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Fuels and Combustion CHAPTER – 4 FUELS AND COMBUSTION 4.1 Introduction
4.2 Requirements of a Good Fuel
Chapter · October 2010
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Fuels and Combustion
4.1
CHAPTER – 4
FUELS AND COMBUSTION
4.1 Introduction
Fuel is a combustible substance, containing carbon as a
main constituent, which on proper burning gives large amount
of heat, which can be used economically for domestic and
industrial purpose.
Example :
Wood, charcoal, coal, kerosene, petrol, diesel, producer gas, oil
gas etc.
During the process of combustion, carbon, hydrogen, etc.,
combine with oxygen with a liberation of heat.
The combustion reaction can be explained as
C + O2
2H2 + O2
CO2 + 94 kcals
2H2O + 68.5 kcals
The calorific value of a fuel depends mainly on the amount of
Carbon and Hydrogen.
4.2 Requirements of a Good Fuel
A good fuel should have the following characteristics:
o
High calorific value.
o
Moderate ignition temperature.
4.2
Applied Chemistry
o
Low contents of non-combustible matters.
o
Low moisture content.
o
Free from objectionable and harmful gases like CO, SOx,
H2S.
o
Moderate velocity of combustion.
o
Combustion should be controllable.
o
Easy to transport and readily available at low cost.
4.3 Classification of Fuels
Fuels are classified into
(i) Primary or Natural fuels - These are found in nature.
(ii) Secondary or Artificial fuels - These are derived from
primary fuels.
Primary and secondary fuels may also be divided into 3
classes namely solid, liquid and gaseous fuels.
Fuels
Primary or
Natural Fuels
Solid Fuels
Ex.: Wood
Secondary or
Artificial Fuels
Liquid Fuels
Ex.: Oil.
Solid Fuels
Ex.: Charcoal
Gaseous Fuels
Ex.: Natural gas
Liquid Fuels
Ex.: Petrol
Gaseous Fuels
Ex.: LPG
Fuels and Combustion
4.3
Fossil fuels
Fossil fuels are those, which have been derived from
fossil remains of plant and animal life.
They are found in the earth’s crust.
All conventional fossil fuels whether solid, liquid or
gaseous (coal, petroleum or Natural gas) contain basically carbon
and / or hydrogen. The fuels on combustion in presence of
oxygen in the air release heat energy.
This heat energy can be utilized for domestic and
industrial purposes.
Advantages of Solid fuels
1. Handling and transportation of solid fuels are easy.
2. Solid fuels are cheap and easily available.
3. They have a moderate ignition temperature
4. This type of fuel can be stored conveniently without any
risk.
Disadvantage of solid fuels:
1. During burning, solid fuels produce a large amount of
ash and disposal of ash is a big problem.
2. The calorific value of solid fuel is comparatively low.
3. Since a lot of air is required for complete combustion, its
thermal efficiency is not so high.
4. A large space is required for storage.
5. Combustion is a slow process and it cannot be easily
controlled.
Advantages of Liquid fuels
1. Liquid fuels do not yield any ash after burning.
2. They require comparatively less storage space.
4.4
Applied Chemistry
3. Calorific value of liquid fuel is higher than that of solid
fuels.
4. Their combustion is uniform and easily controllable.
Disadvantages of liquid fuels:
1. Liquid fuels are comparatively costlier than the solid
fuels.
2. They give
combustion.
unpleasant
odour
during
incomplete
3. Some amount of liquid fuels will escape due to
evaporation during storage.
4. Special type of burners are for effective combustion.
Advantages of gaseous fuels:
1. Gaseous fuels can be easily transported
pipes.
through the
2. They do not produce any ash or smoke during burning.
3. They have high calorific value than the solid fuels.
4. They have high thermal efficiency.
Disadvantages of gaseous fuels
1. They are highly inflammable and hence the chances for
fire hazards are high.
2. Since gases occupy a large volume, they require large
storage tanks.
4.3.1 Solid Fuels
1.Coal and its varieties (or) Ranking of Coal
Coal is an important primary solid fuel that has been
formed as a result of alteration of vegetable matter under
some favourable conditions.
Fuels and Combustion
4.5
The process of conversion of lignite to anthracite is
called coalification (or) metamorphism of coal.
Coal is classified on the basis of its rank. The rank of coal
denotes its degree of maturity. Vegetable matter, under the action
of pressure, heat and anaerobic conditions, gets converted into
different stages of coal namely,
Wood → Peat → lignite → sub-bituminous → bituminous
coal
coal
↓
anthracite
With the progress of coal forming reaction, moisture content and
oxygen content reduces and % of carbon increases. Also calorific
value increases from peat to bituminous. (Table 1.1).
Classification of coal
a) Peat
1. Peat is the first stage in the formation of coal.
2. Its calorific value is about 4000-5400 k cal/kg.
3. It is an uneconomical fuel due to its high proportion of
(80 -90%) moisture and lower calorific value.
4. It is a brown fibrous mass.
b) Lignite
1. Lignite is an intermediate stage in the process of coal
formation.
2. Its calorific value is about 6500-7100 k cal/kg
3. Due to the presence of high volatile content, it burns with
long smoky flame.
4.6
Applied Chemistry
Table 4.1
4.1 Classification of solid fuels & its calorific values
Table
Fuel
Peat
Lignite
Subbituminous
coal
Bituminous
coal
Anthracite
Nature
Highly
fibrous light
brown in
colour
Fibrous,
brown
coloured coal
Black
coloured,
homogenious
smooth mass,
Black, brittle,
burns with
yellow
smoky flame
Hard & most
matured coal,
burns without
smoke
Calorific
value
k.cals/kg
4000 5400
Compo
sition
%
C = 57
H = 06
O = 35
Domestic
fuel, power
generation
6500 7100
C = 67
H = 05
O = 26
C = 77
H = 05
O = 16
Manufacture
of producer
gas & steam
Manufacture
of gaseous
fuels
8000 8500
C = 83
H = 05
O = 10
8500 8700
C = 93
H = 03
O = 03
Power
generation,
coke making,
domestic fuel
Boiler
heating,
metallurgical
furnace
7000 7500
Uses
Here peat is the most immatured coal, hence it is lowest
in rank where as anthracite is the most matured coal, and hence it
is highest in rank
c) Bituminous coal
Bituminous coal is further sub-classified on the basis of
its carbon content into three types as: i) Sub- bituminous coal, ii)
Bituminous coal and iii) semi-bituminous coal.
d) Anthracite
1. Anthracite is the superior grade of coal.
2. Its volatile, moisture and ash contents are very less.
3. Its calorific value is about 8650 k cal/kg.
Fuels and Combustion
4.7
4.4 Analysis of Coal
In order to asses the quality of coal, the following two
types of analysis are made.
I. Proximate Analysis
o It means finding out weight percentage of moisture, volatile
matter, fixed carbon and ash in coal
o This analysis gives the approximate composition of the main
constituents of coal.
o It is useful in deciding its utilization for a particular industrial
use.
Determination of moisture content in coal
About 1 gm of powdered, air dried coal sample is heated
in silica crucible at 100 to 105 °C for one hour. Percentage of
moisture can be calculated from the loss in weight of the coal
sample as
Loss in weight of coal
∴ % of moisture in coal =
×100
Weight of coal initially taken
Determination of Volatile Matter (V.M.) in coal
After the analysis of moisture content the crucible with
residual coal sample is covered with a lid, and it is heated at 950
± 20 °C for 7.0 minutes in a muffle furnace. Percentage of
volatile matter can be calculated from the loss in weight of the
coal sample as
∴ % of volatile matter in coal
Loss in weight of moisture free coal
=
×100
Weight of coal initially taken
4.8
Applied Chemistry
Determination of ash in coal
After the analysis of volatile matter the crucible with
residual coal sample is heated without lid at 700 ± 50 °C for 30
minutes in a muffle furnace.
Percentage of ash content can be calculated from the loss
in weight of the coal sample as
∴ % of ash in coal =
Weight of ash formed
×100
Weight of coal taken
Determination of fixed carbon
It is determined by subtracting the sum of total moisture,
volatile and ash contents from 100.
% of fixed carbon = 100 - % of [moisture + V.M + ash]
Significance (or) Importance of Proximate Analysis
Moisture
High moisture content in coal is undesirable because it,
i)
Reduces Calorific Value of coal
ii)
Increases the consumption of coal for heating purpose
iii) Lengthens the time of heating.
Hence, lesser the moisture content, better is the quality of coal.
Volatile Matter
During burning of coal, certain gases like CO, CO 2, CH4,
H2, N2, O2, hydrocarbons etc. that come out are called volatile
matter of the coal.
Fuels and Combustion
4.9
The coal with higher volatile content,
o Ignites easily (i.e : it has lower ignition temperature)
o Burns with long yellow smoky flame
o Has lower Calorific Value
o Will give more quantity of coal gas when it is heated in
absence of air.
Ash
o
Ash is the combustion product of mineral matters present in
the coal. It consists mainly of SiO2, Al2O3 and Fe2O3 with
varying amounts of other oxides such as Na2O, CaO, MgO
etc.
o
Ash containing oxides of Na, Ca and Mg melt early. (Low
melting ash). During coke manufacture, the low melting ash
forms a fused lumpy - expanded mass which block the
interspaces of the ‘grate’ and thereby obstructing the supply
of air leading to irregular burning of coal and loss of fuel.
o
High ash content in coal is undesirable because it (a)
increases transporting, handling, storage costs, (b) is harder
and stronger, (c) has lower Caloific Value.
Fixed Carbon
It is the pure carbon present in coal. Higher the fixed
carbon content of the coal, higher will be its Caorific Value.
II. Ultimate Analysis
o
It means finding out the weight percentage of carbon,
hydrogen, nitrogen, oxygen and sulphur of the pure coal
free from moisture and inorganic constituents.
o
This analysis gives the elementary constituents of coal.
o
It is useful to the designer of coal burning equipments and
auxiliaries.
4.10
Applied Chemistry
Determination of carbon and hydrogen in coal
A known amount of coal is burnt in presence of oxygen
there by converting carbon and hydrogen of coal into CO2
(C + O2 → CO2) and H2O (H2 + ½ O2 → H2O) respectively. The
products of combustion (CO2 and H2O) are made to pass over
weighed tubes of anhydrous CaCl2 and KOH, which absorb H2O
and CO2 respectively.
The increase in the weight of CaCl2 tube represents the
weight of water formed while increase in the weight of KOH tube
represents the weight of CO2 formed. % of carbon and hydrogen
in coal can be calculated as follows.
Let
X - the weight of coal sample taken
Y - the increase in the weight of KOH tube
Z - the increase in the weight of CaCl2 tube
a) Carbon
C + O2 → CO2
12
32
44
44 gms of CO2 contain 12 gms of carbon.
