0Fuel

It is a combustible substance which contains carbon as the main constituent and produces
significant amount of heat and light energy when burnt in excess of air or oxygen. The energy
obtained is used for domestic and industrial purpose.

Classification of chemical fuels

Chemical fuels are classified as primary and secondary fuels. Fuels, which occur in nature, are
called primary fuels. Fuels, which are derived from primary fuels, are called secondary fuels.
Chemical fuels are further classified as solids, liquids and gases. A complete classification of
fuels with examples is shown in the following Table.

Physical state Primary fuels Secondary fuels

Solid Wood, coal Charcoal, coke
Liquid Petroleum Petrol, diesel, kerosene
Gas Natural Gas LPG

Definition of calorific value of a fuel [Gross calorific value]: It is defined as the amount of
heat liberated when unit quantity (1 kg or 1 m3) of a fuel is completely burnt in air or oxygen and
the products of combustion are cooled to room temperature.
Definition of net calorific value: It is defined as the amount of heat released when unit quantity
of a fuel is completely burnt in air or oxygen and the products of combustion are let off into the
atmosphere.
S. I. unit of calorific value: For solids, calorific value is expressed in J kg-1 (Joules per kg). For
gaseous fuels it is expressed in J m-3 (Joules / m3).
Importance of hydrocarbons as fuels: Fossil fuels contain mainly hydrocarbons. These
hydrocarbons are important sources of energy in daily life. Hydrocarbons are used as energy
sources in cooking, lighting, automobiles, production of electricity in thermal power plants etc.

1
These hydrocarbon fuels meet 80% of the world’s energy demand. Thus hydrocarbons are
important sources of energy.
Determination of calorific value of a solid fuel using Bomb Calorimeter:

Principle: A known mass of the solid sample is burnt completely in excess oxygen. The
liberated heat is absorbed by the surrounding water and the calorimeter. Thus the heat liberated
during combustion of the fuel is equal to the heat absorbed by the water and the calorimeter.

Construction: The bomb calorimeter consists of a cylindrical bomb made up of stainless steel
with an airtight lid. The bomb has an inlet valve for providing oxygen atmosphere inside the
bomb and an electrical ignition coil for starting the combustion of fuel. The bomb is placed in an
insulated copper calorimeter. The calorimeter has a mechanical stirrer for dissipation of heat and
a thermometer for reading the temperature.

Working: A known mass of the solid fuel is placed in a crucible. The crucible is placed inside
the bomb and sealed airtight by using the lid. The bomb is placed inside a copper calorimeter. A
known amount of water is taken in the calorimeter. Water is kept in constant agitation by the
mechanical stirrer and the initial temperature t1 is noted in the thermometer. The bomb is filled
with oxygen and the combustion of fuel is initiated by passing an electric current through the
ignition coil. The fuel burns liberating heat, which is absorbed by the surrounding water and the
copper calorimeter. The maximum temperature attained by the water, t2 , is noted.

Observation and calculations:

w1  w2  s t
Gross calorific value (G C V) = m J kg-1
Where

2
w = w1+ w2
= mass of water in the calorimeter, in kg + water equivalent of the calorimeter, in kg
s = specific heat of water, in J kg-1 oC-1
t = t2-t1 = rise in temperature, in oC
m = mass of the fuel, in kg
(Note: If the mass of fuel is given in grams, convert that into kg. For example, 0.2 g = 0.2  10-3
kg. If specific heat of water is given in J kg-1oC-1, calorific value will be in J kg-1. If the specific
heat is given in kJ kg-1oC-1, then the calorific value will be in kJ kg-1.)
(Note: Specific heat of water is the amount of heat energy required to increase the temperature of
one kg of water by one degree C.)

b) Calculation of N C V

W.k.t. 2 kg of H2 produces 18 kg of steam (from the equation)

H2 + 1/2O2 H2O
 1 kg of H2 produces 9 kg of steam (from the equation)
H% of H2 produces 0.09 H Kg of steam
Heat released when 1 kg of steam condenses = 2454  103 J Kg-1
 Heat released when 0.09xH Kg of steam condenses = 0.09 H 2454x103......J Kg-1
NCV = GCV-0.09  H Latent Heat of steam ……..J Kg-1
NCV = GCV-0.09  H  2454  103J Kg-1

CRACKING
Definition of cracking: Cracking is defined as the process of converting higher molecular
hydrocarbons (high boiling point) into lower molecular hydrocarbons (low boiling point) to
increase the yield of the gasoline. In this process the higher molecular hydrocarbons which are in
lesser demand are converted to higher demand gasoline.

