation and handling.
•highly volatile and easily vapourised.
Gaseous fuels-
1. Natural Gas-It mainly contains 85% metane(CH4) and
hydrogen (H2) along with small amounts of ethane and
ethylene.
•LPG(Liquified petroleum gas) is a mixture of propane and
butane.
2. Blast Furnace gas- It is obtained as by-product during
blast furnace operation.
•calorific value=3800-4200KJ/Kg
3.Producer gas- it is obtained by partial combustion coke and
coal in the presence of air and steam mixture..
•calorific value=5000-6800KJ/Kg.
•it is used in power generation.
4. Water gas-it is obtained by passing steam over candescent
coke.
•Calorific value=10500-2300KJ/Kg.
•It is used in welding and furnaces.

9. CALORIFIC VALUE
• The calorific value of a fuel is defined as the quantity of heat
(expressed in calories or kilo calories) liberated by the complete
combustion of unit weight (1gm or 1kg) of the fuel in air or oxygen,
with subsequent cooling of the products of combustion to the initial
temperature of the fuel.

10. CALORIFIC VALUES OF FUEL
1. Higher Calorific Value(HCV)
• It is the quantity of heat liberated by the complete combustion of unit weight of the fuel with subsequent
cooling of the products of combustion to the initial temperature of the fuel.
• As all fuel contains hydrogen, they produce water vapour during combustion.
• when products of combustion containing water vapours are cooled back to initial
(room) temperature, then all water vapours formed condenses and evolve latent heat.
this adds up to heat liberated by burning the fuel.
• HCV is calculated using Dulong's formula
HCV=1/100[33800C+144000(H-O/8)+9270S]KJ/Kg

11. Lower Calorific Value(LCV)-
• It is defined as the gross calorific value minus the latent heat of
condensation of water (at the initial temperature of the fuel), formed
by the combustion of hydrogen in the fuel.
• Lower heating value (LHV) of fuel refers to the enthalpy of combustion when all
the water (H2O) formed during combustion is in vapour form.
• In most combustion processes, the products of combustion(gas coming out od boiler chimney)
can not be cooled to its initial temperature. Thus water vapours do not condensesand latent
heat of water vapours is lost to the atmosphere The resultant heat liberated by the fuel
excludes latent heat of evaporation of water vapours is known as lower calorific value of
fuel.The heat lost to the atmosphere depends on evaporation pressure and amount of water
vapours formed.Due to difficulty in measuring evaporation pressure, it is assumed that
evaporation takes place at a saturation temperatire of 150C. The latent heat corresponding to
this saturation temperature is 2466KJ/Kg.
LCV= HCV- m* 2466 = HCV- 9H* 2466
• As one part of hydrogen by mass produces nine parts of water vapours by mass, the mass of
water vapours produced is 9H Where H is percentage of hydrogen by weight.

12. Experimental Procedure
to determine the CV of fuel
•Bomb Calorimeter
•Lewis Thomson Calorimeter
•Junker’s gas calorimeter.

13. BOMB CALORIMETER

14. Bomb Calorimeter for determination of Calorific
Value of solid and liquid fuel
• It is used to measure the calorific value (CV) of solid as well
as liquid fuel. But to determine the CV of gas, one need to choose
Junker's calorimeter.
• Calorimeter contain thick walled cylindrical vessel and it consists of lid
which supports two electrodes which are in contact with fuse and fuel
sample of known weight.
• Lid also contain oxygen inlet valve through which high pressure oxygen
gas (at about 25 to 30 atm) is supplied.
• Entire lid with fuel sample is now held in a copper calorimeter containing
known weight of water. Mechanical stirrer is provided to stirred well for
uniform heating of water.
• A thermometer is also provided to measure the change in temperature of
water due to combustion of fuel in Lid

15. Bomb Calorimeter cont
• Procedure of bomb calorimeter experiment
A known quantity of fuel sample is added in crucible.
• Start the stirrer and not the initial temperature of water.
• Start current through crucible and let fuel sample to burn in
presence of oxygen.
• Heat release during combustion of fuel is taken by water and
temperature of it rises.
• Note final steady state temperature of water.
   
f
W
C
m
C
T
T
T
m
m 



 2
1
2
1

16. Junkers Gas calorimeter-

17. Combustion-
During combustion the energy is released by oxidation of fuel
elements such as carbon C, hydrogen H2 and sulphur S, i.e.
high temperature chemical reaction of these elements with
oxygen O2 (generally from air) releases energy to produce
high temperature gases.
These high temperature gases act as heat source.

