ENGINEERING CHEMISTRY: Module 3 [chemical fuels & solar cells]

ENGINEERING CHEMISTRY

1. CHEMICAL FUELS AND PHOTOVOLTAIC CELLS
--------------------------------------------------------------------------------------------------

ENGG.CHEMISTRY [RM] Page 1

MODULE-2 MODULE-3

Definition of a chemical fuel: A chemical fuel is a substance, which produces a significant amount of heat
energy and light energy when burnt in air or oxygen.

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, CNG, Water gas, producer gas

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. These hydrocarbon fuels meet 80% of the world’s energy demand. Thus
hydrocarbons are important sources of energy.

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 m
3
) 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.

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MODULE-2 MODULE-3
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 / m
3
).

Determination of calorific value of a solid fuel using Bomb Calorimeter:

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

Construction: The bomb calorimeter consists of a stainless steel vessel with an airtight lid. This vessel is called bomb.
The bomb has an inlet valve for providing oxygen atmosphere inside the bomb and an electrical ignition coil for starting
of 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. The lid is
closed tightly. The bomb is placed inside a copper calorimeter. A known mass of water is taken in the calorimeter. The
bomb is filled with oxygen at a pressure of 25-30 atm. The temperature t1 in the thermometer is noted.
On passing an electric current through the ignition coil, the fuel gets ignited. The fuel burns liberating heat. The water is
continuously stirred using the stirrer. The maximum temperature attained by the water, t2 , is noted.

Observation and calculations:
b) Calculation of G C V
Gross calorific value (G C V) =
 
m
tsww 
21
kJ kg
-1

Where

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MODULE-2 MODULE-3
w = w1+ w2
= mass of water in the calorimeter, in kg + water equivalent of the calorimeter, in kg
s = specific heat of water, in kJ kg
-1 o
C
-1

t = t2-t1 = rise in temperature, in
o
C
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
-1o
C
-1
, calorific value will be in J kg
-1
. If the specific heat is given in kJ kg
-1o
C
-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

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

NCV = GCV-0.09  % H  2454  10
3
kJ Kg
-1


Problem 1. Calculate the calorific value of a sample of coal from the following data:
Mass of Coal = 0.6 g
Mass of water + water equivalent of calorimeter = 2200 g
Specific heat of water = 4.187 kJ kg
-1o
C
-1

Rise in temperature = 3
o
C
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
w 1 + w2 = 2200 g = 2.2 kg
s = 4.187 kJ kg
-1o
C
-1

t = 3
o
C
Gross calorific value =
 
m
tsww

21
kJ kg
-1


=
3
106.0
3187.42.2



= 46057 kJ/kg

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MODULE-2 MODULE-3

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
o
C. Calculate the gross and net calorific value of the sample. (Given, specific heat
of water = 4.187 kJ kg
-1o
C
-1
and latent heat of steam = 2454 kJ kg
-1
)

Solution: Given m = 0.85 g = 0.85  10
-3
kg
% of hydrogen = 5%
w 1 = 2000 g = 2 kg
w 2= 600 g = 0.6 kg
t = 3.5
o
C
s = 4.187 kJ kg
-1o
C
-1

L = 2454 kJ kg
-1


a) Gross C.V. =
 
m
tsww 
21


=
 
3
1085.0
5.3187.46.02



= 44825kJ kg
-1

b) Calculation of N C V

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

=44825 -0.09  5 2454 =44825 - 1104.3 =43720kJ 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.)