C14H30 C7H16 + C7H14

Fluidized bed catalytic cracking:

Principle: The finely powdered catalyst is kept agitated by the gas streams of vapourised heavy
oil, so that the catalyst can be handled like a fluid system. This also results in a good contact
between the catalyst surface and the reactant leading to efficient cracking.

Construction: A schematic diagram of fluidized bed catalytic cracking method is shown in the
following figure.

3
Optimum conditions:

Feed stock : Heavy oil.

Catalyst used: Al2O3+ SiO2 (Fluidised)
Temperature: 5500 C

Working: The feed stock vapors are passed into cracking chamber consisting of finely divided
catalyst. The vapours of heavy oil are cracked to lower molecular hydrocarbons and sent to the
fractionating column to separate into gasoline and uncracked oil.

Spent catalyst from the cracking chamber is continuously transported into the regeneration
chamber through an air stream. The carbon deposited on catalyst particles is burnt off in
regeneration Chamber. The regenerated catalyst is transported back into the cracking chamber
together with feedstock.

The uncracked oil is cracked in a second stage of cracking to increase the overall yield of
cracked products

REFORMATION
Definition of reformation of petrol: The process in which modification of structure of straight
chain hydrocarbons takes place without change in the no of carbon atoms, to increase the octane
no of petrol is called as reformation.
Thus reformation converts the low octane number petrol into branched chain, cyclic and
aromatic hydrocarbons containing high octane number.

Reformation reaction conditions:

Reactant (feed stock): Gasoline obtained by fractionation of petroleum

Catalyst: Platinum supported on alumina (Pt /Al2O3)
Temperature: about 500 oC
Pressure: 15-50 atm.
Reformation reactions:

The main reformation reactions are:1) Isomerization, 2) Cyclisation (Dehydrogenation) and 3)

Aromatisation (Cyclization and dehydrogenation).

1) Isomerization:
Example: n – Heptane  2- Methyl hexane

4
 H3C CH2 CH2 CH2 CH2 CH2 CH3 H3C CH CH2 CH2 CH2 CH3
CH3
n-Heptane 2-Methyl-hexane

2) Cyclisation: (Dehydrogenation)

Example: Cyclohexane Benzene + Hydrogen

H2
C
H 2C CH 2
+ 3H2
H 2C CH 2
C
H2
3) Aromatisation: (Cyclization and dehydrogenation)
H2 H2
-H2 -3H 2
C C CH 3
H 3C C C
H2 H2
n-hexane cyclohexane benzene

KNOCKING

Definition of knocking: Knocking may be defined as the production of thermal shock wave in
an IC engine as a result of an explosive combustion of fuel-air mixture, leading to a rattling
sound.

Knocking in petrol engines: In petrol engines, the mixture of petrol and air is drawn in to the
cylinder. The fuel-air mixture is compressed by the piston and is ignited by an electric spark. As
the flame front travels in the combustion chamber, rapidly expanding combustion products
compress the remaining unburnt fuel and raise its temperature. If the flame front travels at an
optimum speed, the combustion of unburnt fuel takes place smoothly. On the other hand, if the
flame front travels too slowly, the entire last portion of fuel mixture may get heated up beyond
its ignition temperature and undergo instantaneous explosive combustion. This result in
emission of a characteristic rattling sound called “knocking”.

