2 module 1-21-dec-2018 reference material i-2 fac m1_intro_cv

L.Muruganandam
Fuels & Combustion
(CHE 2006)

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Topics for discussion - Module:1
Fuels: Definition and types (classification) of fuels
Properties of fuels (Solid, liquid, gaseous fuels)
Determination of properties of fuels
Fuel composition analysis (Proximate and Ultimate Analysis)
Calorific value (CV) – Gross and net calorific values(GCV,NCV)
Bomb Calorimetry – empirical equations for CV estimation.
Dulong’s Formula & Experimental methods

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Introduction
Fuel: Anything which burn
to and give heat in
presence of oxygen.
Combustion: The process of burning fuel.
Burning is the intense chemical reaction
with the emission of heat and other exhaust
gases when fuel reacts with oxygen.

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Fuels are defined as “Substance which Undergoes Combustion
in the presence of air or oxygen to produce a large amount of
heat that can be used economically for industrial and domestic
purpose”.
Any chemical process accompanied by the evolution of light and
heat is called combustion. It is simply the reaction of substances
with oxygen and converts chemical energy into heat and light.
Commercially important fuels contain carbon and hydrogen and
their compounds, which provide heating value.
 Wood, Coal, Kerosene,
petrol, Diesel, Natural
gas(LPG) etc.

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Fuel: It is a source for heat and light energy. When the fuel burns in air it
releases energy.
What is combustion?
It is a chemical reaction between the fuel and oxygen (air).
The processes occur with a flame and evolve heat energy.
Fuel + Oxidant  combustion products + energy
The energy, which is produced, is mainly in the form of light and heat.
The overall process is exothermic in nature.

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 Fuels may be solid, liquid or gaseous
 Fuels may be fossil (non-renewable) or biomass (renewable).
 Fossil fuels may be coal, petroleum-crude derived or natural
gas.
 Biomass fuels may be wood, refuse or agricultural residues.
Sources of energy
Wood ; Coal
Natural gas ; Oil
Nuclear ;
Alternative energies Hydropower ; Solar ; Wind
Why we study?
 The knowledge of the fuel properties helps in selecting the right fuel for
the right purpose and efficient use of the fuel.
 Studying combustion of fuel will lead to design combustion system and
control the burning.
 Will help in the research of alternative fuels

APPLICATION
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Fuels play an important role in our everyday life because they are used in
homes, transport and industry for providing energy.
Domestic/house hold Usage: Cooking, lighting, heating, cooling etc. Fuels
like wood, coal, kerosene, cow dung etc are used.
For Transportation: IC engines in cars, buses, trucks etc. Coal, diesel and
petrol are used as fuel for road, sea and air transport in automobiles and
locomotives.
For industry: Boilers, furnaces, reciprocating engines, gas turbine engines,
for power plants, rocket motors etc. Fuels like coal and natural gas are used.
For Air Space Centre:
Specially prepared fuels like hydrazine (Rocket fuels) [NH2-NH2] are used.
Fuel devices:
Candle flames, lighting of match sticks, cigarette burning, wood burning,
gasifier, furnace, Liquefied petroleum gas (LPG) burners for cooking..

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Primary fuels: Fuels which occur naturally such as coal, crude petroleum
and natural gas. Coal and crude petroleum, formed from organic matter
many millions of years ago, are referred to as natural or fossil fuels,
Secondary fuels: Fuels which are derived from naturally occurring ones by
a treatment process. Those which are co-product of regular manufacturing
process . For example coke oven gas from manufacturing of coke and blast
furnace gas from making of iron, gasoline, coal gas etc.
Both primary and secondary fuels may be further classified based upon
their physical state as
[i] solid fuels ; [ii] liquid fuels ; [iii] gaseous fuels.
The quality of fuels depends upon the amount of heat liberated per unit
quantity of the fuel (calorific value) and their efficiency in combustion, the
nature of the product of combustion and safe storage
Classification of fuels

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Classifications of
Fuels
Based on Physical
State
Solid fuel
(e.g., wood,coal)
Liquid fuel (e.g., crude
petroleum, natural
gasoline)
Gaseous fuel
(e.g., natural gas;
CNG; LPG)
Based on
occurrence
Primary or natural
fuels (e.g., wood, coal)
Secondary or prepared
fuels
(e.g., charcoal,
petroleum coke).
Classification of Fuel

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 Wood
 Peat
 lignite
 Crude
oil
 Natural
gas
 Charcoal
 Coke
 LPG
 Petrol
 Biodiesel
 Bio gas
 Coke oven gas
 Coal gas
 Gasification gas
Classification of Fuel

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Fossil fuels are energy-rich substances that have formed from long-buried
plants and microorganisms.
The gasoline that fuels our cars, the coal that powers electrical plants, the
natural gas that heats our homes are all fossil fuels.
High energy density
– 73,890 BTU/ lb of Natural Gas
– 17,400,000 BTU/ton of Lignite Coal
– 138,000 BTU/gal of Fuel oil
Chemical or conventional fuels release their chemical energy by combustion reactions,
whereas nuclear fuels release staggering amounts of energy in a very short time interval,
by fission or fusion of nuclei of atoms.