Problem 3. On burning 0.75g of a solid fuel in a bomb calorimeter the temperature of 2.5kg of water is increased
from 24
0
C to 28
0
C the water equivalent of calorimeter and latent heat of steam are 0.485 Kg and 4.2X587
KJ/Kgrespectively , specific heat of water is 4.2KJ/Kg/
0
C, 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
o
C
s = 4.2 kJ kg
-1o
C
-1

L = 2454 kJ kg
-1

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MODULE-2 MODULE-3
a) Gross C.V.=
 
m
tsww 
21
= KgkJKgJk
X
XX
/66864/
1075.0
2.44985.2
3



b) Calculation of N C V

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

=66864-0.09 2.54.2X587 =66864– 554.715=66309kJ kg

Problem 4. Calculate the gross and net calorific values by data given.
Mass of Coal = 0.7 g = 0.7  10
-3
kg
Mass of water = 2.2 kg
Water equivalent of calorimeter = W2 = 0.25 Kg
Raising temperature = 3.2
0
C
Specific heat of water = 4.187 kJ kg
-1o
C
-1

Latent heat of steam = 580 calories / g [1 calorie = 4.187 kJ]
H2 =2%,
Solution :
m = 0.7  10
-3
kg
w1+W2 = 2.45 kg
t = 3.2
o
C
s = 4.187 kJ/kg/
o
C
L = 580 calories / g=580 X 4.187kJ /Kg = 2428kJ/Kg

Solution:
Gross C.V. =
 
m
tsww 
21
= KgkJ
X
XX
/44689
107.0
187.42.345.2
3



b) Calculation of N C V

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

= 46894- 0.09 2580 X4.187 kJ kg
-1
=46894-437.122 =46456 kJ kg-
1


Problem 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
-3
Kg
W1=3500g=3500X10
-3
Kg
W2=385g= 385X10
-3
Kg

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MODULE-2 MODULE-3
t=2.7
0
C
Latent heat =587 cal/g = 587X4.187 kJ/Kg
S= 4.187 kJ/Kg/
0
C
H2 =5%,
Gross calorific value =
 
KgJk
m
tsww
/
21

=
3
33
1096.0
7.2187.4)10385103500(



XX
= 45749.5 kJ/Kg


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

=45749.5- 0.09 5587X 4.187=44643 kJ/kg

Problem 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 24
O
C to 27.6
O
C.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= 24
O
C
t2=27.6
O
C
s = 4.187kj/Kg
L = 2454Kj/Kg

KgkJ
X
X
KgJ
m
tsww
GCV /56746
1085.0
6.3187.4)1.11.2(
/
)(
3
21







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

=56746- 0.09 22454 =56746-441.72 =56304kJ kg-
1

Problem 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 2.7
o
C. 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%

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MODULE-2 MODULE-3
w 1 = 3.5 kg
w 2 = 0.385 kg
t = 2.7
o
C
s = 4.187 kJ kg
-1o
C
-1

L = 2455 kJ kg
-1


a) Gross C.V. =
 
m
tsww 
21


=
 
3
1096.0
7.2187.4855.3



= 45749 kJ kg
-1


b) Calculation of Net C.V.:

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

=45749-0.09 52455 =45749-1104.3 =44644.7kJ kg-
1


CRACKING
Definition of cracking: Cracking is defined as the process of converting high molecular weight hydrocarbons into
lower molecular weight hydrocarbons to increase the yield of the gasoline.

C
14H
30
C
7H
16 + C
7H
14


Fluidized bed catalytic cracking:

Principle: In fluidized bed catalytic cracking, the powder catalyst is kept agitated by gas streams (cracking fuel) 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.

Construction: A schematic diagram of fluidized bed catalytic cracking method is shown in the following figure.
Optimum conditions:
Feed stock : Heavy oil
Catalyst used: y- type zeolite activated with a rare earth oxide(Al2O3+ SiO2)
Temperature: 450 to 550
0
C
Pressur : 1 to atm
Yield: 70 to 80%

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MODULE-2 MODULE-3
Working: The feed stock vapors are passed
into cracking chamber. The reactants undergo
cracking in the presence of catalyst. The
products are passed to a fractionating
column.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 feed stock.
REFORMATION OF PETROL
Definition of reformation of petrol: Conversion of straight chain hydrocarbons in petrol into branched chain, cyclic and
aromatic hydrocarbons, resulting in up gradation of quality of the petrol is known as reformation.