12
1 gm of CO2 contains
gms of carbon
44
12
Y gm of CO2 contains =
× Y gms of carbon
44
% of C in coal =
12
Y
×
× 100
44
X
b) Hydrogen
H2 + ½ O2
2
16
→
H2O
18
18 gms of water contains 2 gms of hydrogen.
Fuels and Combustion
4.11
1 gm of water contains 2 / 18 gms of hydrogen.
2
× Z gms of Hydrogen.
∴ Z gms of water contains =
18
% of hydrogen in coal
=
2
Z
×
× 100
18
X
Determination of Nitrogen in coal
Nitrogen estimation is done by Kjeldahl’s method. A
known amount of powdered coal is heated with con. H2SO4 and
K2SO4 in a long necked flask (called Kjeldahl’s flask), there by
converting nitrogen of coal to ammonium sulphate.
When the clear solution is obtained (i.e. the whole
nitrogen is converted into ammonium sulphate), it is heated with
50% NaOH solution.
(NH4)2 SO4 + 2NaOH
Na2SO4 + 2NH3
The ammonia thus formed is distilled over and is
absorbed in a known quantity of standard 0.1N HCl solution. The
volume of unused 0.1 N HCl is then determined by titrating
against standard NaOH solution.
Thus, the amount of acid neutralized by liberated
ammonia from coal is determined.
Let,
Volume of 0.1N HCl
Volume of unused HCl
Acid neutralised by ammonia
We know that 1000 ml of 1 N HCl
(A - B) ml of 0.1N HCl
= A ml
= B ml
= (A - B ) ml
= 1 mole of HCl
= 1 mole of NH3
= 14 gms of N2
14 ×( A B ) × 0.1
=
gms of N2
1000 ×1N
4.12
Applied Chemistry
X gms of coal sample contains =
14 x ( A B) x 0.1
gms of N2
1000 x 1N
% of Nitrogen
14 x Volume of Acid consumed x Normality
x 100%
1000 x Weight of coal sample(X)
=
=
1.4 x Volume of Acid consumed x Normality
%
Weight of coal sample(X)
Determination of Sulphur in coal
A known amount of coal is burnt completely in bomb
calorimeter in presence of oxygen. Ash thus obtained contains
sulphur of coal as sulphate, which is extracted with dil HCl. The
acid extract is then treated with BaCl2 solution to precipitate
sulphate as BaSO4.
The precipitate is filtered, washed, dried, and weighed,
from which the sulphur in coal can be computed as follows.
Let,
X = weight of coal sample taken
M = weight of BaSO4 precipitate formed.
S + 2O2
SO4
32
BaSO4
233
233 gms of BaSO4 contains 32 gms of sulphur
1 gm of BaSO4 contains 32 / 233 gms of sulphur
∴ M gms of BaSO4 contains (32 / 233) x M gms of sulphur
% of sulphur in coal =
32
M
x
x 100
233
X
Fuels and Combustion
4.13
Significance (or) Importance of Ultimate Analysis
i) Carbon and Hydrogen
1. Higher the % of carbon and hydrogen, better the quality
of coal and higher is its calorific value.
2. The % of carbon is helpful in the classification of coal.
3. Higher the % of carbon in coal reduces the size of
combustion chamber required.
ii) Nitrogen
1. Nitrogen does not have any calorific value, and its
presence in coal is undesirable.
2. Good quality coal should have very little nitrogen content.
iii) Sulphur
Though sulphur increases the calorific value, its presence in coal
is undesirable because
1. The combustion products of sulphur, i.e, SO2 and SO3 are
harmful and have corrosion effects on equipments.
2. The coal containing sulphur is not suitable for the
preparation of metallurgical coke as it affects the
properties of the metal.
iv) Oxygen
1. Lower the % of oxygen higher is its calorific value.
2. As the oxygen content increases its moisture holding
capacity increases and the calorific value of the fuel is
required.
4.14
Applied Chemistry
Carbonisation of coal
Heating of coal in absence of air at high temperature to
produce a residue coke, tar and coal gas is called as
carbonisation.
i. Caking of coal
When coal is heated strongly, the mass becomes soft and
coherent, then it is called caking of coal.
ii. Coking of coal
Otherwise if the mass produced is hard, porous and
strong then it is called coking of coal.
All the caking coals do not form strong, hard and coherent
residue coke. Hence all the caking coals are not necessarily
coking coal but all the coking coals have to be necessarily caking
in nature.
4.5 Metallurgical Coke
When bituminous coal is heated strongly in absence of
air, the volatile matter escapes out and a lustrous, dense,
strong, porous and coherent mass is left, which is called
metallurgical coke.
Properties or Characteristics of Metallurgical Coke
The most important industrial use of coke is in the
metallurgical industry, especially in the blast furnace. Good coke
for metallurgy should possess the following requirements.
i) Purity
Low moisture and ash content are desirable in
metallurgical coke. It must contain minimum % of P and S.
Fuels and Combustion
4.15
ii) Porosity
High porosity is desirable in furnace cokes to obtain high
rates of combustion.
iii) Strength
The coke should be hard and strong to withstand pressure
of ore, flux etc in the furnace.
iv) Size
Metallurgical coke must be uniform and medium size.
v) Calorific value
The Calorific Value of coke should be high.
vi) Combustibility
It should burn easily.
vii) Reactivity
It refers to its ability to react with O2, CO2, steam and air.
The metallurgical coke must have low reactivity.
viii) Cost
It must be cheap and readily available.
Why is Coke superior as a Metallurgical fuel?
i)
ii)
iii)
Coke is stronger and more porous than coal.
Coke contains lesser amount of sulphur than coal.
Coke does not contain much volatile matter.
4.16
Applied Chemistry
4.5.1 Manufacture of Metallurgical Coke
Mainly two types of ovens are used for metallurgical coke
production. They are
i. Beehive Coke Oven Process.
ii. Otto Hoffman’s by-product method.
i) Beehive Coke Oven Process
The oven is a dome shaped structure and made up of
firebricks. There are two openings, one at the top and the other at
the side. Coal is charged through the top circular opening
whereas coke is removed through the side door after
carbonization. The side door also acts as air inlet.
Its typical dimensions are, height = 2 m; base dia = 3.5 m
and capacity = 5 tons of coal. (Fig .4.1)
First some air is supplied in to the oven to ignite the
coal. The volatile matter present in coal escapes and burns inside.
Therefore heat for carbonization is supplied by burning of
volatile matter and partly by coal itself.
Charging door
Refractory lining
2m
Coal
3.5m
Door for air supply
and coke discharging
Fig
Fig 4.1
4.2 Beehive coke oven
Carbonization proceeds, from top to bottom. It takes
about 3 to 4 days to complete. The hot coke is then removed
through the side door and quenched with water.
Fuels and Combustion
4.17
Demerits
Since volatile matter present in coal escapes into
atmosphere as waste, we cannot recover any byproduct. Large
coking time and coke yield of only 60% are other demerits.
ii) Otto Hoffman’s by-product Coke Oven
In order to increase the thermal efficiency of the
carbonization process and recover the valuable by products (like
coal gas, ammonia, benzene, etc.) Otto Hoffman developed
modern by-product coke oven.
The oven consists of a number of silica chambers. Each
chamber is about 10 - 12 m long, 3 - 4 m height and 0.4 - 0.45 m
wide. Each chamber is provided with a charging hole at the top, it
is also provided with a gas off take valve and iron door at each
end for discharging coke.
A, B, C, D - Coke Ovens
1,2,3,4 - Heat re-generators
To Chimney
Coal gas
To Chimney
Producer gas
Air
4.2 Otto Hoffman’s by-product coke oven
Fig 4.3
Coal is introduced in to the silica chamber and the
chambers are closed. The chambers are heated upto 1200°C by
burning pre heated air and the producer gas mixture in the
interspaces between the chambers.
4.18
Applied Chemistry
The air and gas are preheated by sending them through 2 nd
and 3 hot regenerators. Hot flue gases produced during
carbonization are allowed to pass through 1st and 4th regenerators
are heated by hot flue gases, the 2 nd and 3rd regenerators are used
for heating the incoming air and gas mixture.
rd
For economical heating, the direction of inlet and flue
gases are changed frequently. The above system of recycling the
flue gases to produce heat energy is known as the regenerative
system of heat economy. When the process is complete, the coke
is removed and quenched with water.
Time taken for complete carbonisation is about 12 - 20
hours. The yield of coke is about 70 %.
The valuable by products like coal gas, tar, ammonia, H2S
and benzene, etc are removed from the flue gas.
Recovery of by products
i) Tar
The coke oven gas is first passed through a tower in
which liquor ammonia is sprayed. Tar and dust get collected in a
tank below, which is heated by a steam coil to recover back the
ammonia sprayed.
ii) Ammonia
The gas is then passed through the other tower where
water is sprayed. Ammonia gets converted to NH4OH.
iii) Benzene and other aromatic compounds
The gas is then passed through the next tower in which
creosite oil is sprayed. Benzene and other aromatic compounds
are dissolved in the oil and recovered.
Fuels and Combustion
4.19
iv) Hydrogen sulphide
The gas then enters into a purifying chamber packed with
Fe2O3, which removes any sulphur compound present in coal gas.
Advantages of Otto Hoffman’s process
o
High thermal efficiency and carbonization time is less.
o
Valuable by products (like coal gas, ammonia, benzene,
etc. are recovered as by products.
o
Heating done externally by producer gas.
4.6 Liquid Fuels
Petroleum
Petroleum or crude oil is a naturally occurring brown to
black coloured viscous oil formed under the crust of earth, on
shore or off shore. Chemically it is a mixture of various
hydrocarbons with small amounts of N, O, S compounds.
The approximate composition of petroleum is
C = 80 - 84%
H = 10 - 14 %
S = 0.1 - 0.5 %
N+ O = Negligible
Classification
Petroleum is classified on the basis of various types of
hydrocarbons.
i)
Paraffin based oil - Contains mainly n - alkanes
(Ex : Pennsylvanian and gulf coast oil)
4.20
ii)
iii)
Applied Chemistry
Asphalt base oil - Contains aromatic and alicyclic
hydrocarbons.
(Ex: Californian oil)
Mixed base oil - Contains higher proportion of aromatics
and naphthenes (cyclo alkanes).
(Ex : Mexican oil)
4.6.1 Refining of Petroleum (or) Crude Oil
Definition
The process of removing impurities and separating out
the oil into various fractions having different boiling points is
known as refining of petroleum.