18. Why Perform Combustion Analysis?
Combustion analysis to Improve Fuel Efficiency ,Reduce Emissions, Improve
Safety. It involves the measurement of gas concentrations, temperatures and
pressure for boiler tune-ups, emissions checks and safety improvements. Parameters
that are commonly examined
include:
• Oxygen (O2)
• Carbon Monoxide (CO)
• Carbon Dioxide (CO2)
• Exhaust gas temperature
• Supplied combustion air temperature
• Draft
• Nitric Oxide (NO)
• Nitrogen Dioxide (NO2)
• Sulfur Dioxide (SO2)

19. COMBUSTION OF FUEL
1. C + 02 → CO2
12kg C + 32kg 02 → 44 kg CO2
1kg C + 8/3 kg 02 → 11/3 kg CO2
2. 2H2 + 02 → 2H20
4 kg H2 + 32 kg 02 → 36 kg H20
1 Kg H2 + 8 kg 02 → 9 kg H20
3. S + 02 → SO2
32 kg S + 32 kg 02 → 64 kg SO2
1 kg S + 1 kg 02 → 2 kg SO2

20. 4. 2 C + 02 → 2 CO
24 kg C + 32 kg O2 → 56 kg CO
1 kg C + 4/3 KG 02 → 7/3 kg CO
5. 2 CO + 02 → 2 CO2
56 kg CO + 32 kg O2 → 88 CO2
1 kg CO + 4/7 kg O2 → 11/7 kg CO2

21. Atmospheric Air
•For the purposes of combustion calculations the composition of
air is approximated as a simple mixture of oxygen and nitrogen
by volume
•Oxygen 21% by volume
•Oxygen /23.3 by mass
• Nitrogen 79% by volume
• Nitrogen /76.7 by mass
21

22. Minimum air required to burn unit Kg of Fuel
• Minimum Oxygen required = [8/3 C +8 H2 + S -O2]=11.63kg/kg of fuel
•Minimum air required= 100/23[8/3 C +8 H2 + S -
O2]=11.63kg/kg of fuel

23. Theoretical air and Excess air
• Theoretical air : The minimum amount of air that supply sufficient
oxygen for the complete combustion of all the carbon ,hydrogen and
any other elements in the fuel that may be oxidize is called the
theoretical air
• Excess air : The excess air is the amount of air supplied over and above
the theoretical air .
• In practice it is found that complete combustion is not likely to achieve
unless the amount of air is some what greater than the theoretical
amount For example 150% theoretical air means that air actually
supplied is 1.5 times the theoretical air . Thus 150 percent theoretical
air is equivalent to 50% excess air.

24. Equivalence Ratio
• It is sometimes convenient to use term excess air ratio,
defined as:
/
/
actual A F ratio
stoichiometric A F ratio
24

25. Excess Air
• In practice it is impossible to obtain complete combustion under stoichiometric conditions.
Incomplete combustion is a waste of energy and it leads to the formation of carbon monoxide,
an extremely toxic gas, in the products.
• Excess air is expressed as a percentage increase over the stoichiometric
requirement and is defined by:
Excess air will always reduce the efficiency of a combustion system
25
/ /
100%
/
actual A F ratio stoichiometric A F ratio
stoichiometric A F ratio



26. Ultimate Analysis
It gives information about
• Chemical elements of coal
• Ash
• Moisture.
•We know that C + H + O + N + S + M + A = 100% by mass
For Example-
1. Bituminous Coal
C=81%, H2=5%, O2=8%, N2=1.5%, S=1%, Ash=3.5%
2. Anthracite Coal
C=91%, H2%=3, O2=2.5%, N2=0.5% , S=0.5%, Ash= 2.5%

27. Proximate Analysis
• It gives behavior of coal when heated.
• It indicates- percentage by weight of fixed carbon, volatile matter, ash and
moisture in coal.
• FC+VM+M+A = 100 % by mass
• Measurement of Moisture- When 1 gm of coal sample is heated for period of
1 hour at temperature of 1050C .Then loss of weight =Moisture content
• Volatile matter- When 1 gm sample of coal is heated for 7 minutes at
temperature of 9500C then loss of weight = volatime matter
• Measurement of Ash = when 1 gm of sample of coal is completely burnt then
remaining is ash.
• Measurement of fixed carbon-
FC = 100- (% of moisture + Volatile + Ash)

28. Adiabatic flame temperature:
•Temperature that could be attained by the products of
combustion when the combustion reaction is carried
out in limit of adiabatic operation of combustion
chamber.
•This is the maximum temperature which can be
attained in a combustion chamber .