How reformation improves the quality of petrol?

The octane number for straight chain hydrocarbons is low. For branched chain, cyclic and aromatic hydrocarbons, the
octane number is high. Thus reformation converts the low octane number petrol into high octane number petrol.
Reformation reaction conditions:

Reactant (feed stock) : Gasoline obtained by fractionation of petroleum + H2
Catalyst: Platinum supported on alumina (Pt / Al2O3 )
Temperature: about 500
o
C
Pressure: 15-50 atm
Reformation reactions:

The main reformation reactions are:1) Isomerization, 2) Dehydrogenation and 3) Cyclization and dehydrogenation.

1) Isomerization:

Example: n – Heptane  5 - Methyl hexane

H
3C CH
2
CH
2
CH
2
CH
2
CH
2
CH
3
n-Heptane
H
3C CH
CH
2
CH
2
CH
2
CH
3
CH
3
2-Methyl-hexane


2) Dehydrogenation:

Example: Cyclo hexane Benzene + Hydrogen

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MODULE-2 MODULE-3

H
2C
H
2C
C
H
2
CH
2
CH
2
H
2
C
+ 3H
2


3) Cyclization and dehydrogenation

H
2C
H
2C
C
H
2
CH
2
CH
2
H
2
C
H
3C
H
2
CH
2
C
H
2
C
H
2
C CH
3
-H
2 -3H
2


Synthetic petrol
Synthesis of petrol by Fischer-Tropsch process:

Franz Fisher- Hans Tropsch showed that carbon monoxide and hydrogen in the presence of catalyst at 180-250
o
C form a
mixture of aliphatic hydrocarbons ranging from methane to wax. The natures of the products formed however depend on
the temperature, pressure and catalyst used.

Process: In this process a mixture of [CO+H2] is preheated and passed through a compressor. Then the mixture is send to
catalyst reactor contains a iron based or cobalt based catalyst at 200-300
0
C, produces a mixture of hydrocarbons. The
petroleum (Mixture of hydrocarbons) so formed is next fractionated to yield gasoline, kerosene, heavy oil and paraffin
wax.
Diagram :

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MODULE-2 MODULE-3
n CO + 2nH
2
C
nH
2n + nH
2O
Unsaturated hydrocarbons

n CO + (2n + 1)H
2
C
nH
2n + 2 + nH
2O
Saturated hydrocarbons


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

FOUR STROKE INTERNAL COMBUSTION ENGINE (CAR ENGINE / PETROL ENGINE):
 1st Stroke INTAKE: piston moves down, air + diesel fuel mixture drawn in
 2nd Stroke COMPRESSION: piston moves up, air + fuel mixture compressed then ignited by sparking plug
 3rd Stroke POWER: piston pushed down by exploding air + fuel mixture
 4th Stroke EXHAUST: piston moves up, spent gases pushed out through exhaust


Compression ratio =
?????? �? ?ℎ? ??? ?? ?ℎ? ????�?? ??????
?????? �? ?ℎ? ??? ?? ?ℎ? ????????�?? ??????


Mechanism of knocking in chemical terms:
A. Under normal conditions there is smooth oxidation of the fuel. It involves following reaction
Oxygen combines with hydrocarbon molecule (for example, ethane) forming carbon dioxide and water
C
2H
6 + 31/2O
2 2CO
2 + 3H
2O


B. Under knocking conditions, the rate of combustion is very high. It involves following steps.

1. Oxygen combines with hydrocarbon molecule forming peroxides.
H
3C CH
3+ O
2
H
3C O O CH
3
Ethane
Ethane peroxide(unstable)

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MODULE-2 MODULE-3
2. The peroxides decompose readily to give a number of gaseous products. For example,

H
3C O O CH
3
Ethane peroxide
CH
3
CHO+ H
2O
Acetaldehyde


CH
3
CHO
Acetaldehyde
+ 11/2O
2
HCHO + CO
2 + H
2O
Formaldehyde


Formaldehyde
HCHO + O
2 H
2O + CO
2

Formation of peroxide, number of gaseous product and rapid increase of pressure results in explosive combustion &
knocking in IC engine .