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Characteristics of a Good Fuel
 Ignite easily. The temperature of the fuel at which ignition starts and
continues to burn without further addition of heat is called ignition
temperature. It should be moderate for a good fuel. Very low ignition
temperature leads to fire hazard and very high ignition temperature
delay the starting of fire.
 Give out a lot of heat, that is, its specific heat should be high.
 Have low smoke and combustible matter such as ash. It should not give
out harmful combustion products. This property depends on the nature
of elements present in thefuel.
 Inexpensive and readily available.
 Easy to store and transport.
 Have low ash content. Ash reduces the calorific value of the fuel,
causes hindrance to the flow of air and heat, reduces the specific heat
and leads to unwanted disposable problems.

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 High Calorific value.
 Moderate Ignition Temperature.
 Low Moisture Content.
 Low Non-Combustible Matter Content.
 Moderate velocityof combustion.
 Product should not be Harmful.
 Low cost.
 Easy of Transport.
 Combustion Should be Easily Controllable.
Characteristics of a Good Fuel

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The autoignition temperature or kindling point of a substance
is the lowest temperature at which it spontaneously ignites in
normal atmosphere without an external source of ignition, such as a flame
or spark.
The activation energy needed for the combustion to initiate is supplied by
increasing the temperature.
This ignition temperature at which a chemical ignites decreases as the
pressure or oxygen concentration increases.
It is usually applied to a
combustible fuel mixture.
Substances Temp
Calcium 790 °C
Diesel 210 °C
Petrol 247–280 °C
Hydrogen 536 °C
paper 218–246 °C
White
phosphorous
34 °C

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To measuring the flammability and combustibility of a substance.
(1) Flash point ; (2 ) Fire point and (3) ignition temperature.
The flash point of a volatile material is the lowest temperature at which a
substance vaporises into a gas, which can be ignited with the introduction of an
external source of fire. At this temperature the vapor may cease to burn when
the source of ignition is removed.
The fire point is the lowest temperature at which vapors of the material will burns
continuously for atleast 5 seconds after being ignited and the ignition source
removed
The autoignition temperature, which is the temperature at which the vapor
ignites spontaneously without an ignition source.
Fuel Flash point
Boiling Points at
Atmospheric
Pressure
Fire point
(10 °C greater
than Flash Point)
Autoignition
temperature
Gasoline
(petrol)
−43 °C 95°C - 33 260 °C
Diesel (2-D) >52 °C 150°C 75°C 210 °C

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Advantages of liquid fuels over solid fuels - Boiler application
Advantages :
1) Liquid fuel having higher calorific value.
2) Less space is required for storage.
3) Easy control of combustion by stopping supply of fuel.
4) It is very clean fuel, dust free.
5) Reduction in cost of handling.
6) Easily transported through pipes.
7) During burning it does not form ash.
Disadvantages :
1) Cost of liquid fuel is high.
2) The storage tank specially designed.
3) It has higher cost.
4) Danger of explosion.
5) Liquid fuels mostly we import from other countries. So we
depends on other countries.

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Advantages of gaseous fuels :
1. They are free from solid and liquid impurity.
2. Maximum complete combustion of gaseous fuel is possible.
3. The rate of combustion and temperature in the combustion
chamber can be easily controlled.
4. For complete combustion less amount of excess air is
required.
5. Do not produce ash and smoke.
6. Large amount of heat and temperature is obtained at a
moderate cost.
Disadvantages :
1) They are readily inflammable.
2) They require large storage capacity.
3) The cost of gaseous fuel are more.

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Advantages of solid fuel :
1. Solid fuel can be stored conveniently without any risk of
explosion.
2. They can easily transported.
3. They have moderate ignition temperature.
4. Leakage problem is not takes place.
Disadvantages :
1. Rate of combustion of solid fuel can’t be easily controlled.
2. Large amount of heat is wasted.
3. The ash content of solid fuel is very high.
4. The cost of handling of solid fuel is high.
5. After burning it produce large quantity of smoke.

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Merits of liquid fuels over gaseous fuels:
1. Required less space for storage.
2. Higher calorific value.
3. Easy control of consumption.
4. Easy handling & transportation.
5. Absences of danger from spontaneous combustion.

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Natural liquid fuel Artificial liquid fuel
It is obtained from
reservoirs in the earth.
It is obtained by distillation process of
crude oil
Raw material of oil
industries.
This is the final product of oil
industries
Impure form of fuel. Pure form of fuel.
It is cheap. It is costly.
Crude petroleum. Gasoline, Kerosene, Diesel,
Lubricating oil, and grease.
Solid Fuels Gaseous Fuel
Required Large space Required Large space
Low calorific value Low calorific value
For combustion more air is
required
For combustion less air is required
Produce ash & smoke after
combustion
Do not Produce ash & smoke after
combustion
Low cost High cost
Impure form Pure form

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Candle Flame
LPG and Wood

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Where does the energy comes for the combustion reactions?
The energy stored in the chemical bonds that hold the carbon and
hydrogen atoms together, releases when bonds are broken and atoms
are rearranged.
How does this energy stored in fuel?
This energy stores in fuel by the process of photosynthesis which converts
sunlight into chemical energy and storing it in the bond of sugar. Plants
needs Carbon dioxide, Water and sunlight to make sugar. The overall
reaction is:
6CO2 + 6H2O + Sunlight -----> C6H12O6 + 6O2
These plants when became dead, buried under the soil, as more and
more soil deposited over them for thousands of years, they became
compressed and then under high temperature and pressure converted
into fossil fuel.