29. •Enthalpy of Formation:
Enthalpy of formation of a compound is the energy
released or absorbed when compound is formed from its
elements at standard reference state.
•Enthalpy of Combustion:
Enthalpy of combustion of fuel is defined as the
difference between the enthalpy of the products and
enthalpy of reactants

31. Standard Reference State
• It refers to thermodynamic state at which the enthalpy datum can be set
for study of reacting systems.
• At standard reference state, zero value is assigned arbitrarily to the
enthalpy of stable elements. Generally, standard reference state is taken
as 25°C and 1 atm.
Tref = 25°C = 298.15 K, Pref = 1 atm

33. Dissociation:
• It refers to the combustion products getting dissociated and thus
absorbing some of energy. Such as, the case of carbon dioxide getting
formed during combustion and subsequently getting dissociated can be
explained as below
•Dissociation: Heat + CO2 → C + 02
• Combustion: C + 02 → CO2+ Heat

34. INTRODUCTION to flue gas analysis
To have proper control on combustion process, an idea about complete
or incomplete combustion of fuel is made by the analysis of flue
gas. Thus,
(i) if the gases contain considerable amount of carbon monoxide, it
indicates that incomplete combustion is occurring (i.e. considerable
wastage of fuel is taking flue).
Also indicates the short supply of oxygen for combustion
(ii) if the flue gases contain a considerable amount of oxygen, it
indicates the oxygen supply is in excess, though the combustion may
be complete.
The analysis of flue gases made with the help of ORSAT’S
APPARATUS.

35. FLUE GAS ANALYSIS
ORSAT APPARATUS
Introduction
Construction
Working
Includes three steps.

36. ORSAT APPARATUS

37. ORSAT APPARATUS

39. CONSTRUCTION
The absorption bulbs have solutions for the absorption of CO2, O2 and CO respectively.
First bulb has ‘potassium hydroxide’ solution (250g KOH in 500mL of
boiled distilled water), and it absorbs only CO2.
Second bulb has a solution of ‘alkaline pyrogallic acid’ (25g pyrogallic
acid+200g KOH in 500 mL of distilled water) and it can absorb O2.
Third bulb contains ‘ammonical cuprous chloride’ (100g cuprous
chloride + 125 mL liquor ammonia+375 mL of water) and it can absorb
CO.
Hence, it is necessary that the flue gas is passed first through potassium
hydroxide bulb, where CO2 is absorbed, then through alkaline pyrogallic
acid bulb, when only O2 will be absorbed ( because CO2 has already
been removed) and finally through ammonical cuprous chloride bulb,
where only CO will be absorbed.

40. Volumetric and gravimetric analysis
•Volumetric Analysis: Combustion analysis when carried out
based upon percentage by volume of constituent reactants
and products is called volumetric analysis.
•Gravimetric analysis: Combustion analysis carried out based
upon percentage by mass of reactants and products is called
Gravimetric analysis.

41. Gravitational Analysis
Let us see how air required for complete burning of a fuel having 85.5% carbon,
12.3%hydrogen and 2.2% ash is calculated;
Mass of constituents
per kg of fuel (A)
Oxygen required per
kg of constituent (B)
Oxygen required per
kg of fuel (C) = (A) x
(B)
Mass of air required,
(theoretical, for air
having 23% oxygen,
77% nitrogen)
= ∑C x100/ 23
C = 0.855
H2 = 0.123
Ash = 0.022
8/3
8
2.28
0.984
= 3.264 x 100/23 =
14.19
∑(C) = 3.264

42. NUMERICAL 1 P.N0-72 HEGDE
1. The percentage composition of a fuel on mass basis is as
follows: C=90% , H2 = 3.5%, O2 =1% , S=0.5% and ash=5%
Calculate the
i) minimum air required for complete combustion of 1kg of fuel
ii) composition of dry flue gases on mass basis if 50 % excess
air is supplied.
solution-
minimum air required= 100/23[8/3C +8 H2 + S -
O2]=11.63kg/kg of fuel

43. Combustion Reactions
• CH4 + 2O2 → CO2 + 2H2O
16 + 64 → 44 + 36 On mass basis
1 kg (CH4) + 4 kg (O2) → 11/ 4 kg (CO2) + 9 /4 kg (H2O) On mass basis
1 mol (CH4) + 2 mol (O2) 1 mol (CO2) + 2 mol (H2O)
On mole basis

44. Combustion Reactions
• 2C2H6 + 7O2 → 4CO2 + 6H2O
60 + 224 → 176 + 108 On mass basis
1 kg (C2H6) + 56/15 kg (O2) → 44/15 kg (CO2) + 27/15 kg (H2O)
•2 mol (C2H6) + 7 mol (O2) → 4 mol (CO2) + 6 mol (H2O)
On mole basis

45. Volumetric Analysis
• In general, for any hydrocarbon’s complete combustion,
• a . C8H18 + b .O2 → d . CO2 + e . H2O
• Equating C, H and O on both sides of above equation,
• a x 8 = d or, d =8a
• a x 18 = e x 2 or, e =9a
• b x2 = (dx 2) + e or, 2b =2d + e,
• or 2x b = 16a + 9a
• Above yield, b = 12.5a, d = 8a, e = 9a
2 mol (C8H18) + 25 mol (O2) 16 mol (CO2) + 18 mol (H2O)