Antiknocking agent for Petrol engine: Antiknocking agent is a chemical substance added to petrol to improve the
Antiknocking property of the petrol. For example, tetra ethyl lead (TEL). 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 ]

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 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 being tested.

CH
3 C
CH
3
CH
3
CH
2CH
CH
3
CH
3
2,2,4 Tri Methyl Pentane
H
3C CH
2
CH
2
CH
2
CH
2
CH
2
CH
3
n - Heptane


Knocking in diesel engine:
In diesel engine there is no spark plug.

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MODULE-2 MODULE-3
The air is passed into the cylinder and it is compressed by the piston till it attains
ignition temperature of the fuel .As soon as the fuel is injected in to the cylinder, it
absorbs heat from the heated air and burns on attaining ignition temperature
spontaneously.
Ignition delay is the time lag between the start of the fuel injected and its ignition.
Straight chain hydrocarbon has shorter ignition delay whereas branched and
aromatic hydrocarbon has longer ignition delay.
Ignition delay causes accumulation of more hydrocarbons and results in explosive
combustion is called as ‘diesel knocking’.

Antiknocking agent for diesel engine: Antiknocking agent is a chemical substance added to diesel to improve the
Antiknocking property of the diesel.
Ex. Alkyl nitrates and Amyl nitrates

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 being tested.

H
3C (CH
2)
14
CH
3
n-cetane
= C
16H
34 = Hexadecane
C
10H
7
CH
3
Methyl naphthalene


POWER ALCOHOL
Definition of power alcohol: When ethyl alcohol is used as an additive to motor fuel(petrol) to act as a fuel for
internal combustion engines, it is called power alcohol.

The importance’s of power alcohol as fuel are:
(Characteristics of alcohol-blended petrol are: )
1. The power output is good.
2. Addition of alcohol reduces the emission of carbon monoxide and volatile organic compounds in to the atmosphere.
3. Alcohol improves the antiknocking property of the fuel.
4. Petrol-alcohol mixture has the same lubrication as the pure petrol has.
5. Since ethanol is produced from agricultural products, it can be a sustainable fuel.
6. Ethanol is biodegradable

Biodiesel: Biodiesel is obtained by the transesterification of the vegetable oil or animal fat or blue green algae and
used as a fuel in the internal combustion engine.

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MODULE-2 MODULE-3

Production:
Biodiesel is produced by the trans-esterification of the
vegetable oil or animal fat or blue green algae with
methanol and NaOH catalyst at 60
O
C.


CH
2OCOR
1
CHOCOR
2
CH
2OCOR
3
+3CH 3OH
NaOH
60
o
C
CH
3OCOR
1
+
CH
3OCOR
2
+
CH
3OCOR
3
+
CH
2OH
CHOH
CH
2OH
Biodiesel
GlycerolTriglyceride oil


NOTE: Soap formation is avoided by esterification of
free fatty acid in presence of acid catalyst fallowed by
trans-esterification

Advantages:
1. Biodiesel is a renewable source of energy.
2. It is an alternative fuel for fossil fuels.
3. It refers to a vegetable oil or animal fat based diesel.(Green fuel).
4. It is biodegradable.
5. It has a high flash point and low ignition point than diesel.
6. Addition of alcohol reduces the emission of carbon monoxide and volatile organic compounds in to the
atmosphere.
7. It is nontoxic.

Solar Energy

Direct solar energy: Direct solar energy involves only one step transformation into a usable form.
Examples: Solar water heater, Solar cooker, Photovoltaic cells.
Indirect solar energy: Indirect solar energy involves more than one transformation to reach a usable
form.
Examples: Photosynthesis, Ocean thermal energy

Photovoltaic Cells:
Photovoltaic cells are semiconductor device which convert solar energy into electrical energy.
(Photovoltaic cell is based on the principle of photoelectric effect).