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During the process of combustion of a fuel, the atom of carbon,
hydrogen etc. combine with oxygen with simultaneous liberation of
Heat at rapid rate. This energy is liberated due to the “rearrangement
of valence electrons” in these atoms, resulting in the formation of
Compounds like CO2 and H2O. These compounds have less energy (heat
content) as compared to reactants.
FUEL + O2 PRODUCTS + HEAT
More heat energy content Lesser heat energy content
29
C (s) + O2(g) CO2(g) + DH= -94.1Kcal/mole

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Fuels (electron donor)
Exhaust!
Oxidizers(electron acceptor)
Heat
Products
Reactants
Combustion is a chemical process in which a substance (fuel) reacts
rapidly with oxygen and gives off thermal energy - Heat
Components that exist before the
chemical reaction
Components that exist after
the chemical reaction
Rapid oxidation
generating heat
Combustion is a field which involves fluid mechanics, heat transfer, thermodynamics
and chemical kinetics. Also, knowledge of mathematics is essential
for establishing their inter-relationships

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CalorificValue
Units of Energy : Joule (J); Calorie (cal); British Thermal Unit (BTU)
A British Thermal Unit may defined as, the heat required to raise the temperature of one
pound of water from 60 F to 61 F.
Calorie, a unit of heat, may be defined as, the heat required to raise the temperature of one
gram of water from 15 C to 16 C. The unit is also called the centigrade heat unit. ( CHU)
Conversion Factors:
1 calorie = 4.186 J ; = 0.003968 BTU
1 kcal = 3.968 BTU = 1.162 Wh
1 BTU = 1055 J = 0.252 kcal = 252 cal
1 MJ = 238.89 kcal = 947.86 BTU
1 kWh = 3.6 x 106 J = 860.42 kcal = 3412 BTU = 1.341 hp
Calorific values:
1 BTU/lb = 0.5556 kcal/kg
1 kcal/kg = 1.8 BTU/lb
1 kJ/kg = 0.239 kcal/kg = 0.4299 BTU/lb
1 k.cal/m3 = 0.1077 BTU/ft3
1 BTU/ft3 = 9.3 k.cals/m3
Solid/Liquid Fuels
Gas Fuels

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Thermochemic
al calorie
≡ 4.184 J
≈ 0.003964 BTU
≈ 1.162×10−6 kWh
≈ 2.611×1019 eV
the amount of energy equal to exactly 4.184 joules
4 °C calorie ≈ 4.204 J
≈ 0.003985 BTU
≈ 1.168×10−6 kWh
≈ 2.624×1019 eV
the amount of energy required to warm one gram of air-
free water from 3.5 to 4.5 °C at standard atmospheric
pressure.
15 °C calorie ≈ 4.1855 J
≈ 0.0039671 BTU
≈ 1.1626×10−6 kWh
≈ 2.6124×1019 eV
pressure. Experimental values of this calorie ranged from
4.1852 to 4.1858 J.
20 °C calorie ≈ 4.182 J
≈ 0.003964 BTU
≈ 1.162×10−6 kWh
≈ 2.610×1019 eV
pressure.
Mean calorie ≈ 4.190 J
≈ 0.003971 BTU
≈ 1.164×10−6 kWh
≈ 2.615×1019 eV
1⁄100 of the amount of energy required to warm one gram
of air-free water from 0 to 100 °C at standard atmospheric
pressure.

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1 Watt = 1 joule per second.(Energy per second)
it mean if any working device used or convert 1 joule per second or 10 joules in 10
seconds then the power generated by that object is equals to 1 Watt.
Watt(W) - SI unit of Power
Joule(J) - SI unit of Energy
Power is the energy transferred per unit time
Watts are units of Power, whereas Joules are units of Energy.
Power is Energy in accordance to time: Power = Energy / Unit Time
1 watt = 1 joule / 1 second
W = J / s.
So One Watt of Power is equal to one Joule per second.
The unit of energy is the "Joule" - a Watt is just a flow of one Joule per second.
The other popular unit of power is the "horsepower".
The conversion is that one horsepower = 756 Watts.