The efficiency of the power output in internal combustion engine increases with increase
in CR along with that the tendency of knocking also increases.

cylinder volume at the suction stroke

Compression ratio = when the piston is at the bottom (V1)
(CR) cylinder volume at the compression stroke
when the piston is at top (V2)
The compression ratio is always greater than 1, i.e V1 > V2. When the compression ratio is 5-10
then the engine is said to possess good efficiency.
The graph depicting efficiency verses CR is as shown below

5
Knocking Mechanism

C2H6 + 1/2 O2 2CO2 + 3H2O Smooth combustion

C2H6 + O2 CH3 O O CH3 Ethane peroxide f ormation
(Knocking)

Adverse effects of petrol knocking

1. Fuel consumption is increased
2. Power output is decreased
3. Causes mechanical damage to engine parts such as spark plug, piston and engine walls
4. Produce undesirable rattling noise
5. Driving becomes unpleasant

Remedial measures to control knocking

1. By using gasoline of high octane no
2. By using critical compression ratio
3. By using anti-knocking agents

Definition of octane number: The octane number of a gasoline (petrol) is the percentage
volume of iso-octane in a mixture of iso-octane and n-heptane, which has the same knocking
property as the gasoline under test.
CH3
CH 3

CH 3 C CH2 CH CH3 H3 C CH2 CH 2 CH 2 CH 2 CH 2 CH3

CH 3 n - Heptane

2,2,4 Tri Methyl Pentane

(isoctane)
Octane no = 100 Octane no = 0
Antiknocking agent: Antiknocking agent is a chemical substance (organometallic compound)
added to petrol to improve the Antiknocking property of the petrol.
Eg: tetra ethyl lead (TEL).
6
Addition of TEL to petrol increases the octane number of petrol. TEL is normally used along
with ethylene dibromide or ethylene dichloride. [Formula of tetra ethyl lead = (C2H5)4Pb]
TEL is converted into non volatile Pb and PbO, which gets deposited in engine parts and causes
mechanical damage. Presence of ethylene dibromide or ethylene dichloride converts them into
volatile PbCl2 or PbBr2, which are exhausted along with other gases.

Unleaded petrol: Unleaded petrol is the petrol where tetra ethyl lead (TEL) is not used as an
antiknocking agent.

Addition of TEL to petrol leads to release of PbBr2 and PbCl2 into the atmosphere through
automobile exhaust. This results in air pollution. To prevent air pollution, unleaded petrol is
preferred. In unleaded petrol, methyl-t-butyl ether (MTBE) and ethyl-t-butyl ether (ETBE) are
used as antiknocking agents.

Definition of Cetane number: The cetane number of a diesel fuel is the percentage by volume
of n-cetane in a mixture of n-cetane and -methyl naphthalene which has the same knocking
characteristic as the diesel under test.

H 3C (CH2)14 CH3

Hexadeecane (n-cetane) 1-methyl naphthalene

cetane no = 0 cetane no = 100
Power Alcohol:
The fuel obtained by mixing petrol with ethyl alcohol, thus increasing the octane no of the
mixture is called power alcohol. Generally petrol and ethyl alcohol are mixed in the ratio of 4:1.
Advantages of Power alcohol:
1. The octane no of the power alcohol is high, so it has excellent antiknocking property.
2. Power output is high
3. Addition of alcohol reduces the emission of carbon monoxide and volatile organic
compounds into the atmosphere.
4. Air required for complete combustion is low
5. Ethanol is obtained from agricultural products, hence renewable and biodegradable

Biodiesel: It is an alternative fuel which is obtained from the vegetable oil or animal fat by
simple chemical process.

Chemically Biodiesel is a mixture of monoalkyl esters of long chain fatty acid obtained by the
trans-esterification of the vegetable oil or animal fat or blue green algae with excess of alcohol in

7
presence of a base catalyst .Vegetable oil or animal fat consists of of triglycerides of long chain
alkyl fatty acids (lipids)

CH2OCOR1
CH3OCOR1 CH2OH
NaOH +
CHOCOR2 + 3CH3OH CH3OCOR2 + CHOH
60oC +
CH2OCOR3 CH3OCOR3 CH2OH
Triglyceride oil Biodiesel Glycerol
Where R1, R2 and R3 are the long chain alkyl groups
Glycerol is highly soluble in water and so can be easily removed from the reaction mixture.
Instead of MeOH, Ethanol can also be used. Since MeOH is cheaper, it is generally used in the
production of biodiesel.