Construction & Working of Photovoltaic Cell:

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MODULE-2 MODULE-3

 A typical silicon photovoltaic cell consists of p-n junction made of silicon.
 It consists of two electrical contacts, one of which is in the form of metallic grid and allows light
to fall on the semiconductor between the grid lines.
 The cell’s other electrical contact is formed by a metallic layer on the back of the solar cell.
 An antireflective layer avoids the reflection of light.
 When light radiation falls on the p-n junction diode, electron-hole pairs are generated by the
absorption of the radiation.
 � = ℎ� = ℎ
?
�

 The electrons are made to move across the external circuit.
 Thus photoelectric current is produced and available for use.

Advantages:

 Fuel source is vast and essentially infinite.
 No emissions, no combustion or radioactive residues for disposal. Does not contribute to global
change or pollution.
 Low operating cost (no fuel).
 No moving parts and so no wear and tear.
 High reliability in modules.
 Can be integrated into new or existing building structures.
 High public acceptance and excellent record.

Disadvantages:

 Sun light is a diffuse, i.c., it is relatively low density energy.
 High installation cost.
 Poor reliability of auxiliary elements including storage.
 Energy can be produced only during the day time.

Metallic grid
Layer of silver

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MODULE-2 MODULE-3
Importance of Photoelectric cell:

 Photovoltaic cell provides enormous amount of energy from sun which is unlimited,
inexhaustible and renewable.
 Photovoltaic cells can serve for both off grid and on grid application.
 Photovoltaic cell produces no pollution so it is environment friendly.
 Photovoltaic cell energy conversion is highly modular. This is important with respect to the
development of electricity supply systems in many rural and remote areas, where grid
extension is economically not feasible.
 It provides power for space craft and satellite.
 Photovoltaic can be used as roof integrated systems, providing power and also serving as
optical shading elements for the space below and preventing overheating in the summer.
 Developments in the field of Photovoltaic cells will boost the semiconductor industry and
storage battery industries.

PHOTOVOLTAIC MODULE , PANEL & ARRAY


Photovoltaic module: A number of Photovoltaic cells electrically connected to each other and mounted
in a support structure or frame is called photovoltaic module.(0.5-0.6V)

Photovoltaic panel: Multiple modules electrically connected to each other and mounted in a support
structure or frame is called photovoltaic panel.(27-36 cells, 12-18V)

Photovoltaic arrays: Photovoltaic array is the complete power generating unit, It consist of number of
photovoltaic panels, electrically connected to each other and mounted in a frame.



Physical properties of silicon relative to photovoltaics:

Physical properties of silicon are as follow:
 Silicon is a semiconductor with band gap of 1.2eV at 25
o
C.
 At atmospheric pressure, silicon crystallizes to diamond cube like structure.
Panel
Array

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MODULE-2 MODULE-3
 Silicon contracts when melted and expand when solidify.
 High refractive index limits the optical applications of silicon.
 The absorption and transmission properties in the 0.4-0.5 µm wavelength spectra are important in
the performance of photovoltaic cells.

Chemical properties relevant to photovoltaics:
 Silicon is stable in the tetravalent and has a strong affinity for oxygen, forming stable oxides and
silicates.
 Silicon and carbon form a strong Si-C bond and stable products.
 Silicon forms hydrides, and monosilane (SiH4) is key chemical compound for the production of
amorphous silicon and the purification of silicon to semiconductor grad.
 Silicon forms trichlorosilane and tetrachlorosilane with chlorine, which are the intermediates and
the by-products of the purification process in metallurgical grade silicon to semiconductor grade.

Production of solar grade silicon:

Solar grade silicon has impurities in the ppb level is required for polysilicon used in semiconductor
industry. It can be prepared by using two methods.