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1 J/s = 1 Watt
Or otherwise 1 J = 1 watt.sec = 1 Ws
1 Watt. sec = 1 J
1kilo Watt .sec = 1000 J
1 kilo Watt.hour = 3600000 J
1kWh = 3600 kJ = 3600 kWs
1kWh = one UNIT in INDIA
One kWh represents the amount of energy needed by a 1000-Watt device to
operate continuously for one hour. The kilowatt hour (symbol kWh, kW⋅h or kW h)
is a unit of energy equal to 3.6 megajoules.
If the energy is being transmitted or used at a constant rate (power) over a period
of time, the total energy in kilowatt hours is equal to the power in kilowatts
multiplied by the time in hours.
A unit (as mentioned on the electricity bills) is represented in kWH or Kilowatt
Hour. This is the actual electricity or energy used. If you use 1000 Watts or 1
Kilowatt of power for 1 hour then you consume 1 unit or 1 Kilowatt-Hour (kWH) of
electricity.
745.6 W = 1 hp
1kW = 1000/745.6 = 1.341 hp

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Analysis:
5 kg of Solid fuel: (let 10,000 kJ/kg)
50,000kJ = kWs
Boiler
60%
Device
Steam
80%
Turbine
90%
Coal gasifier 80% Turbine
90%
Gasifier
80%
Separator
100%
Fuel Cell
95%
Power ?
Power ?
Power ?
Power ?
36000kWs
= 10 kWh
Water heater of 1000W Time ?
Kettle/cooking 250 W Time ?
Fan/Computer/Light Time ?

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Calorific values are of two types :
1. High or Gross Calorific Value.
2. Low or Net Calorific Value.
High Calorific Value is defined as “the total amount of heat
produced when one unit of the fuel has been burnt completely
and the products of combustion have been cooled to 16 C or 60
F.”
Low Calorific Value is defined as “the net heat produced when
unit mass or volume of fuel is completely burnt and products of
combustion are allowed to escape into the atmosphere.”
The calorific value of fuels is determined theoretically by Dulong
formula or I.A. Davies formula.

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Gross and net calorific Value
Gross Calorific Value: It is the total amount of heat generated when a unit
quantity of fuel is completely burnt in oxygen and the products of combustion are
cooled down to the room temperature.
As the products of combustion are cooled down to room temperature, the steam
gets condensed into water and latent heat is evolved. Thus in the determination
of gross calorific value, the latent heat also gets included in the measured heat.
Therefore, gross calorific value is also called the higher calorific value.
The calorific value which is experimentally determined by Bomb calorimeter gives
the higher calorific value (HCV)
Net Calorific Value: It is defined as the net heat produced when a unit quantity of
fuel is completely burnt and the products of combustion are allowed to escape.
When heat absorbed (or) carried away by the products of combustion is not
recovered and steam is formed during combustion is not condensed then amount of
heat obtained per Kg of fuel is known as net (or) lower calorific value.
The water vapour do not condense and escape with hot combustion gases. Hence,
lesser amount than gross calorific value is available. It is also known as lower
calorific value (LCV).
Theoretical calculation of Calorific value of a Fuel:
The calorific value of a fuel can be calculated if the percentages of the constituent
elements are known

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STP stands for Standard Temperature and Pressure.
STP is set as 0°C and 100 kPa or 1 bar.
NTP stands for Normal Temperature and Pressure.
NTP is set at 101.325 kPa [ 1atm] but uses 20°C as the
temperature.
Calorific value of fuels :
The calorific value (or) heating value of solid (or) liquid fuel may be defined as
amount of heat given out by complete combustion of 1 Kg. of fuel ( if soild); [ (or)
of 1 m3 of fuel (if Gas/liquid)]
It is expressed in Kcal/Kg. of fuel at N.T.P. in S.I. system J/Kg.
0°C = 273.15K
20°C = 293.15K
Usually fuels are compared based on the net calorific value.
Calorific values are measured in MJ/kg for solid fuels and MJ/nm3 for gases.
For liquids MJ/L is often used, even though liter is a metric (not SI) unit of volume.
In SI unit MJ/dm3 can be used, because 1dm3 is equal to one liter.

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Calorific Value is the amount of heat energy produced on complete combustion of
1kg of fuel. Hence we can use the following formula to calculate Calorific value.
Calorific Value = Heat Produced / Amount of Fuel used for burning in KJ/Kg.
For Example, If 5Kg of fuel is completely burnt and amount of heat produce d is
1,90,000 KJ, What is the calorific value?
Calorific Value = 1,90,000 KJ / 5.0 Kg
= 38,000KJ/Kg
Hence Calorific value of fuel is calculated to 38,000 KJ/Kg
The calorific value of petrol and diesel – Comparison.
The calorific value of diesel fuel is roughly 45.5 MJ/kg (megajoules per kilogram),
slightly lower than petrol which is 45.8 MJ/kg. However, diesel fuel is denser than
petrol and contains about 15% more energy by volume (roughly 36.9 MJ/litre
compared to 33.7 MJ/litre).
Calorific Value of Coal: It indicates the amount of heat that is released when the
coal is burned. The Calorific Value varies on the geographical age, formation,
ranking and location of the coal mines. It is expressed as kJ/kg in the SI unit system.
Power plant coals have a Calorific Value in the range of 9500 kJ/kg to 27000 kJ/ kg.