Uncondensed gases
Loose cap
Petroleum ether
Gasoline
Chimney
Naptha
Tray
Kerosene
Crude oil
Diesel oil
Lubricating oil
Heavy oil
Furnace at 400° C
Fractionating Column
Fig
Fig.4.3
4.4 Fractional distillation of crude oil
Fuels and Combustion
4.21
i) Removal of Impurities
The impurities present in the oil are the fine water
droplets, NaCl, MgCl2, Sulphur etc. The crude oil is an extremely
stable emulsion of oil and salt water. Water is separated from the
oil by Cottrell’s process using ring electrodes. In this method, the
crude oil is allowed to flow between two highly charged
electrodes. The colloidal water droplets combine to form large
drops, which are then separated from oil.
Modern techniques like electrical desalting are used to
remove NaCl and MgCl2 from oil.
Sulphur is removed by treating the oil with copper oxide
and separated by filtration.
ii) Fractional Distillation
The purified crude oil is heated in a furnace called oil
heater where the temperature will be around 400o C. Here the oil
gets vapourised. The hot vapours are then sent to the
fractionating column (Fig. 1.3 ).
It is a tall cylindrical tower consisting of a number of
horizontal stainless steel tray at short distances. Each tray is
provided with a small chimney, which is covered with a loose
cap. The tower will be hot at the lower end and comparatively
cooler at the upper end.
When the oil vapours go up in the tower, they become
cool and condense. The heavier compounds having higher boiling
points get cooled first and condensed in the trays whereas the
fractions having lower boiling points condense near the top of the
tower.
4.22
Applied Chemistry
Lower fractions are used after purification while the high
boiling point fractions are subjected to cracking operation to get
more useful lower fractions.
The gasoline obtained by this fractional distillation is called
straight-run gasoline. Various fractions obtained at different trays
are given in Table 1.2.
Table 4.2 Various fractions, Compositions and their
uses
Sl.No
1.
2.
3.
4.
Name of the Boiling
fraction
Range oC
Uncondensed
Below 30
gases
Petroleum ether 30-70
Gasoline
or 40-120
petrol
Naphtha
or 120-180
solvent spirit
5.
Kerosene oil
180-250
6.
Diesel oil
250-320
7.
Heavy oil
320-400
Range of Uses
C-Atoms
C1-C4
As a fuel
under the
name
of
LPG
C5-C7
As a solvent
C5-C9
Fuel for IC
engines
C9-C10
As a solvent
in
paints
and in dry
cleaning
C10-C16
Fuel
for
stoves and
jet engines
C15-C18
Diesel
engine fuel
C17-C30
Fuel
for
ships and
for
production
of gasoline
by cracking.
Fuels and Combustion
4.23
Heavy oils on refraction gives
S.No
1.
2.
3.
4.
Name of the Fraction
Lubricating oil
Petroleum jelly or
Vaseline
Grease
Paraffin wax
5.
Pitch at above 4000C
Uses
as lubricants
Used in medicines and
cosmetics
Used as lubricant
Used in candles, boot polishes
etc.,
Used for making roads, water
proof roofing etc.
Some important fractions of petroleum
i) Petrol (or) Gasoline (C5-C9)
o
It is a low boiling fraction of petroleum obtained
between 40 - 120o C.
o It is a mixture of hydrocarbons pentane to nonane (in
terms of carbon atoms C5 - C9).
o
Its calorific value is about 11,250 kcals/kg.
o It is used as fuel in ICE of automobiles and aero planes.
o Its antiknock value can be improved by the addition of
Tetra Ethyl Lead (TEL).
Uses : It is used as a fuel in IC engine and also used in dry
cleaning and as a solvent.
ii) Naphtha (C9-C10)
It is a colourless, light fraction obtained between
120 C to 1800C during fractional distillation of petroleum. It
is a mixture of hydrocarbons such as nonane and decane.
Uses
Naphtha is also called as white spirit, which is generally
used in dry cleaning and as thinner for varnish, floor and
0
4.24
Applied Chemistry
furniture polishes etc. The lightest portion of the distillate is used
as solvent for fats and rubbers, whereas the heaviest portion of
the same is used as a fuel.
iii) Kerosene( C10-C16)
It is relatively a high boiling fraction obtained between
180-250oC during fractional distillation of petroleum. It is a
misture of hydrocarbons such as decane to hexadecane.s
approximate composition is C = 84%, H = 16%, > 0.1% S. Its
calorific value is about 11,100 kcal/kg.
Uses:
It is mainly used as a domestic fuel in stoves and lamps. It
is also used as jet engine fuel and for making oil gas.
iv) Diesel (C15-C18)
It is also a high boiling fraction obtained between 250320oC during fractional distillation of petroleum. It is a mixture
of hydrocarbons such as C15H32 to C18H38. Its calorific value is
about 11,000 kcal/kg.
Uses
It is used as a very good diesel engine fuel.
v) Heavy oil or Residual fuel oil (C17 – C30)
The left over portion of petroleum after distilling off all
the lighter fractions are called Fuel Oil. The approximate
composition of fuel oil is C = 86%, H = 12%, S = 1%, H2O =
0.6%; sediments = 0.4%. Its calorific value is about 9200 kcal/kg.
The following fractions are obtained on further vacuum
distillation of the fuel oil.
i) Light fuel oil = 350 -420oC
ii) Heavy neutral oil = 420-500oC
Fuels and Combustion
4.25
Uses:
It is used as fuel for ships and also used in metallurgical
furnaces. Gasoline is also obtained from oil by cracking process.
vi) Asphalt
Asphalts are obtained by
i) The oxidation of residual heavy oil in presence of air at higher
temperature.
ii) The deep vacuum distillation of residual heavy oil.
Asphalts are available in the market in liquid, semi-solid
and solid forms.
Uses:
It is used for road making and making water-proofing
roofs. It is also used for the manufacture of water proofing
concrete and water proofing paints.
4.7 Cracking
Cracking is defined as “the decomposition of high
boiling hydrocarbons of high molecular weight into smaller,
low boiling hydrocarbons of low molecular weight”
C10H22
C5H12
Decane
n-Pentane
B.Pt : 174°C
+
C5H10
Pentene
B.Pt : 36°C
The crude oil on fractional distillation yields only about
15 - 20 % gasoline. This is known as Straight Run Gasoline.
The quality of straight run gasoline is not so good. It
contains mainly straight chain paraffin’s, which ignite readily and
more rapidly than any other hydrocarbons and hence it produces
knocking (unwanted sound) in IC engines.
4.26
Applied Chemistry
In order to overcome these difficulties and also to
improve the quality of gasoline, high boiling fractions are
cracked into more valuable low boiling fractions suitable for SI
engines. The gasoline obtained by cracking is called Cracked
Gasoline.
During cracking
o Straight chain alkanes are converted into branched chain
hydrocarbons.
o Saturated hydrocarbons are converted into mixture of
Saturated and Unsaturated hydrocarbons.
o Aliphatic alkanes are converted into cyclic alkanes.
o All hydrocarbons obtained by cracking have lower
boiling point than the parent hydrocarbons.
Types of cracking
There are two kinds of cracking
1. Thermal cracking
2. Catalytic cracking
1. Thermal Cracking
When cracking is carried out a higher temperature and
pressure without any catalyst, it is called Thermal cracking.
There are two types of thermal cracking:
i) Liquid phase Thermal cracking
In this method, the heavy oil is cracked at a temperature
of 475-5300C under high pressure of 100 kg/cm2 to keep the
reaction product in liquid state. The cracked products are then
separated into various fractions in a fractionating column. The
yield of gasoline is about 50-60% and the octane number is
65-70.
Fuels and Combustion
4.27
ii) Vapour phase thermal cracking
In this method, the heavy oil is first vapourised and then
cracked at a temperature of 600-6500C under a lower pressure of
10-20 kg/cm2. The yield of gasoline is about 70%. This process is
suitable only for those oils which are readily vapourised.
Catalytic Cracking
When cracking is carried out at lower temperature and
pressure in the presence of suitable catalyst, it is called Catalytic
Cracking. The Catalyst used are aluminium silicate or alumina.
There are two types of catalytic cracking.
1. Fixed bed catalytic cracking
The heavy oil vapour is heated to 420-450oC in a
preheated chamber. The hot vapours are then passed through a
catalytic chamber, maintained at 425-450oC and 1.5 kg/cm2
pressure, where catalyst ( artificial clay mixed with zirconium
oxide), are kept in fixed beds. During this passage through the
catalytic chamber about 40% of the heavy oil is converted into
gasoline and about 2-4% carbon is formed.
The carbon gets adsorbed on the catalyst bed. The cracked
vapours are then passed through the fractionating column, where
heavy oil gets condensed at the bottom. The vapours of gasoline
are then sent through the cooler where gasoline gets condensed
along with some gases. The gasoline containing some dissolved
gases is then sent to a stabilizer, where the dissolved gases are
removed and pure gasoline is recovered.
After 8-10 hours, the catalyst loses its activity due to the
deposition of carbon. It is reactivated by burning off the
deposited carbon.
4.28
Applied Chemistry
Fig 4.4 Fixed bed catalytic cracking
2. Moving bed or Fluid bed catalytic cracking
In this process, the solid catalyst is finely powdered, so
that it behaves as a fluid, which can be circulated in oil vapour
(Fig 1.5) .
The heavy oil vapour is heated to 420-450oC in a
preheater and it is mixed with the catalyst powder. Then this
mixture is forced into the reactor, which is maintained at a
temperature of 500oC and a pressure of 5 kg/cm2, where cracking
takes place. Near the top of the reactor, there is a centrifugal
separator ( called cyclone), which allows only the cracked oil
vapours to pass on to the fractionating column leaving behind the
catalyst powder in the reactor itself.
The catalyst powder gradually becomes heavier, due to
coating of carbon and it settles down at the bottom of the reactor.
Then it is forced into the regenerator maintained at 600 oC, where
carbon is burnt and the regenerated catalyst is again recirculated
along the heavy oil vapour.
Fuels and Combustion
4.29
Fig 4.5 Moving bed catalytic cracking
From the reactor, the cracked oil vapours are passed into
the fractionating column, where heavy oil settles down and the
vapours are then passed through the cooler where gasoline
condenses along with some gases. The dissolved gases are
separated from gasoline by passing it through a stabilizer.
4.8 Manufacture of Synthetic Petrol
Petrol can be synthesized by any one of the following
methods.
I. Polymerization
a. Thermal Polymerization
b. Catalytic Polymerization
II. Hydrogenation of Coal
a. Bergius Process (or) Direct Process.
b. Fisher Tropsch Process (or) Indirect Process.
III. Alkylation
4.30
Applied Chemistry
I. Polymerisation
The gases produced in cracking contain C3 and C4 olefins
(iso propylene, iso butylene etc) and alkanes (methane, ethane,
propane, butane). These gases undergo polymerisation in
presence of catalyst, (H3 PO4) at suitable temperature and pressure
to give gasoline (Polymer petrol), rich in branched chain
hydrocarbons.
Hence, polymerisation is mainly for the production of
superior gasoline and is complementary to catalytic cracking.