Advantages:
1. Produced from renewable sources
2. Nontoxic and eco-friendly
3. Biodegradable and reuces the depletion of petroleum sources
4. Creates lot of self employment and encourage farming activities

Biogas: The combustible gas produced by anaerobic digestion of aqueous organic matter with
bacterial culture involving methane producers is called biogas and the process is known as
biomethanation. This process involves 3 stages.
In the 1st stage Larger molecules are broken down to smaller molecules by the enzymes
such as cellulase protease and lipase.
In the 2nd stage smaller molecules are further broken down to acetic acid, CO2 & H2. In
this stage anaerobic condition is produced which is required for 3rd stage called methanogensis.
In the 3rd stage the anaerobes converts acetic acid, CO2 & H2 to Methane (CH4) & CO2.
Biogas composition: 60-70% CH4, + 30-40% CO2 + traces of H2S, ammonia and water vapour.
The overall chemical reaction occurring in biogas synthesis is

C6H12O6 anaerobic bacteria CO2 + CH4

Advantages of Biogas:
1. Produces more energy
2. Free from dust smoke and environment friendly
3. Utilises bio-waste so avoids pollution
4. Slurry produced in biogas is good quality manure free from weeds, foul smell and pathogens.

Problems

Problem 1. Calculate the calorific value of a sample of coal from the following data:
Mass of Coal = 0.6 g

8
Mass of water + water equivalent of calorimeter = 2200 g
Specific heat of water = 4.187 kJ kg-1oC-1
Rise in temperature = 6.52 oC

Solution:
(Note: In solving the problem, follow the steps given below:
1. Write the given quantities and convert them into appropriate units.
2. Write the equation.
3. Substitute the values.
4. Simplify using calculator if necessary.
5. Write the answer.
6. Write the units.)

Given: m = 0.6 g = 0.6  10-3 kg

w1 + w2 = 2200 g = 2.2 kg
s = 4.187 kJ kg-1oC-1 = 4.187  103 J kg-1oC-1
t = 6.52 oC

w1  w2  s t
Gross calorific value = m J kg-1

2.2  4.187  10 3  6.52

= 0.6  10 3

= 1.001 108 J kg-1

Problem 2. A 0.85 g of coal sample (carbon 90 %, H2 5%, and ash 5% ) was subjected to
combustion in a bomb calorimeter. Mass of water taken in the calorimeter was 2000 g and the
water equivalent of calorimeter was 600 g. The rise in temperature was 3.5 oC. Calculate the
gross and net calorific value of the sample. (Given, specific heat of water = 4.187 kJ kg-1oC-1 and
latent heat of steam = 2454 kJ kg-1)

Solution: Given m = 0.85 g = 0.85  10-3 kg

% of hydrogen = 5%
w1 = 2000 g = 2 kg
w2= 600 g = 0.6 kg
t = 3.5 oC
s = 4.187 kJ kg-1oC-1 = 4.187  103 J kg-1oC-1
L = 2454 kJ kg-1 = 2454  103 J kg-1

9
 w1  w2  s t 2  0.6  4.187  10 3
 3.5
m 0.85  10 3
a) Gross C.V. = = = 4.4825 107 J kg-1
b) Calculation of N C V

NCV = GCV-0.09  H Latent Heat of steam ……..J Kg-1

=4.4825 107-0.09  5 2454  103 J kg-1
=4.4825 107 - 1.104300 106 J kg-1
=4.3720107J kg-1
(Note: Latent heat of steam is the amount of heat energy liberated when one kg of steam
is converted into one kg liquid water.)

3. On burning 0.75g of a solid fuel in a bomb calorimeter the temperature of 2.5kg of water is
increased from 240C to 280C the water equivalent of calorimeter and latent heat of steam are
0.485 Kg and 4.2X587 KJ/Kg, specific heat of water is 4.2KJ/Kg/0C.H2 2.5%,
Solution: Given m = 0.75 g = 0.75  10-3 kg
w1+W2 = (2.5+0.485) kg = 2.985 kg
t = t2-t1=28-24=4 oC
s = 4.2 kJ kg-1oC-1 = 4.2  103 J kg-1oC-1
L = 2454 kJ kg-1 = 2454  103 J kg-1

w1  w2  s t 2.985 X 4 X 4.2 X 10 3

3
J / Kg  6.66 X 10 7 J / Kg
a) Gross C.V. = m = 0 .75 X 10
b) Calculation of N C V
NCV = GCV-0.09  H Latent Heat of steam ……..J Kg-1
=6.68107-0.09 2.54.2X587 103 J kg-1
=6.68107 - 5.54715 105 J kg-1
=6.63107J kg