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According to Dulong, the calorific value of a fuel is the sum of the calorific
values of all the elements present. The calorific values of different elements
are given as under:
 Calorific Value of C = 8080 cal/g
 Calorific Value of H = 34500 cal/g
 Calorific Value of S = 2240 cal/g
Determination of Calorific Value
If oxygen is also present, it combines with hydrogen to form H2O. Thus the
hydrogen in the combined form is not available for combustion and is called fixed
hydrogen.
Amount of hydrogen available for combustion = Total mass of hydrogen-hydrogen
combined with oxygen(fixed hydrogen).
Fixed Hydrogen = Mass of oxygen in the fuel
Therefore, mass of hydrogen available for combustion
= Total mass of hydrogen-(1/8) mass of oxygen in fuel
= H-O/8

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where C, H2, O2 & S are in mass % . In this formula, oxygen is assumed to be present
in combination with hydrogen as water.
L.C.V. (N.C.V.) = H.C.V. – Latent heat of water vapour formed.
Since 1g of hydrogengives 9 g of water, this equationreduces to
NCV = GCV – (weight of hydrogenx 9 x latent heat of steam)
L.C.V. (N.C.V.) = H.C.V. – 0.09H x 587
Dulong’s formula for calculating the calorific value is given as:
Gross calorific Value (HCV)
 Calorific Value of C = 8080 cal/g
 Calorific Value of H = 34500 cal/g
 Calorific Value of S = 2240 cal/g
H.C.V. of Coal = [8080 C+34500 (H2 - O2/8) + 2240 S]/100 kcal/Kg.
L.C.V. = H.C.V. – 0.09 H2 x 587 kcal/kg
Net Calorific value (LCV)

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Theoretically Determination
Dulong’s formula for calculating the calorific composition of the fuel
may be written asfollows:
Determination of Calorific Value

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Calorific value of a fuel can be determined using bomb calorimeter. The procedure
is explained below. The crucible of the calorimeter is filled with the known mass of
fuel, and then it is ignited. This heats the surrounding water, and the initial and
final temperatures are recorded using a thermometer.
The calorific value can be determined using the heat balance.
Heat given by the fuel is equal to the heat gained by the water.
LHS: Mass of fuel × calorific value.
RHS: mass of water × specific heat ×change in temperature.
Equate LHS and RHS to get the calorific value.
Example:
Calculate the calorific value of newly invented fuel called "X".
The test recordings are given below.
Mass of surrounding water:3 kg
Mass of fuel:5 g
Rise in temprtature:5°C
Solution:
CV=(3 × 4.18 × 5)/(0.005)=12540 kJ/kg

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Experimentally Determination Bomb calorimeter
For calorific values of solid and
Non-volatile liquid fuels
A known amount of the fuel is burnt in
excess of oxygen and heat liberated is
transferred to a known amount of water.
The calorific value of the fuel is then
determined by applying the principle of
calorimeter
i.e. Heat gained = Heat lost

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Determination of Calorific value
1. Determination of calorific value of solid and non volatile liquid fuels:
It is determined by bomb calorimeter.
Bomb Calorimeter :
It is a apparatus used for finding the higher calorific value of solid and liquid fuels.
In this calorimeter, as shown in Fig. the fuel is burnt at a constant volume and
under a high pressure in a closed vessel called bomb.
Construction :
The bomb is made mainly of acid-resisting stainless steel, machined from the
solid metal, which is capable of withstanding high pressure (up to 100 bar),
heat and corrosion. The cover or head of the bomb carries the oxygen valve
for admitting oxygen and a release valve for exhaust gases. A cradle or carrier
ring, carried by the ignition rods, supports and Silica crucible, which in turn
holds the sample of fuel under test. There is an ignition wire of Tungston,
Platinum or Chrome which dips into the crucible. It is connected to a battery,
kept outside, and can be sufficiently heated by passing current through it so as
to ignite the fuel. The bomb is completely immersed in a measured quantity of
water. The heat, liberated by the combustion of fuel, is absorbed by this water,
the bomb and copper vessel. The rise in the temperature of water is measured
by a precise thermometer, known as Beckmann thermometer which reads up
to 0.01o C.

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Procedure :
A carefully weighed sample of the fuel (usually one gram or so) is placed in
the crucible. Pure oxygen is then admitted through the oxygen valve, till
pressure inside the bomb rises to 30 atmosphere. The bomb is then
completely submerged in a known quantity of water contained in a large
copper vessel. This vessel is placed within a large insulated copper vessel
shown in the figure to reduce loss of heat by radiation. When the bomb and
its contents have reached steady temperature (this temperature being
noted), fuse wire is heated up electrically. The fuel ignites, and continues to
burn till whole of it is burnt. The heat released during combustion is
absorbed by the surrounding water and the apparatus itself. The rise in
temperature of water is noted.