Polymerisation is of two types
i) Thermal Polymerisation
Polymerisation of cracked gases is carried out at 500 600 C and 70 - 350 kg/cm2 pressure. The product is the gasoline
and gas oil mixture, from which gasoline is separated by
fractionation.
o
ii) Catalytic Polymerisation
This process is carried out in presence of catalyst like
H3PO4 . By this method isobutylene can be polymerised to give
higher olefin’s which is hydrogenated to gasoline hydrocarbons.
II. Hydrogenation of Coal
Coal contains 4.5 % of Hydrogen, where as petrolium
contains 18 % of hydrogen. So coal is a hydrogen deficient
compound, if coal is heated with hydrogen at high temperature
and high pressure, it is converted in to gasoline. The preparation
of liquid fuels from solid coal is called Hydrogenation of coal.
Petrol can be synthesised by destructive hydrogenation of
coal (Bergius process) and liquification of coal (Fischer Tropsch process).
Fuels and Combustion
4.31
i) Bergius Process (or) Direct Process
The raw materials used in this process are coal dust,
heavy oil and nickel oleate or tin oleate.
A coal paste is prepared by mixing coal dust with heavy
oil and catalyst. It is then pumped into the converter where the
paste is heated to 450 - 500o C under 200 - 250 atm in presence of
hydrogen. (Fig. 4.6)
Cooler
Powdered coal
Catalyst
Gases
Gasoline
Paste
Middle
oil
Heavy oil
H2
Heavy oil
Converter
Fractionating Column
Fig
Fig.4.6
1.5 Bergius Process
The reaction products mainly contain mixture of hydrocarbons.
Coal dust
suspended in
heavy oil
+ H2
450 °C
Mixture of hydrocarbons
200-250 atm
Condensation
Crude oil
4.32
Applied Chemistry
Since the reaction is exothermic the vapours leaving the
converter are condensed in the condenser to give synthetic
petroleum or crude oil.
The oil is then fractionally distilled to give
(i) Petrol
(ii) Middle oil
(iii) Heavy oil.
Middle oil is again hydrogenated in presence of solid
catalysts to produce more amount of gasoline. Heavy oil is used
for making paste with fresh coal dust. Yield is about 60 %.
II Fischer - Tropsch Process
The raw materials used in this process are hard coke,
steam to produce water gas i.e., water gas is obtained, by passing
steam over red hot coke.
H2
Water gas
Cooler
Gases
Fe2O3
Gasoline
Middle
oil
Heavy oil
Fe2O3+
Na2CO3
Compressor Converter
Cracking
Fractionating Column
Fig. 4.7 Fisher Tropsch Process
Fuels and Combustion
4.33
C + H2O
1200oC
CO + H2
Water gas
The first step in this process is purification of gas. To
remove H2S, the gas is passed through Fe2O3 and to remove
organic sulphur compounds, the gas is again passed through a
mixture of Fe2O3 and Na2CO3.( Fig. 4.7 )
The purified gas is compressed to 5 - 25 atm. and passed
over a catalyst bed containing oxides of Th, Co and Mg at 200 300o C. While passing the purified gas through this catalyst bed, it
is converted to straight chain paraffin and olefins.
n CO + (2n + 1) H2
n CO + 2n H2
CnH2n+2 + n H2O
n – paraffins
CnH2n + n H2O
olefins
Since the reactions are exothermic, the vapours leaving
the vessel are condensed in the condenser to give petroleum. It is
fractionally distilled to yield petrol and heavy oil.
III. Alkylation
Replacement of hydrogen atom from a hydrocarbon by an
alkyl group is known as Alkylation.
Example:
Anhydrous
H3 C
CH CH 3
+
H2C C CH3
CH3
CH 3
Iso butane
Iso butene
HF
H3C
H3 C
CH CH2
H 3C
C
CH3
Iso octane
CH 3
4.34
Applied Chemistry
The reaction of isobutane and isobutene in the presence of
anhydrous HF at room temperature gives isooctane. This process
is highly useful for the production of high quality petrol.
Table 4.3 Comparison of straight run, cracked and polymer
gasoline
Straight run
petrol
1. Obtained from
straight distillation
of crude oil.
2.Contains only
n-alkanes.
Cracked petrol
Polymer petrol
Obtained
heavy oil.
from Obtained from low
molecular
weight
gaseous fraction.
Contains
Contains
more
of
isoparaffins and branched chain hydroaromatics.
carbons
Narrow range
Very narrow range.
3.Composition
range C5 – C9
4. Low octane Higher
number
number
octane Higher octane number
Purification of Petrol (or) Gasoline
Gasoline or petrol obtained either from crude oil or
synthetic process may contain some undesirable impurities. They
are mainly unsaturated olefins, colouring matters, sulphur
compounds etc.
The unsaturated olefins get oxidised and polymerised
there by causing gum and sludge formation on storing
On the other hand, sulphur compounds lead to corrosion
of ICE and also affect tetra ethyl lead (TEL) which is added to
gasoline to obtain good petrol. So these undesirable contents
must be removed from gasoline. Unsaturated olefins and
colouring matter are removed by using adsorbent like Kieselguhr,
fuller’s earth.
Fuels and Combustion
4.35
Sulphur containing petrol is known as sour spirit.
The process of desulphurisation of petrol is called
sweetening of petrol.
Desulphurisation
It can be done by Doctor’s treatment. i.e.; petrol is treated
with an alkaline solution of sodium plumbite (Doctor’s solution)
and little of sulphur.
2RHS + Na2PbO2
Sodium Plumbite
(R-S-S-R) Pb + 2NaOH
Lead mercaptide
The resulting disulphide ( lead mercaptide) in gasoline are
extracted by using suitable solvent. After refining of gasoline,
some inhibitors or antioxidants are added in order to retard the
oxidation of olefins (olefin peroxide) which cause the formation
of gum on storage.
4.9 Knocking
Fractions like petrol and diesel oil are used as engine
fuels. Piston engines can be divided into spark ignition (SI) and
compression ignition (CI) engines. The former consumes petrol
and the latter operates on diesel oil.
SI Engines
In a four stroke SI engine, petrol vapour is mixed with air
in the carburetor. It is sucked into the cylinder during the suction
stroke. The mixture is compressed by the piston in the
compression part of the cycle. Then the compressed mixture is
ignited by an electric spark. The product of combustion increases
pressure and pushes the piston out, providing an output of power.
4.36
Applied Chemistry
In the last part of the cycle, the piston ascends and expels the
exhaust gases from the cylinder.
Knocking in SI Engines (Petrol Engines)
Normally the fuel - air mixture should burn smoothly and
rapidly by sparking. In some cases, as a result of compression,
the fuel-air mixture may get heated to a temperature greater than
its ignition temperature and spontaneous combustion occurs even
before sparking. This is called pre-ignition.
Further, the spark also is emitted which makes the
combustion of the rest of the mixture faster and explosive. So, we
have a sudden, badly controlled burning and explosion results a
characteristic metallic or rattling sound from the engine. This is
called knocking or detonation or pinking. Knocking lowers the
efficiency of engine which results in loss of energy.
Chemical Structure and Knocking
The knocking tendency decreases as follows :
n-alkanes
isoparaffins
olefins
naphthenes
aromatics
n-alkanes have lowest antiknock value. So the presence of
maximum quantity of aromatics and minimum quantity of
n-alkanes is desirable in petrol.
Octane number (Measurement of knocking in SI engines)
Octane number expresses the knocking characteristics of
petrol. n - heptane (a constituent of petrol) knocks very badly, so
its anti-knock value has been given zero. On the other hand, isooctane (also a constituent of petrol) gives very little knocking, so
its anti-knock value has been given 100.
Fuels and Combustion
4.37
Percentage of iso-octane present in iso octane &
n-heptane mixture, which matches the same knocking
characteristics of gasoline mixture test sample.
If a petrol sample behaves like a mixture of 60%
iso-octane and 40% n-heptane, its octane number is taken as 60.
Leaded Petrol (or) Improvement of Anti-knock Value
Adding some additives in it increases octane number of
petrol. In motor fuel about 1.0 to 1.5 ml tetra ethyl lead (TEL) is
added per litre of petrol. Petrol to which TEL is added is called
leaded petrol.
Mechanism of knocking
Knocking follows free radical mechanism, leading to a
chain growth. If the chains are terminated before their growth,
knocking will cease. TEL decomposes thermally to form ethyl
free radicals, which combines with the free radicals of knocking
process and thus the chain growth is stopped.
Disadvantage of using TEL
TEL forms lead oxide, which deposits on spark plug and
creates problems. So, to remove it, ethylene dibromide is added.
During burning lead bromide is formed which evaporates away in
the heat engines and goes out together with exhaust gases. This
creates atmospheric pollution for human beings. Hence, at
present aromatic phosphates are used instead of TEL.
CH2 - Br
PbBr2 + CH2 = CH2
Pb +
CH2 - Br
4.38
Applied Chemistry
4.10 Diesel
o
It is relatively a high boiling point fraction of petroleum
obtained between 250 - 320o C.
o It is a mixture of hydrocarbons in terms of carbon atoms
C15 - C18
o Its calorific value is about 11,000 kcals/kg. It is used as fuel
for compression ignition engine.
o Its antiknock value can be improved by doping with isoamyl
nitrate.
CI Engines
In a CI engine, air is alone compressed. This raises the
cylinder temperature as high as 300o C. Then the oil is injected or
sprayed, which must ignite spontaneously. Now combustion
products expand and power stroke begins.
Knocking in CI Engines
Some times, even after the compression stroke is over and
even after the diesel oil is sprayed, burning may not start. So,
more and more fuel is injected automatically and sudden ignition
may occur and burn the whole of the oil. This delayed ignition
results an uncontrolled, excessive combustion produces ‘diesel
knock’.
So in SI - engine, knocking is due to premature or too
early ignition in CI - engines, knocking is due to delayed ignition
or ignition lag.
Cetane number (or) Cetane Rating
Cetane number expresses the knocking characteristics of
diesel.
Cetane (C16 H 34) has a very short ignition delay and hence
its cetane number is taken as 100. On the other hand, - methyl
Fuels and Combustion
4.39
napthalene has very large ignition delay and hence its cetane
number is taken as zero.
Cetane number is defined as “the percentage of cetane
present in a mixture of α –methyl naphthalene and cetane”.
CH3
CH3 - (CH2)14 - CH3
n-cetane
(cetane number = 100)
n-Hexadecane
Chemical Structure and Knocking
The knocking tendency in CI - engines increases as follows,
n-alkane
iso paraffins
Olefins
naphthenes
aromatics
Octane numbers decreases in this order. Therefore an oil
of high octane number has a low cetane number and vice-versa.
Consequently, petroleum crude gives petrol of high octane
number and diesel of low cetane number.