4. Calculate the gross and net calorific values by data given.

b) Mass of Coal = 0.7 g = 0.7  10-3 kg
c) Mass of water = 2.2 kg
d) Water equivalent of calorimeter = W2 = 0.25 Kg
e) Raising temperature = 3.2 0C
f) Specific heat of water = 4.187 kJ kg-1oC-1 = 4.187  103 J kg-1oC-1
g) Latent heat of steam = 580 calories / g [1 calorie = 4.18 joule]
h) H2 =2%,
m = 0.7  10-3 kg
w1+W2 = 2.45 kg
t = 3.2 oC
s = 4.187  103 J kg-1oC-1
L = 580 calories / g=580 X 4.18 X 103 J /Kg = 2424 X 103 J/Kg

10
 Solution:
w1  w2  s t 2.45 X 3.2 X 4.187 X 10 3
 3
 4.689 X 10 7 J / Kg
Gross C.V. = m = 0.7 X 10
b) Calculation of N C V

NCV = GCV- 0.09  H Latent Heat of steam ……..J Kg-1

=4.689107-0.09 2580 X 4.18 X 103 J kg-1
=4.689107 -4.36392 105 J kg-1
=4.68107J kg-1

5. On burning 0.96g of a solid fuel in bomb calorimeter the temperature of 3500g of H2O
increased by 2.70C water equivalent of colorimeter and latent heat of steam are 385 g and
587 cal/g respectively. If the fuel contains 5% H2 calculate its gross and net calorific value.
Given:
m=0.96g=0.96X10-3Kg
W1=3500g=3500X10-3Kg
W2=385g= 385X10-3Kg
t=2.70C
Latent heat =587 cal/g = 587X4.18X103 J/Kg
S= 4.187X103J/Kg/0C
H2 =5%,
w1  w2  s t
J / Kg
Gross calorific value = m
(3500 X 10 3  385 X 10 3 )  4.187  10 3  2.7
= 0.96  10 3
= 4.537X 106 J/Kg =

NCV = GCV- 0.09  H Latent Heat of steam ……..J Kg-1

=4.537 106- 0.09 5587X 4.18 X 103 J kg-1 =3.43 10

6. On burning 0.85 ×10-3 kg of a solid fuel in a bomb calorimeter ,the temperature of 2.1 kg
wateris raised from 24OC to 27.6OC.the water equivalent of calorimeter and latent heat of steam
are 1.1 kg and 2454 kj/kg respectively. Specific heat of water is 4.187 kj/kg. if the fuel contains
2% hydrogen ,calculate its gross and net calorific values.

m = 0.85 X 10-3 kg
W1=2.1Kg
W2 = 1.1 Kg
t1= 24OC

11
 t2=27.6OC
s = 4.187kj/Kg=4.187 X 103 J/ kg
L = 2454Kj/Kg=2454 X 103 J/Kg

( w1  w2 ) st (2.1  1.1) 4.187 X 10 3 X 3.6

GCV  J / Kg  3
 5.6746 X 10 7 J / Kg
m 0.85 X 10

NCV = GCV- 0.09  H Latent Heat of steam ……..J Kg-1

=5.6746107-0.09 2580 X 4.18 7X 103 J kg-1
=5.6746107 -4.37122 105 J kg-1
=5.63107J kg-1

7. On burning 0.96 X 10-3 Kg of a solid fuel, in a bomb calorimeter, the temperature of 3.5
Kg of water was increased by 2.7oC. Water equivalent of the calorimeter and latent heat of
steam are 0.385 Kg 2455 KJ/Kg, respectively. If the fuel contains 5% hydrogen, calculate
its gross and net calorific value.

Solution: Given m = 0.9610-3 kg

% of hydrogen = 5%
w1 = 3.5 kg
w2 = 0.385 kg
t = 2.7 oC
s = 4.187 kJ kg-1oC-1 =4.187  103 J kg-1oC-1
L = 2455 kJ kg-1 = 2455 103 J kg-1

w1  w2  s t
a) Gross C.V. = m
3.855 4.187  10  2.7
3

= 0.96  10 3 = 4.5749 107 J kg-1

b) Calculation of Net C.V.:

NCV = GCV- 0.09  H Latent Heat of steam ……..J Kg-1

=4.5749107-0.09 52455X 103 J kg-1
=4.5749107-1.104750 106 J kg-1
=4.46107J kg-1

Solar Cell / Photovoltaic cell: It is a device which converts solar energy into electrical energy
by Photovoltaic effect.