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Let
mf = Mass of fuel sample burnt in the bomb in Kg.
H.C.V. = Higher calorific value of the fuel sample in Kcal/Kg.
mw = Mass of water filled in the calorimeter in Kg.
me = Water equivalent of apparatus in kg.
t1 = Initial temperature of water and apparatus in oC &
t2 = Final temperature of water and apparatus in oC.
We know that heat liberated by fuel = mf x H.C.V. (i)
and heat absorbed by water and apparatus = (mw + me) x Cw x (t2 – t1) (ii)
Since the heat liberated is equal to the heat absorbed (neglecting losses),
therefore equating equations (i) and (ii),

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Theoretical (or) Minimum mass of air required for
complete combustion.
We know proper supply of oxygen is very essential for the complete
combustion of a fuel, for obtaining maximum amount of heat from a fuel. The
theoretical or minimum mass (or) volume of oxygen required for complete
combustion of 1 kg of fuel may be calculated from chemical analysis of the
fuel. The mass of oxygen, required by each of the constituents of the fuel, may
be calculated from the chemical equation.
Now consider 1 kg of a fuel.
Let, Mass of the carbon = C kg
Mass of the hydrogen = H2
Mass of sulphur = S kg
we know that
1 kg of carbon requires 8/3 kg of oxygen for its complete Combustion.
1 kg of hydrogen requires 8 kg of oxygen for its complete Combustion and
1 kg of sulphur requires 1 kg of oxygen for its complete combustion.
Total oxygen required for complete combustion of 1 kg of fuel.
= 8 C + 8 H2 + S kg -------------------1
3

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= 8 C + 8 H2 + S kg -------------------1
3
If some oxygen (say O2 kg) is already present in the fuel, then total
oxygen for complete combustion of 1 kg of fuel.
Total oxygen required for complete combustion of 1 kg of fuel.
Normally oxygen has to be obtained from atmospheric air which mainly
consist of nitrogen & oxygen along with of rare gases like argon, neon and
krypton etc. But for all calculations the compositions of air is taken as;

L.Muruganandam51
Excess air supplied :-
We know, minimum air required for complete combustion but many times for
complete combustion and rapid combustion of fuel. Some quantity of air in form
of excess is supplied.
If just minimum amount of air is supplied a part of the fuel may not burn
properly.
The amount of excess air supplied varies with the type of fuel and firing
conditions. It may approach to a value of 100 percent; but in modern days it uses
25 to 50% excess air.

L.Muruganandam52
Mass of excess air supplied :-
IMPORTANT FORMULAES :

General information
L.Muruganandam54
Bomb calorimeter
Entire unit – constant volume system. ( no volume change after the process)
Any combustion zone/chamber – always constant pressure processes
(the heat of reaction/combustion is always at atmospheric pressure ).
Dn refers to the change in the number of moles of gases during the reaction.
Calibration is done using known heat value of benzoic acid. And pre filled water
molecules.

L.Muruganandam55
Calculations ( different Unit)
Let weight of the fuel sample taken = x g
Weight of water in the calorimeter = W g
Water equivalent of the Calorimeter, stirrer, bomb,
thermometer = w g
Initial temperature of water = t1 Co
Final temperature of water = t2 Co
Higher or gross calorific value = C cal/g
Heat gained by water = W .Dt . specific heat of water
= W (t2-t1) . 1 cal
Heat gained by Calorimeter = w (t2-t1) cal
Heat liberated by the fuel = x C cal
Heat liberated by the fuel = Heat gained by water and calorimeter
x.C = (W+w) (t2-t1) cal

L.Muruganandam56
Net Calorific value:
Let percentage of hydrogen in the fuel = H
Weight of water produced from 1 gm of the fuel = 9H/100 gm
Heat liberated during condensation of steam = 0.09H x 587 cal.
Net (Lower calorific value) = GCV-Latent heat of water formed
= C- 0.09Hx 587 cal/gm
Corrections: For accurate results the following corrections are also
incorporated:
(a) Fuse wire correction: As Mg wire is used for ignition, the heat generated by
burning of Mg wire is also included in the gross calorific value. Hence this
amount of heat has to be subtracted from the total value.
(b) Acid Correction: During combustion, sulphur and nitrogen present in the fuel
are oxidized to their corresponding acids under high pressure and temperature.

L.Muruganandam57
The corrections must be made for the heat liberated in the bomb by the
formation of H2SO4 and HNO3. The amount of H2SO4 and HNO3 is analyzed by
washings of the calorimeter.
For each ml of 0.1 N H2SO4 formed, 3.6 calories should be subtracted.
For each ml of 0.01 HNO3 formed, 1.43 calories must be subtracted.
(C) Cooling correction: As the temperature rises above the room temperature,
the loss of heat doesoccur due to radiation, and the highest temperature
recorded will be slightly less than that obtained. A temperature correction is
therefore necessary to get the correct rise in temperature.
If the time taken for the water in the calorimeter to cool down from the maximum
temperature attained, to the room temperature is x minutes and the rate of
cooling is dt/min, then the cooling correction = x .dt. This should be added to the
observed rise in temperature.
Therefore, Gross calorific value
C = (W+w)(t2-t1+Cooling correction)-[Acid+ fuse corrections]
Mass of the fuel

L.Muruganandam59
Boy’s calorimeter for Gas and volatile liquid fuels
1. Gas or volatile liquid burns at constant rate.
2. Water flowing at constant rate absorbs the heat produced.
3. Calorific value is calculated from volume of water, increase in
temperature and volume of gas/liquidburnt.