Diesel Index
The quality of a diesel oil is indicated by diesel index
number using the following formula
Diesel index number
Specific gravity (API) x Aniline point in o F
100
Aniline point and specific gravity is noted from API (American
Petroleum Institute) Scale.
=
4.40
Applied Chemistry
Table 4.4 : Comparison between Petrol and Diesel in Internal
Combustion Engines (ICE)
S.No
PETROL
DIESEL
1.
Low
boiling
petroleum
(C5 – C10)
Fuel for SI engine
Knocks due to
premature ignition
Knocking tendency
measured in octane
rating
Anti-knocking is
improved
through
addition of TEL
Lower thermal
Efficiency
More consumption
High boiling fraction of
petroleum containsC15– C18
hydrocarbons
Fuel for CI engine
Knocks due to ignition delay
2.
3.
4.
5.
6.
7.
Knocking tendency measured
in cetane rating.
Anti-knocking is improved by
doping with isoamyl nitrate
Higher thermal efficiency
Less consumption
Improvement of cetane number
Centane number can be increased by adding additives
called dopes.
E.g.: ethyl nitrate, isoamyl nitrate
4.11 Aviation gasoline
Low molecular weight petroleum fraction, which is used
as a fuel in aeroplanes is known as Aviation Gasoline. It
condenses at 30o -120o C during fractional distillation of crude
oil. Its octane number is 100 or even more.
Aviation gasoline may also be produced by the
polymerisation , alkylation, isomerisation and reforming process.
Fuels and Combustion
4.41
Generally aviation gasoline is made up of higher perecntage of
isoparaffins and smaller percentage of naphthalenes an aromatics.
Aviation gasoline possesses very good stability, higher
volatility and antiknock value than ordinary gasoline. And also in
some cases TEL is mixed with avaiation gasoline.
4.12 Natural Gas
Natural gas is always found above the oil in the oil wells.
It is also called Marsh gas. It conssists of methane and other
saturated hydrocarbons. The average composition of natural gas
is as follows :
Its calorific value varies from 12,000 to 14,000 kcal/m3.
Constituents
Methane
Ethane
Propane
Butane
Pentane
Percentage(%)
88.5
5.5
4
1.5
0.5
If natural gas contains lower hydrocarbons like methane
and ethane it is called lean or dry gas. In the natural gas contains
higher hydrocarbons like propane, butane along with methane it
is called rich or wet gas.
Uses
1. It is used as a domestic and industrial fuel.
2. It is used as a raw material for the manufacture of carbon
black and hydrogen.
3. It is also used for the generation of electricity by using it in
fuel cells.
4.42
Applied Chemistry
4.13 WATER GAS (or) BLUE GAS.
Water gas essentially a mixture of combustible gases like
CO, H2 with a little non combustible gases like CO 2 and N2
Calorific Value
Its Calorific Value is 2800 kcals/m3.
Composition
The average composition is
H2 = 48 %
CO = 44 %
CO2, N2 and CH4 = rest
Manufacture
A Water gas generator is a steel cylindrical vessel. At the
top, it is provided with a hopper for adding coke. Water gas outlet
is provided near the top. At the bottom, it is provided with an
arrangement of taking out ash formed (Fig. 1.8 ).
It is obtained by the action of steam on a bed of coke
heated to 1000o C.
C + H2O + 28 k.cal
CO +
H2
Since the above reaction is endothermic, the coal cools
down after a few minutes and the reaction proceeds in a different
way to form CO2 and H2 instead of water gas (CO + H2).
C+ 2 H2O
CO2 + 2H2 - 19 kcals
Fuels and Combustion
Cup and cone
feeder
4.43
Coke
Water gas
outlet
Red hot coke
at 900-1000oC
Grate
Air supply
Steam
supply
Refractory
brick lining
Ash outlet
Ash
Fig
Fig4.8
: 7 Water gas production
In order to avoid the above undesirable reaction, the blow
of air replaces the blow of steam. The following reaction now
occurs.
C + O2
CO2 + 97 kcals
2C + O2
2CO + 59 kcals
Due to exothermic reactions, the temperature of the bed
rises and when the temperature increases to 1000° C, air entry is
stopped and steam is again passed. Thus steam and air are blown
alternatively. Therefore, the manufacture of water gas is
intermittent.
Properties
It has high Calorific Value & burns with non- luminous
flame. Its flame is short but very hot.
Uses
i) It is used for the manufacture of ammonia by Haber’s process.
ii) Carbureted water gas (water gas + oil gas) is used for lighting
4.44
Applied Chemistry
and heating purposes.
iii) Water gas is also used for welding purposes.
iv) Used for the manufacture of methyl alcohol.
v) Used for the manufacture of synthetic petrol.
4.14 Producer Gas
It is prepared by passing air mixed with a little steam over
a red hot coal (or) a coke bed maintained at above 1100 ° C in a
special reactor called ‘‘ Gas Producer’’.
Composition
It is a mixture of CO and N2 .
Its average composition is
N2 = 50 %
CO = 30 %
H2 = 10 %
CO2 and CH4 = rest.
Calorific Value
Its Calorific Value is 1800 kcals/m3 .
Manufacture
The furnace used for the manufacture of producer gas is
known as producer. It consists of large airtight mild steel
cylindrical tower, lined inside with refractory bricks. At the
bottom, it is provided with pipe for blowing air and an
arrangement for removing ash. Coal is added through a hopper at
the top and producer gas comes out from an exit near the top.
a. Combustion zone
Fuels and Combustion
4.45
When a mixture of air and little steam is passed through a
bed of red hot coal, carbon (of the coal) combines with oxygen
(of the air) in the lower part of the furnace to form CO2
C + O2
Distillation zone
Reduction zone
Combustion zone
Air mixed
with a little
steam
CO2 + 97 kcals
Coke
Cup and cone
feeder
Producer gas outlet
Refractory
brick lining
CoalO
1000C
Exit for
ash
Fig 4.8
4.9 Gas Producer
Fig
b. Reduction zone
CO2 so formed rises up through the red-hot coal and gets
reduce to CO during its passage
CO2 + C
2CO - 39 kcals
N2 of air remain unaffected throughout the process. Thus
a mixture of CO and N2 with traces of CO2 and hydrocarbons
comes out through the exit at the upper end of the producer.
Properties
i) It is a poisonous gas.
ii) Insoluble in water.
iii) Heavier than air.
4.46
Applied Chemistry
Uses
i.
It is used as fuel for heating open - hearth furnaces (used in
steel and glass manufacture), muffle furnaces (used in coal
gas and coke manufacture).
ii. It provides reducing atmosphere in certain metallurgical
operations.
4.15 Liquified Petroleum Gas (LPG)
It is a mixture of propane and butane.
It is obtained as one of the top fractions in the fractional
distillation of petroleum. It is easily liquified and so can be
economically stored and transported in cylinders.
Composition
Its approximate composition is
n-Butane
= 70 %
Isobutane
= 17 %
n-Propane
= 11 %
Butylene and Ethane = rest.
Calorific Value
Its Calorific Value is 27,000 kcals/m3.
Uses
i.
It is used as a fuel for domestic cooking.
ii. Used for heating industrial furnaces.
iii. Used as an alternate for Gasoline in automobiles.
Fuels and Combustion
4.47
4.16 Biogas
These gases generally produced by the fermentation of
bio wastes, sewage wastes etc., by anaerobic bacteria.
For example, natural gas is a Biogas, which results after a
long periodic decay of animal and vegetable matters burried
inside the earth.
The cheapest and easily obtainable biogas is Gobar gas,
which is produced by anaerobic fermentation of cattle dung. The
biogas is burnt to raise steam, which can drive turbines to
produce electricity.
Steel gas holder
Inlet
tank
Ground
level
Dung +
water
slurry
Gas outlet
to kitchen
Outlet tank
for manure
10 cm pipe
10 cm pipe
Digestion
well
Masonry
work
Fig.4.10
4.9 Gobar gas plant
Fig
1.16.1 Gobar Gas
i)
It is essentially methane.
4.48
Applied Chemistry
ii)
iii)
Its Calorific Value is 5400 kcals/m3.
Its average composition is
CH4
CO2
H2
= 60 %
= 30 %
= 10 %
It is obtained by fermentation of Gobar (dung) in absence
of air. In a typical Gobar gas plant, the dung in the form of slurry
is introduced into a brick - lined well called fermentation well.
An inverted drum is placed air tightly on the well. It acts
as gasholder and it can be moved up and down with the help of
pulleys (Fig. 1.10).
Formation of gas starts in a week. The optimum
temperature for this fermentation is 34 - 48o C. As the gas starts
collecting in the drum, the drum begins to rise and float. The gas
is taken out from the exit provided at the top of the drum. It is
used as a domestic fuel in villages. It is used for domestic heating
and small pump running.
4.17 Combustion
Combustion is an exothermic chemical reaction, which is
accompanied by development of heat and light at a rapid rate,
temperature rises considerably. For example, combustion of
carbon in oxygen:
C(s) +O2(s)
CO2 (g) + 97 kcal
For proper combustion, the substance must be brought to
its kindling (or) ignition temperature, which may be defined as
Fuels and Combustion
4.49
the minimum temperature at which the substance ignites and
burns without further addition of heat from outside.
Factors affecting the rate of combustion
The rate of combustion depends on the following factors:
1. The concentration of the fuel and air.
2. The nature of the combustable substance
3. The temperature
4. With increase in pressure or surface area of the fuel the
rate of combustion can be increased.
5. It increases with in increase in pressure of air.
6. It Increases with preheating of fuel and air.
4.18 Calorific Value
It is the most important characteristic property of any fuel.
Calorific value may be defined as “the amount of heat liberated
by the complete combustion of a unit mass of a fuel”.
The quantity of heat can be measured by the following
units.
i. Calorie
ii. Kilocalorie
iii. British thermal units
iv. Centigrade heat units
Calorie
The amount of the heat required to raise the temperature
of 1gm of water through 1OC (15 to 16 OC)
HIGHER AND LOWER CALORIFIC VALUE
i) Gross (or) High Calorific Value (GCV or HCV)
4.50
Applied Chemistry
The total heat generated when a unit quantity of fuel is
completely burnt and the products of combustion are cooled to
room temperature.
For example, when a fuel containing hydrogen is burnt, it
under goes combustion and will be converted to steam. If the
combustion product is cooled to room temperature, the steam gets
condensed into water and the latent heat is evolved. Therefore
the latent heat of combustion of condensation of ‘steam’ so
liberated is included in gross calorific value.
Dulong’s formula (Theoretical calculation)
Dulong’s formula for the theoretical calculation of
calorific value is
1
O
GCV (or) HCV =
(8080 C + 34500 [ H
] + 2240 S ) kcal/kg
100
8
where, C, H, O & S represent the % of the corresponding
elements in the fuel.