The Union Carbide process:

This process involves the following steps:

STEP – 1: The hydrogenation of tetrachlorosilane through a bed of metallurgical silicon is carried out in
a fluidized bed reactor
Si + 4HCl SiCl4+2H2 at 300
0
C
SiCl4+H2 HSiCl3+HCl at 1000
0
C

STEP 2: The trichlorosilane is separated by distillation while the unreacted tetrachlorosilane is recycled
back to the hydrogenation reactor.
The purified trichlorosilane is passed through a fixed bed column filled with quaternary ammonium ion
exchange resin acting as catalyst. Trichlorosilane gets converted into dichlorosilane.
2SiHCl
3 SiH
2Cl
2 + SiCl
4

STEP 3: The products are separated by distillation, tetrachlorosilane is recycled to the hydrogen reactor
and dichlorosilane is passed through a second fixed bed column filled with quaternary ammonium ion
exchange resin. Dichlorosilane is converted into silane.

3SiH
2Cl
2
SiH
4 + 2SiHCl
3

STEP 4 : The above products are separated by distillation and trichlorosilane is recycled to the first bed
column. Silane is further purified by distillation and then pyrolized to produce polysilicon onto heated
silicon seed rods mounted in a metal bell-jar reactor
SiH
4
2H
2 + Si

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MODULE-2 MODULE-3
Purification of Silicon by Zone refining Process
Zone refining method is used to get ultra-pure silicon..This process is used where the impurities are
soluble in molten state. When Si rod is melted, impurities tend to concentrate in the molten.
Process:
The impure silicon rod surrounded by RF coil is clamped to vertical zone refiner and it is placed in an
inert argon atmosphere. The silicon rod is heated near to its melting point using RF coil, when the RF
coil is moved slowly from top to bottom ,the dissolved impurities in silicon will also moves along with
the coil [since impurities are more soluble in its molten state] by leaving pure silicon at the upper portion
.The rod is subjected for several zone process to get ultra-pure silicon.


Doping of Silicon:

It is the method of addition of impurity atom to a pure semiconductor to get desired extrinsic (p or n)
semiconductors.
In the preparation of solar grade silicon, impure silicon is converted into vapors this vapors is condensed
on the silicon rod. The vapor of doping material is passed in to the chamber, doping can be accompanied
by simultaneous depositing a dopant with the semiconductor material.

Diffusion Technique:
It is a process of diffusing the impurity atoms into a silicon wafer by heating the thin wafer just below its
melting point in an atmosphere of impurity.

Doping of Silicon by Diffusion Technique:
In this technique, a region of a semiconductor 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, a n-type

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MODULE-2 MODULE-3
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 Boran. The extent of diffusion is regulating by temperature and
concentration of the impurity atom.
Si + PH
3
N-Dopped silicon
Diffusion
Si + BH
3
B
2H
6
P-Dopped silicon



Questions :
1. What is chemical fuel ? write a note on classification of fuel
2. Explain the determination of calorific value of a solid or a liquid fuel using bomb calorimeter
3. 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
o
C. Calculate the gross and net calorific
value of the sample. (Given, specific heat of water = 4.187 kJ kg
-1o
C
-1
and latent heat of steam =
2454 kJ kg
-1
)
4. Define the terms : gross calorific value & net calorific value.
5. What is cracking ?Explain the process of cracking using fluidized bed catalytin cracking (FCC)
6. What is reformation? Give any three reaction
7. What is knocking in IC engine? Explain the mechanism of knocking in chemical terms.
8. Explain the following terms : a. octane number b. cetane number c. antiknocking agents
9. Explain the following terms : a. Power alcohol b. biodiesel
10. What is solar cell? Explain the construction & working of PV cell
11. Explain the production of Si by union carbide process
12. Explain photovoltaic module, panel & array
13. Explain the purification os Si by zone refining process
14. What is doping? Explain doping of Si by diffusion technique.