L.Muruganandam60
Junker’s calorimeter
1. Control of rate of burning of
gaseous/liquid fuel and water
circulation is maintained.
2. The combustible products are
released at nearly the
atmospheric pressure.
3. Calorific value is calculated
from amount of water passed,
volume of gas burnt, the
steady rise in temperature and
mass of the condensed water
flowing out.

L.Muruganandam62
Calculations
Calculation of theoretical air for combustion of a fuel requires the
following points:
1. Percentage of oxygen in air by volume is 21% and 23.2% by weight.
2. Stoichiometric equations involved in combustion
Combustion

L.Muruganandam63
Flue Gas Analysis
It comprises the gaseous products of combustion of fuel. Its
analysis helps in finding out the correct quantity of air to be
supplied in a furnace.
Orsat’s apparatus

Tutorials
L.Muruganandam65
Ex.1 A sample of coal has following composition on mass basis Carbon 82%,
Hydrogen 8%, Sulphur 2%, Oxygen 4% and Ash 4%.
Calculate using Dulong’s formula higher and lower calorific value of fuel.
Soln : Given
Composition of coal on mass basis.
composition Wt %
Carbon (C) 82
Hydrogen (H2) 8
Sulphur (S) 2
Oxygen (O2) 4
Ash 4
Total 100
We know Dulong’s formula.
H.C.V. of Coal = (8080 C+34500 (H2 - O2/8) + 2240 S)/100 kcal/Kg.
Putting above values in formula.
= [8080 x 82 + 34500 (8 – 4/8) + 2240 x 2]/100
H.C.V. of coal = 9257.9 kcal/kg = 38735 KJ/Kg.
L.C.V. = H.C.V. – 0.09 H2 x 587
= 9257.9 – 0.09 x (8) x 587
= 8835 kcal/kg
L.C.V. = 36966.73 KJ/Kg.

L.Muruganandam66
Ex.2 A sample of coal has the following composition by mass, carbon 76%,
Hydrogen 5%, Oxygen 8.5%, Nitrogen 2%, Sulphur 1.5% and Ash 7%
calculate higher and lower calorific value of fuel per Kg.
Soln : Given
Composition of coal by mass.
composition Wt %
Carbon (C) 76
Hydrogen (H2) 5
Oxygen (O2) 8.5
Nitrogen (N2) 2
Sulphur (S) 1.5
Ash 7
Dulong’s formula.
1) H.C.V. of Coal = (8080 C+34500 (H2 - O2/8) + 2240 S)/100
kcal/Kg
= 7532.84 kcal/kg
= 31517.4 KJ/Kg.
2) L.C.V. = H.C.V. – 0.09 H2 x 587
= 7268.68 kcal/kg
= 30412 KJ/Kg.

L.Muruganandam67
Ex.3 A sample of coal has the following composition by mass Carbon 75%,
Hydrogen 6%, Oxygen 8%, Nitrogen 2.5%, Sulphur 1.5% and Ash 7%
calculate higher and lower calorific value of per Kg.
composition Wt fraction
Carbon (C) 0.75
Hydrogen (H2) 0.06
Oxygen (O2) 0.08
Nitrogen (N2) 0.025
Sulphur (S) 0.015
Ash 0.07
Dulong’s formula.
1) H.C.V. of Coal = 78186. kcal/kg
H.C.V. = 32713 KJ/Kg.
2) L.C.V. of Coal = 7501 kcal/kg
L.C.V. = 31386 KJ/Kg.
L.C.V. = H.C.V. – 0.09 H2 x 587 kcal/kg

L.Muruganandam68
Ex.4 A sample of coal has the following composition by mass Carbon 60%,
Hydrogen 10%, Oxygen 15%, Nitrogen 4.5%, Sulphur 3.5% and
remaining is ash calculate H.C.V. and L.C.V. of per Kg.
1) H.C.V. of Coal = 7729 kcal/kg
2) L.C.V. of Coal = 7201 kcal/kg
Ex 5. A coal has the following composition by mass Carbon 80%, Hydrogen 5%,
Oxygen 6%, Nitrogen 2.5%, Sulphur 1.5% and Ash 5%. Calculate HCV and
LCV per kg of coal.
L.C.V. = H.C.V. – 0.09 H2 x 587 kcal/kg

L.Muruganandam71
TYPE 2 :
Ex.1 The following is the percentage composition of a sample of coal on mass basis.
C = 85, H2 = 4, O2 = 10 and remaining is ash find minimum mass of air required
for complete combustion of 1 Kg. of coal.
Given, Composition of coal on mass basis composition Wt fraction
Carbon (C) 0.85
Hydrogen (H2) 0.04
Oxygen (O2) 0.10
Minimum mass of air required for complete combustion of
1 Kg. of fuel.
= 100/23 (2.67 C + 8 H2 + S – O2) Kg.
= 100/23 (2.67 x 0.85 + 8 + 0.04 + 0 – 0.10)
= 100/23 (2.2695 + 0.32 – 0.1)
= 40.82 Kg. per Kg. of Coal burnt.
Ex.2 The following is the percentage composition of coal on mass basis.
C = 80, H2 = 3.3, O2 = 4 and S = 0.9 and remaining is ash. Calculated
theoretical air required to 1 Kg. of coal completely.
Minimum mass of air required for complete combustion of 1
Kg. of fuel.
= 100/23 (2.67 C + 8 H2 + S – O2) Kg.
= 100/23 (2.67 x 0.80 + 8 x 0.033 + 0.009 – 0.004)
= 100/23 (2.136 + 0.264+ 0.009 – 0.004)