It is based on the assumption that the calorific values of
C, H & S are found to be 8080, 34500 and 2240 kcal, when 1 kg
of the fuel is burnt completely. However, all the oxygen in the
fuel is assumed to be present in combination with hydrogen in the
ratio H : O as 1 : 8 by weight. So the surplus hydrogen available
O
for combustion is H
.
8
ii) Net (or) Lower Calorific Value (NCV or LCV)
The net heat produced when a unit quantity of fuel is
completely burnt and the products of combustion are allowed to
escape.
NCV = GCV – Latent heat of condensation of steam produced
1 part by weight of H2 produces 9 parts by weight of H2O as
follows.
The latent heat of steam is 587 cal/gm.
Fuels and Combustion
H2 +
2gms
1
4.51
½ O2
16gms
8
H2O
18gms
9
Thus,
9
H ×587 kcal/kg
100
NCV = GCV – 0.09 H ×587 kcal/kg
NCV = GCV –
where ,
H = % of H2 in the fuel.
4.19 Determination of Calorific value using
Bomb Calorimeter
The calorific value of a solid or liquid fuel can be
determined by using bomb calorimeter ( Fig 4.11 ).
4.52
Applied Chemistry
Fig 4.11 Bomb Calorimeter
Fuels and Combustion
4.53
4.54
Applied Chemistry
Fuels and Combustion
ii) Acid correction
4.55
4.56
Applied Chemistry
Problems based on Calorific Value
1. Calculate the Gross and Net calorific values of a coal having
the following compositions, C = 80 %, H2 =08 %, O2 = 08 %, S
= 2 % and ash=2. Latent heat of steam is = 587 cal/gm.
Solution
(i)Gross Calorific Value (GCV)
1
O
(8080 C + 34500 [ H
] + 2240 S )
100
8
1
8
(8080 × 80 + 34500 [8 – ] + 2240 × 2 ) kcal/kg
=
100
8
1
(646400 + 241500 + 4480 ) kcal/kg
=
100
GCV =
Fuels and Combustion
=
4.57
1
(892380 ) kcal/kg
100
= 8923.8 kcal / kg.
(ii) Net Calorific Value (NCV)
9
H ×587 kcal/kg
100
9
= 8923.8 –
×8×587 kcal/kg
100
= 8923.8 – 422.64 kcal/kg
= 8501.16 kcal / kg
= GCV –
2. Calculate the Gross and Net calorific values of a coal having
the following compositions, C = 63 %, H2 = 19 %, O2 = 03 %,
S = 13 % and ash=2. Latent heat of steam is = 587 cal/gm.
Solution
(i) Gross Calorific Value (GCV)
1
O
(8080 C + 34500 [ H
] + 2240 S ) kcal/kg
100
8
1
3
(8080 × 63 + 34500 [19 ] + 2240 ×13 ) kcal/kg
=
100
8
1
(509040 + 64562 + 29120 ) kcal/kg
=
100
1
(1180722 ) kcal/kg
=
100
GCV =
= 11807.22 kcal / kg.
(ii) Net Calorific Value (NCV)
4.58
Applied Chemistry
9
H ×587 kcal/kg
100
9
= 11807.22 –
× 19 ×587 kcal/kg
100
= 11807.22 – 1003.77 kcal/kg
= 10803.45 kcal / kg
= GCV –
3. Calculate the Gross and Net calorific values of a solid fuel
having 80% of carbon & 20% of hydrogen. Latent heat of steam
is = 587 cal/gm.
Solution
(i) Gross Calorific Value (GCV)
1
O
GCV =
(8080 C + 34500 [ H
] + 2240 S ) kcal/kg
100
8
Here, the % of H2 and S are Zero.
1
0
=
(8080 × 80 + 34500 [20 ] + 2240 ×0 ) kcal/kg
100
8
1
[646400 + 690000 ] kcal/kg
=
100
1
[1336400 ] kcal/kg
=
100
= 13364 kcal / kg.
(ii) Net Calorific Value (NCV)
9
H ×587 kcal/kg
= GCV –
100
9
= 13364 –
× 20 ×587 kcal/kg
100
= 13364 – 1056.6 kcal/kg
= 12307.4 kcal / kg
Fuels and Combustion
4.59
4. A coal sample on analysis gives C = 75%,H2= 6 %,O2 = 3.5 %
S = 03 % and the rest ash. Calculate the Gross and Net calorific
values of the fuel. Latent heat of steam is = 587 cal/gm
Solution
(i) Gross Calorific Value (GCV)
GCV =
1
O
(8080 C + 34500 [ H
] + 2240 S ) kcal/kg
100
8
1
3.5
(8080 × 75 + 34500 [6
] + 2240 ×3 ) kcal/kg
100
8
1
=
[606000 + 191906 + 6720 ] kcal/kg
100
1
=
[804626 ] kcal/kg
100
=
= 80462.6 kcal / kg.
(ii) Net Calorific Value (NCV)
9
H ×587 kcal/kg
= GCV –
100
9
= 80462.6 –
× 6 ×587 kcal/kg
100
= 80462.6 – 316.98 kcal/kg
= 80145.62 kcal / kg
5. On analysis, a coal sample has the following composition by
weight; C = 75 %, O2 = 04 %, S = 05 %, and ash = 3%. Net
calorific value of the fuel is 9797.71kcal / kg. Calculate the
percentage of hydrogen and gross calorific value of coal.
Solution
4.60
Applied Chemistry
(i) Gross Calorific Value (GCV)
We know that,
GCV = [NCV + 0.09H x 587 ] kcal / kg
= [ 9797.71 + 0.09H + 587 ] kcal / kg
= [ 9797.71 + 52.8 H ] kcal / kg ………….(1)
1
O
GCV =
(8080 C + 34500 [ H
] + 2240 S ) kcal/kg
100
8
1
4
(8080 × 75 + 34500 [H ] + 2240 ×5 ) kcal/kg
100
8
1
=
[606000 + 34500 H 17250 + 11200 ] kcal/kg
100
= [6060 + 345H – 172.5 + 112] kcal / kg
=
=
5999.5 + 345 H kcal / kg
…………..(2)
Equation (2) is substituted in equation (1)
9797.71 + 52.8 H = 5999.5 + 345 H
9797.71 - 5999.5 = 345 H - 52.8 H
3798.21 = 292.2 H
H=
3798 .21
292 .2
% of H2 = 12.99 (i.e 13 % )
Substituting the value of H2 in the GCV equation
1
4
(8080 × 75 + 34500 [13 ] + 2240 ×5 ) kcal/kg
GCV =
100
8
Fuels and Combustion
4.61
1
[606000 + 431250 + 11200 ] kcal/kg
100
1
=
[1048450 ] kcal/kg
100
GCV =
= 10484.5 kcal / kg.
4.20 Theoretical Calculation of Minimum Air
required for Combustion
In order of achieve efficient combustion of fuel, it is
essential that the fuel is brought into intimate contact with
sufficient quantity of air to burn all the combustible matter under
appropriate conditions.
The correct conditions are
i). Intimate mixing of air with combustible matter and
ii). Sufficient time to allow the combustion process to be
completed.
If these factors are inappropriate, inefficient combustion occurs.
The elements usually present in common fuels which
enter into the process of combustion are mainly C, H, S and O.
Nitrogen, ash and CO2 (if any) present in the fuel are
incombustible matters and hence they do not take any oxygen
during combustion.
Air contains 21% oxygen by volume and 23% of oxygen
by weight.
4.62
Applied Chemistry
Hence from the amount of oxygen required by the fuel,
the amount of air can be calculated.
From the combustion reaction equations, we can calculate
the quantity of oxygen by weight or volume and from this, the
weight or volume of air required can be calculated.
For example,
i) Combustion of Carbon
C + O2
12
32
CO2
44 (by weight)
12 parts by weight of carbon require 32 parts by weight of
oxygen for complete combustion.
(or)
1 part by volume of carbon requires 1 part by volume of oxygen
for complete combustion.
C parts by weight of carbon require =
32C
parts by weight of O2
12
ii) Combustion of Hydrogen
Oxygen when present in the fuel is always in combination
with hydrogen. So, the quantity of hydrogen in combination with
oxygen, which is present in the fuel, will not take part in the
combustion reaction. Therefore, the quantity of hydrogen in
combination with oxygen is deduced from the total hydrogen in
the fuel.
Now, the quantity of hydrogen available for combustion
O
reaction will be, H
where H is the total quantity of hydrogen
8
and O is the total quantity of oxygen in the fuel. (In water the
quantity of hydrogen in combination with oxygen is one-eighth
of the weight of oxygen).
Fuels and Combustion
4.63
2H2 + O2
2 x 2 32
2H2O
36 (by weight)
4 parts by weight of hydrogen require 32 parts by weight of
oxygen for complete combustion.
(or)
2 parts by volume of hydrogen require 1 part by volume of
oxygen for complete combustion.
∴ H parts by weight of hydrogen requires
32H
=
parts by weight of O2
4
But, some of the hydrogen is present in the combined
form with oxygen (i.e, as H2O). This combined hydrogen does
not take part in the combustion reaction. Therefore, the quantity
of combined hydrogen must be deduced from the total hydrogen
in the fuel.
O
part by weight of hydrogen requires
∴ H
8
O
[ H ] ×32
O
8
= 8[ H ] parts by weight of O2
4
8
iii) Combustion of Sulphur
S + O2
32
32
SO2
64 (by weight)
32 parts by weight of sulphur requires 32 parts by weight of
oxygen for complete combustion.
(or)
1 part by volume of sulphur requires 1 part by volume of oxygen
for complete combustion.
4.64
Applied Chemistry
∴ S parts by weight of sulphur requires
32 × S
=
= S parts by weight of O2
32
Consequently, theoretical amount of oxygen required for the
complete combustion of 1kg of solid or liquid fuel.
Theoretical minimum O2 = [
32
O
×C + 8(H
) + S ] kg
12
8
Since mass % of O2 in air is 23, the amount of air required
theoretically for combustion of 1 kg of the
× fuel is,
Air (theoretical) =
100 32
O
[
) + S ] kg
×C + 8(H
23 12
8
Volume of Air Required for Complete Combustion of
Gaseous Combustible Matters
i).
H2(g) + 1/2O2
H2O
1 vol.
0.5vol.
1 volume of H2 (g) requires 0.5 vol. of oxygen
ii).
CH4 + 2 O2
CO2 + 2 H2O
1 vol 2 vol
1 volume of CH4 requires 2.0 volume of oxygen
iii).
CO(g) + ½ O2
CO2
1 vol 0.5 vol
1 volume of CO requires 0.5volume of oxygen
Fuels and Combustion
4.65
iv).
C2H4(g) + 3O2
2CO2 + 2H2O
1 vol
3 vol
1 volume of C2H4 (g) requires 3.0 vol. of oxygen
Amount of O2 required by the fuel will be given by
subtracting the amount of O2 already present in the fuel from the
total or theoretical amount of O2 required by the fuel.