12
Physical Properties of Silicon Relevant to Photovoltaic’s:
1. Silicon is a semiconductor with a band gap Eg of 1.12 eV at 25◦C.
2. Silicon doped with trivalent or pentavalent atoms ionizes at lower temperature. Therefore
doped silicon can be used as a photovoltaic material at room temperature.
3. Silicon has high refractive index and reflects most of the solar radiation falling on it, so they
are coated with antireflective layer.
4. Doped silicon has a minority carrier life of 50- 300 microseconds. Minimum minority carrier
life required for a photovoltaic material is 25 microseconds. Minimum minority carrier is the
time elapsed before an ejected electron recombines with a hole.
5. Pure silicon has a negative temperature coefficient of resistance, since the number of free
charge carriers increases with temperature.

Chemical Properties Relevant to Photovoltaic’s:

1. Silicon is stable in the tetravalent state and has a strong affinity for oxygen, forming stable
oxides and silicates.
2. Elemental silicon immediately oxidizes, forming a thin protective film of silica of less than
100 A°, which prevents further oxidation.
3. Silicon and carbon form a strong Si–C bond and stable products
4. Silicon reacts with hydrogen to form silanes.
5. Silicon reacts with chlorine to form chlorosilanes which are highly volatile and can be easily
separted from other impurities based on their boiling point difference.

DOPING:
Doping of Silicon by Diffusion Technique:
In this technique, a region of a semi conductor material is incorporated with dopant atoms by the
different impurity atom into the crystal of the material without actually melting it. By this
technique the extent of impurity penetration can be controlled to a very small thickness of the
material. For example, n-type silicon can be obtained by heating a silicon wafer below its
melting point in an atmosphere of n-type impurity such as Phosphorus. The impurity atoms
condense on the surface of the wafer diffuse into the crystal. P-type silicon can be obtained by
heating a silicon wafer below its melting point in an atmosphere of P-type impurity such as
Boron.
Diffusion
Si + PH3 N-Doped silicon
Diffusion
Si + BH3 B2H6 P-Doped silicon

Construction & working of Photovoltaic cell

Construction: A p-n junction is created by diffusing an n-type and p-type semiconductor
Sunlight
materials. It has two electrical contacts, one is a metallic grid over n-type and second is a layer of

13 Antireflective layer

Metal grid
n-type layer
silver metal at the back of p-type semiconductor thus completely covering it. An antireflection
layer of silicon nitride or TiO2 silicon oxide is coated in between the metal grids to prevent
reflection of solar light.

Working: The PV cells are fixed side by side in parallel and in series to form a module. The
good number of modules are placed together to form an array or panel. When light falls on the n-
type surface, the electron-hole pairs are created. The electrons are drifted to and collected at the
n-type, whereas holes are drifted to and collected at the p-type end. When the two ends are
electrically connected, the electrons move through the load from n-type and finally reaches the p-
type to get neutralized. Thus the solar is converted into electrical energy.

Advantages of PV cells:
1. Operating and maintenance costs for PV panels are considered to be low.
2. PV panels have no mechanically moving parts, hence have less maintenance.
3. Residential solar panels are easy to install on rooftops without any interference to residential
lifestyle.
4. No emissions of toxic gases, hence do not contribute to pollution.
5. Solar energy is supplied by nature, it is free and abundant.
6. Long component life.

Disadvantages:
1. Solar energy panels require additional equipment (inverters) to convert direct electricity (DC)
to alternating electricity (AC) for power network.
2. Though PV panels have no maintenance or operating costs, they are fragile and can be
damaged easily.
3. For a continuous supply of electric power, Photovoltaic panels require not only Inverters but
also storage batteries, thus increasing the investment cost for PV panels considerably.
4. Solar panels efficiency levels are relatively low (between 14%-25%).
5. No electricity during night and also cloudy weather.
6. Low power output.

14