L.Muruganandam72
Ex.3 During a boiler trial the coal analysis on mass basis was reported as C
= 62.4%, H2 = 4.2%, O2 = 4.5%, Moisture = 15% and Ash
13.9%. Calculated minimum air required to burn 1 Kg. of coal also
calculate H.C.V. & L.C.V.
Given : Composition of coal on mass basis
composition Wt
Percentage
Wt
Fraction
Carbon (C) 62.4 0.624
Hydrogen (H2) 4.2% 0.042
Oxygen (O2) 4.2% 0.045
Moisture 15% 0.15
Ash 13.9 0.139
Minimum mass of air required for complete
combustion of 1 Kg. of fuel.
= 100/23 (2.67 C + 8 H2 + S – O2) Kg.
= 100/23 (2.67 x 0.624 + 8 x 0.042 + 0 – 0.044)
= 100/23 (2.136 + 0.264 + 0.009 – 0.004)
= 100/23 (1.666 + 0.336 – 0.045)
We know Dulong’s formula.
H.C.V. of Coal = 6585 kcal/kg
H.C.V. = 27551 KJ/Kg.
L.C.V. of Coal = 6363 kcal/kg
= 26623 KJ/Kg
L.C.V. = H.C.V. – 0.09 H2 x 587 kcal/kg

L.Muruganandam74
Sample Question bank
How fuels are classified ?
Define Calorific value of the fuel.
Define L.C.V. & its unit
What is H.C.V. & L.C.V. ?
Define fuel and state the type of fuel.
Enlist any four types of gaseous fuels.
List the properties of fuel.
List out the merit of liquid fuel over gaseous fuels.
State requirement of good fuel.
Differentiate between Natural and Artificial liquid fuel.
State & explain Dulong’s formula for theoretical determination of calorific value of fuel.
Give the significance of ultimate analysis of fuel. How is % of carbon & hydrogen
determined in this analysis.
Define calorific value of fuel. Differentiate between H.C.V. and L.C.V. of the fuel. State which
value is used in calculation and why?
Explain Ultimate analysis and proximate analysis of coal Explain H.C.V. and L.C.V. of the
fuels.
Describe with neat sketch construction and working of Bomb calorimeter. Write Dulong’s
formula and state it’s use.
Compare i) Solid fuel and Gaseous fuel ii) Ultimate analysis and proximate analysis.

L.Muruganandam
Rank of Coal: It denote the maturity of coal. So peat the most immature coal has
a lowest rank while the anthracite the most mature coal has highest rank.
Metamorphism of coal: The process of conversion of lignite to anthracite is
called metamorphism of coal or coalification.
Carbonization of coal: Heating of coal in absence of air at high temperature to
produce coke, tar and gases is called carbonization of coal.
Gasification of coal: Heating of coal in insufficiently less amount of air plus
steam to produce a gas rich in CO and H2 is called gasification of coal.
Proximate analysis of coal: Finding out the weight percent of moisture, volatile
matter, fixed carbon and ash content in coal. The analysis is useful in deciding
the utilization for a particular purpose.
Ultimate analysis: Finding out the weight percent of carbon, hydrogen, nitrogen,
oxygen and sulphur of pure coal free from moisture and inorganic constituents.
The analysis is useful in designing of coal burning equipments.
Fundamentals / Definitions

L.Muruganandam
Calorific value: The quantity of heat liberated by combustion of unit quantity of
fuel is called its calorific value.
Gross calorific value: Where the heat obtained from condensation of water
vapours in the flue gases is also include.
Net calorific value: Where the heat obtained from condensation of water
vapours in the flue gases is not include.
Flue gas: The gaseous product of combustion of a fuel.
Heat capacity: Amount of heat required to raise the unit weight of substance by
one degree.
Specific heat ratior: It is the ratio of heat capacity of a substance to the heat
capacity of water (Cp > Cv).
Ignition temperature: It is the minimum temperature at which the fuel ignites.
Flash point: It is the minimum temperature at which the fuel give enough
vapours which produces a momentary flash when exposes to flame.
Pour point: It is the minimum temperature at which fuel keeps its flowing nature
when cooled under specific conditions.

L.Muruganandam78
General:
Heat capacity of one gram of water = 1 cal/(g.K)
= 4.184 Joules/(g.K)
= 4.184 Joules/(g.C)
Heat capacity of one mole of water = 4.184* 18
= 75.312 Joules /( gmole.K)

2 module 1-21-dec-2018 reference material i-2 fac m1_intro_cv

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