∴ Net amount of O2 required = Total amount of O2 required
– O2 already present in the fuel.
Air contains 21 % of O2 by volume and 23 % of O2 by
weight. Hence from the amount of O2 required by the fuel, the
amount of air required can be calculated.
100
x minimum O2
23
o
Minimum weight of air required =
o
Minimum volume of air required =
o
Molecular mass of air is taken as 28.94 g/mol
o
Density of air at NTP = 1.29 kg/cm2
o
22.4 litres (or 22,400 ml) of any gas at NTP (i.e 0°C and
760 mm of Hg) has a mass equal to its 1 mol (gram
molecular weight)
o
Thus 22.4 litres of CO2 at NTP will have a mass of 44 g (44
is the molecular weight of CO2 )
100
x minimum O2
21
4.66
Applied Chemistry
Excess air for combustion
It is necessary to supply excess air for complete
combustion of the fuel. It is found out from the theoretical
amount of air as follows.
The amount of air required if excess air is supplied.
Theoretical amount of air
x [ 100 + % of excess air ]
100
4.21 Flue Gas Analysis [ Orsat’s Method ]
The mixture of gases (like CO2, O2, CO, etc.,) coming out
from the combustion chamber is called flue gases. The analysis
of a flue gas would give an idea about the complete or incomplete
combustion process. The analysis of flue gases is carried out
using orsat’s apparatus.
Description of orsat’s apparatus
It consists of a horizontal tube. At one end of this tube,
an U-tube containing fused CaCl2 is connected through a 3-way
stop cock. The other end of the tube is connected with a
graduated burette. The burette is surrounded by a water jacket to
keep the temperature of gas as a constant. The lower end of the
burette is connected to a water reservoir by means of a rubber
tube. The level of water in the burette can be raised or lowered by
raising or lowering the reservoir.
The horizontal tube is also connected with three different
absorption bulbs I, II and III for absorbing CO2, O2 and CO.
Fuels and Combustion
4.67
I - bulb : It consists of ‘potassium hydroxide’ solution and it
absorbs only CO2.
II - bulb : It consists of ‘alkaline pyrogallol’ solution and it
absorbs only CO2 and O2.
III - bulb : It consists of ‘ammoniacal cuprous chloride’ solution
and it absorbs CO2, O2 and CO.
Seperating funnel
Gas burette
Flue gas
I
III
II
I
CaCl2 for drying
Absorption
bulb for CO2 Absorption Absorption
bulb for O2 bulb for CO
Water Jacket
Fig 4.12
Working
The 3-way stop cock is opened to the atmosphere and the
reservoir is raised, till the burette is completely filled with water
and air is excluded from the burette. The 3-way stop cock is now
connected to the flue gas supply and the flue gas is sucked into
the burette and the volume of flue gas is adjusted to 100 cc by
4.68
Applied Chemistry
raising and lowering the reservoir. Then the 3-way stop cock is
closed.
a) Absorption of CO2
The stopper of the absorption bulb-I, containing KOH
solution, is opened and all the gases is passed into the bulb-I by
raising the level of water in the burette. The gas enters into the
bulb-I, where CO2 present in the flue gas is absorbed by KOH.
The gas is again sent to the burette. This process is
repeated several times to ensure complete absorption of CO 2. The
decrease in volume of the flue gas in the burette indicates the
volume of CO2 in 100 cc of the flue gas.
b) Absorption of O2
Stop cock of bulb-I is closed and stop cock of bulb-II is
opened. The gas is again sent into the absorption bulb-II, where
O2 present in the flue gas is absorbed by alkaline pyrogallol. The
decrease in volume of the flue gas in the burette indicates the
volume of O2.
c) Absorption of CO
Now the stop cock of bulb-II is closed and stop cock of
bulb-III is opened. The remaining gas is sent into the absorption
bulb-III, where CO present in the flue gas is absorbed by
ammoniacal cuprous chloride. The decrease in volume of the flue
gas in the burette indicates the volume of CO. The remaining gas
in the burette after the absorption of CO2, O2 & CO is taken as
nitrogen.
Fuels and Combustion
4.69
Significance (or) uses of flue gas analysis
1. Flue gas analysis gives an idea about the complete or
incomplete combustion process.
2. If the flue gases contain considerable amount of CO, it
indicates that incomplete combustion is occurring and it also
indicates that the short supply of O2.
3. If the flue gases contain considerable amount of O 2, it
indicates that complete combustion is occurring and also it
indicates that the excess of O2 is supplied.
Ignition Temperature
It is defined as “the lowest temperature to which the fuel
must be heated, so that it starts burning smoothly”.
The ignition temperature of coal is about 300°C. In the
case of liquid fuels, the ignition temperature is called the flash
point, which ranges from 200 – 400°C. For gaseous fuels, the
ignition temperature is in the order of 800°C.
Spontaneous Ignition Temperature (SIT)
It is defined as the minimum temperature at which the
fuel catches fire (ignites) spontaneously without external
heating.
If the ignition temperature of a fuel is low it can catch fire
very quickly. On the other hand if the ignition temperature is high
it is difficult to ignite the fuel. If the heat evolved in a system is
unable to escape, temperature of the system goes on increasing
and when SIT is reached, the system burns on its own.
4.70
Applied Chemistry
Explosive Range (or) Limits of Inflammability
Most of the gaseous fuels have two percentage limits
called upper limit and lower limit. Those limits represent
percentage by volume of fuel present in fuel-air mixture. The
range covered by these limits is termed as explosive range of the
fuel.
For continuous burning the amount of fuel present in the
fuel-air mixture should not go below the lower limit or above the
upper limit.
For example, the explosive range of petrol is 2 – 4.5. This
means that when the concentration of petrol vapour in petrol-air
mixture is between 2 and 4.5 by volume, the mixture will burn on
ignition. When the concentration of petrol vapour in petrol-air
mixture is below 2% (lower limit) or above 4.5% (upper limit) by
volume, the mixture will not burn on ignition.
Calorific Intensity (or) Flame Temperature
It is the minimum temperature reached when the fuel is
completely burnt in the theoretical amount of air.
Problems Based on Combustion
1. Calculate the volume of air (Volume % of O2 in air = 21)
required for complete combustion of 1 litre of CO
Solution
The combustion equation of CO is written as follows
CO + ½ O2
CO2
1 vol
0.5 vol
One volume ( litre) of CO requires 0.5 volume (litre) of O2 for
complete combustion.
We know that,
Fuels and Combustion
4.71
21 litres of O2 is supplied by 100 litres of air.
100 × 0.5
lit of air
∴ 0.5 litres of O2 is supplied by
21
= 2.38 litres of air
Result
The volume of air required for the
complete combustion of 1 lit of CO = 2.38 litres
2.What is the volume of air required for the complete
combustion of 1m3 of mixture containing 75 % of CH4 and
25 % of C2H6 ?
Solution
i) 1m3 of the mixture contains
75
100
= 0.75 m3 of CH4
25
= 0.25 m3 of C2H6
100
ii) The combustion equations of CH4 and C2H6 are written as
follows
CH4 + 2 O2
1 vol 2 vol
CO2 + 2 H2O
i.e., 1 volume ( or m3 ) of CH4 requires 2 volume ( or m3 ) of O2
for complete combustion.
2 × 0.75
∴ 0.75 m3 of CH4 requires =
= 1.5 m3 of O2
1
C2H6 +
1 vol
7
/2 O2
3.5 vol
2CO2 + 3H2O
i.e., 1 volume ( or m3 ) of C2H6 requires 3.5 volume ( or m3 ) of
O2 for complete combustion.
4.72
3.5 × 0.25
1
iii) Total volume of O2 required
∴ 0.25 m3 of C2H6 requires =
Applied Chemistry
= 0.875 m3 of O2
= 1.5 + 0.875 m3
= 2.375 m3
We know that, 21 m3 of O2 is supplied by 100 m3 of air
2.735 ×100
= 11.31 m3 of air.
∴ 2.735 m3 of O2 is supplied by
21
Result
The volume of air required for the complete
combustion of 1m3 of mixture
= 11.31 m3
3.Calculate the minimum volume of air required for the
complete combustion of 1m 3 of a gaseous fuel containing the
following composition by volume. CO = 25 %, H2 = 10 %,
CH4 =08 %, CO2 = 5 %, N2 = 50 % and O2 = 2 %.
Solution
1. 1m3 of fuel contains
25
a)
= 0.25 m3 of CO
100
10
b)
= 0.10 m3 of H2
100
8
c)
= 0.08 m3 of CH4
100
5
d)
= 0.05 m3 of CO2
100
50
e)
= 0.50 m3 of N2
100
2
f)
= 0.02 m3 of O2
100
N2 and CO2 are non combustible constituents. They do not burn
and not require any oxygen.
The combustion equations of the remaining constituents are
written as follows.
Fuels and Combustion
4.73
CO + 1/2 O2
1 vol 0.5 vol
CO2
i.e., 1 volume (or m3 ) of CO requires 0.5 volume (or m3 ) of O2
for complete combustion.
0.5 x0.25 3
m of O2
∴ 0.25 m3 of CO requires =
1
= 0.125 m3 of O2
H2 + 1/2 O2
1 vol 0.5 vol
H2 O
i.e., 1 volume (or m3 ) of H2 requires 0.5 volume (or m3 )
of O2 for complete combustion.
0.5 x0.1 3
m of O2
1
= 0.05 m3 of O2
∴ 0.10 m3 of H2 requires =
CH4 + 2 O2
1 vol 2 vol
CO2 + 2 H2O
i.e., 1 volume (or m3 ) of CH4 requires 2 volume (or m3 )
of O2 for complete combustion.
0.08 x 2 3
∴ 0.08 m3 of CH4 requires =
m of O2
1
= 0.16 m3 of O2
Total volume of O2 required = 0.125 + 0.05 + 0.16 m3
= 0.335 m3 of O2
∴ Net volume of O2 required = Total volume of O2 required O2 already present in the fuel.
= 0.335 – 0.02 m3 of O2
= 0.315 m3 of O2
We know that
4.74
Applied Chemistry
21 m3 of O2 is supplied by 100 m3 of air
∴ 0.315 m3 of O2 is supplied by
0.315 × 100
21
= 1.5 m3 of air.
=
Result
The volume of air required for the
complete combustion of 1m3 of the = 1.5 m3
gaseous fuel
4. A fuel contains C = 60 %, H2 = 09 %, O2 = 08 %, S = 13 %,
remaining ash. Calculate the minimum quantity of air required
for the complete combustion of 1kg of a fuel.
Solution
1. 1 kg of fuel contains
60
a)
= 0.60 kg of Carbon
100
10
= 0.09 kg of Hydrogen
100
08
c)
= 0.08 kg of Oxygen
100
13
d)
= 0.13 kg of Sulphur
100
The combustion equations of the various elements present in the
fuel are as follows.
b)
C +
O2
12 kg
32 kg
CO2
i.e., 12 kg of carbon requires 32 kg of oxygen for complete
combustion.
32x0.6
∴ 0.60 kg of carbon requires
kg of O2 = 1.6 kg of O2
12
Fuels and Combustion
H2 +
1
2 kg
16 kg
/2 O2
4.75
H2 O
i.e., 2 kg of hydrogen requires 16 kg of oxygen for complete
combustion.
16 × 0.09
∴ 0.09 kg of H2 requires
kg of O2 = 0.72 kg of O2
2
S
32 kg
+
O2
SO2
32 kg
i.e., 32 kg of sulphur requires 32 kg of oxygen for complete
combustion.
32 × 0.13
kg of O2 = 0.13 kg of O2
∴ 0.13 kg of S requires
32
Total amount of O2 required = 1.60 + 0.72 + 0.13 kgs
= 2.45 kg of O2
But the amount of O2 already present in the fuel = 0.08 kg.
∴ Net amount O2 required = Total amount of O2 required O2 already present in the fuel.
= 2.45 – 0.08 kg of O2
= 2.37 kg of O2
We know that
23 kg of O2 is supplied by 100 kg of air
2.37 × 100
∴ 2.37 kg of O2 is supplied by
= 10.30 kg of air.
23
Result
The minimum amount of air required
for the complete combustion of 1kg of the fuel = 10.30 kg.
4.76
Applied Chemistry
5. Calculate the minimum amount of air required for the
complete combustion of 50 kgs of coal containing C = 75 %,
H2 = 10 %, O2 = 02 %, S = 05 %, and the rest nitrogen by
weight.
Solution
1. 1 kg of fuel contains
75
a)
= 0.75 kg of Carbon
100
10
b)
= 0.10 kg of Hydrogen
100
2
c)
= 0.02 kg of Oxygen
100
5
d)
= 0.05 kg of Sulphur
100
N2 is a non combustible constituent. They do not burn and do not
require any oxygen.
The combustion equations of the various elements present
in the coal are as follows.
C +
O2
12 kg
32 kg
CO2
i.e., 12 kg of carbon requires 32 kg of oxygen for complete
combustion.
32 × 0.75
∴ 0.75 kg of carbon requires
kg of O2 = 2.0 kg of O2
12
H2 +
1
2 kg
16 kg
/2 O2
H2 O
i.e., 2 kg of hydrogen requires 16 kg of oxygen for complete
combustion.
16 × 0.10
∴ 0.10 kg of H2 requires
kg of O2 = 0.80 kg of O2
2
Fuels and Combustion
S
32 kg
+
O2
4.77
SO2
32 kg
i.e., 32 kg of sulphur requires 32 kg of oxygen for complete
combustion.
32 × 0.05
∴ 0.05 kg of S requires
kg of O2 = 0.05 kg of O2
32
Total amount of O2 required = 2.00 + 0.80 + 0.05 kgs
= 2.85 kg
But the amount of O2 already present in the fuel = 0.02 kg.
= Total amount of O2 required ∴ Net amount O2 required
O2 already present in the fuel.
= 2.85 – 0.02 kg
= 2.83 kg
We know that
23 kg of O2 is supplied by 100 kg of air
2.83 × 100
∴ 2.83 kg of O2 is supplied by
= 12.30 kg of air.
23
The amount of air required for 1 kg of coal = 12.30 kg
∴ The amount of air required for 50 kg of coal = 12.30 × 50 kg
= 615 kg
Result
The amount of air required for the
complete combustion of 50 kgs of the coal = 615 kg.
6. Calculate the minimum amount of air required for the
complete combustion of 150 kgs of fuel containing 70 %
Carbon, 15 % hydrogen, 5 % sulphur and the rest nitrogen by
weight.
Solution
1. 1 kg of fuel contains
70
a)
= 0.70 kg of Carbon
100
4.78
Applied Chemistry
15
= 0.15 kg of Hydrogen
100
05
c)
= 0.05 kg of Sulphur
100
N2 is a non combustible constituent. They do not burn and do not
require any oxygen.
The combustion equations of the various elements present
in the coal are as follows.
b)
C +
O2
12 kg
32 kg
CO2
i.e., 12 kg of carbon requires 32 kg of oxygen for complete
combustion.
0.7 × 32
0.70 kg of carbon requires
kg of O2 = 1.87 kg of O2
12
H2 +
1
2 kg
16 kg
/2 O2
H2O
i.e., 2 kg of hydrogen requires 16 kg of oxygen for complete
combustion.
0.15 x16
∴ 0.15 kg of H2 requires
kg of O2 = 1.20 kg of O2
2
S
+ O2
32 kg
SO2
32 kg
i.e., 32 kg of sulphur requires 32 kg of oxygen for complete
combustion.
0.05 × 32
∴ 0.05 kg of S requires
kg of O2 = 0.05 kg of O2
32
Total amount of O2 required = 1.87 + 1.20 + 0.05 kgs
= 3.12 kg.
We know that
Fuels and Combustion
4.79
23 kg of O2 is supplied by 100 kg of air
3.12 × 100
∴ 3.12 kg of O2 is supplied by
23
= 13.57 kg of air.
The minimum amount of air required for the
complete combustion of 1 kg of fuel
= 13.57 kg
∴ The minimum amount of air required for
the complete combustion of 50 kg of fuel = 13.57 × 150 kg
= 2035.5 kg
Result
The minimum amount of air required
for the complete combustion of
= 2035.5 kg
150 kg of the fuel
7. Calculate the weight and volume of air required for the
complete combustion of 1 kg of coke (or) carbon.
Solution
The combustion equation of the carbon is as follows.
C +
O2
12 kg
32 kg
CO2
12 kg of carbon requires 32 kg of O2 for complete combustion.
32 × 1
∴ 1 kg of coke requires
= 2.67 kg of O2
12
We know that,
23 kg of O2 is supplied by 100 kg of air.
2.67 × 100
∴ 2.67 kg of O2 is supplied by
= 11.61 kg of air.
23
We know that,
1 gm mole of any substance occupies 22.4 litres
(or)
32 kg of O2 occupies 22.4 m3 at NTP [1 m3 = 1000 litres ]
4.80
Applied Chemistry
11.61 kg of O2 occupies
22.4 x 11.61
32
= 8.127 m3
We know that,
21 m3 of O2 is supplied by 100 m3 of air
8.127 x 100
= 38.70 m3 of air.
21
∴ 8.127 m3 of O2 is supplied by
Result
1. The amount of air required for complete
combustion of 1 kg of coke
=
11.61 kg
2. The volume of air required for complete
combustion of 1 kg of coke
=
38.70 m3.
8. Calculate the minimum theoretical quantity of air needed for
the combustion of 10 kg of coal which is 95 % pure
Solution
The combustion equation of carbon is as follows.
C + O2
12 kg 32 kg
CO2
i.e., 12 kg of carbon requires 32 kg of oxygen for complete
combustion.
32 x 10
∴ 10 kg of carbon requires
kg of O2
12
= 26.67 kg of O2
We know that
23 kg of O2 is supplied by 100 kg of air
∴ 26.67 kg of O2 is supplied by
26.67 x 100
kg of air
23
= 115.96 kg of air.
Fuels and Combustion
4.81
The amount of air required for the combustion
of 10 kg of coal which is 100 % pure
= 115.96 kg
The amount of air required for the combustion
of 10 kg of coal which is 95 % pure
=
115 .96 × 95
100
= 110.16 kg
Result
The amount of air required for the
combustion of 10 kg of coal
which is 95 % pure
= 110.16 kg
9. A sample of coal was found to have the following percentage
composition C = 75 %, H2 = 5.2 %, O2 = 12.8 %, N2 = 1.2 % and
the rest ash. Calculate the amount of air needed for the
complete combustion if 1 kg of the coal is burnt with 30 %
excess air.
Solution
1. 1 kg of coal contains
75
a)
= 0.75 kg of Carbon
100
5 .2
b)
= 0.052 kg of Hydrogen
100
12 .8
c)
= 0.128 kg of Sulphur
100
12 .8
d)
= 0.128 kg of Oxygen
100
1 .2
e)
= 0.012 kg of Nitrogen
100
N2 and ash are non combustible constituents. They do not
burn and do not require any oxygen.
4.82
Applied Chemistry
The combustion equations of the remaining constituents
present in the fuel are as follows.
C +
O2
12 kg
32 kg
CO2
i.e., 12 kg of carbon requires 32 kg of oxygen for complete
combustion.
32 × 0.75
∴ 0.75 kg of carbon requires
kg of O2 = 2.0 kg of O2
12
H2 +
1
2 kg
16 kg
/2 O2
H2O
i.e., 2 kg of hydrogen requires 16 kg of oxygen for complete
combustion.
16 × 0.052
∴ 0.052 kg of H2 requires
kg of O2 = 0.416 kg of O2
2
S
+ O2
32 kg
SO2
32 kg
i.e., 32 kg of sulphur requires 32 kg of oxygen for complete
combustion.
32 × 0.128
∴ 0.128 kg of S requires
kg of O2 = 0.128 kg of O2
32
Total amount of O2 required = 2.00 + 0.416 + 0.128 kgs
= 2.544 kg of O2
But the amount of O2 already present in the fuel = 0.128 kg.
∴ Net amount O2 required
= Total amount of O2 required O2 already present in the fuel.
= 2.544 – 0.128 kg of O2
= 2.416 kg of O2
We know that
Fuels and Combustion
4.83
23 kg of O2 is supplied by 100 kg of air
100 × 2.416
= 10.50 kg of air.
∴ 2.416 kg of O2 is supplied by
23
If 30 % excess air is used
=
10.5 x [100 + 30]
100
= 13.65 kg of air.
Result
The volume of air required for the
complete combustion of 1kg of the
fuel if 30 % excess air is used.
= 13.65 kg.
Review Questions:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
What is coal ? how is it formed ? what are its types ?
Describe proximate analysis.Bring out its importance
What is metallurgical coke ?
Describe Otto Hoffmann by product coke oven method
Discuss Bergius process, What is knocking ? Explain
How will you improve octane number of a fuel
What is water gas ? How is it manufactured ?
Write the uses of water gas.
How is producer gas manufactured ?
What are the uses of producer gas ?
Write a brief note on CNG
What is calorific value ? mention it’s units
Describe Orsat’s apparatus
How is flue gas analysed using Orsat’s apparatus
Discuss the calculation of minimum air requirement for the
complete combustion.
16. What is flue gas ?. Discuss Dulong’s formula
17. Define i) Gross calorific value ii) Net calorific value
18. Write the importance of flue gas analysis.
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