Fuels Ppts
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12/13/2010 Dr P B Dwivedi, NMIMSUniversity, Mumbai
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CHEMISTRY OF FUELS
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Fuels: substances which undergo combustion in
the presence of air to produce a large amount of
heat that can be used economically for domestic
and industrial purpose.
This definition does not include nuclear fuel because
it cannot be used easily by a common man.
The various fuels used economically are wood, coal,
kerosene, petrol, diesel gasoline, coal gas, producer
gas, water gas, natural gas (LPG) etc.
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Classifications
Fuels can be broadly classified by origin as,
(i)Primary or natural fuels: coal, wood etc
(ii)Secondary or artificial or derived fuels: petrol, diesel
On the basis of physical state, as :
(i) Solid fuels
(ii) Liquid fuels
(iii) Gaseous fuels
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Basis Origin Physical
State
Source Natural or primary Artificial or Secondary or
Derived
Wood, peat, lignite,
coal
Semi coke, charcoal Solid fuels
Petroleum Petrol, kerosene,
gas oil, coal tar,
etc.
Liquid
fuels
Producer gas, coke-
oven gas, water
gas, blast furnace
gas, compressed
butane gas.
Gaseous
fuels
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Characteristics of Fuels
The physical properties for which fuels are tested and
their ideal requirements are listed below :
(i) Calorific value or specific heat of combustion.(ii) Ignition temperature
(iii) Flame temperature
(iv) Flash and Fire point.
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(v) Aniline point
(vi) Knocking.
(vii) Specific gravity
(viii) Cloud and Pour point
(ix) Viscosity
(x) Coke number.
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The chemical properties include the compositional
analysis of fuel.
For solid and liquids fuels :
(i) Percentage of various elements such as C, H, O,
N, S, etc.
(ii) Percentage of moisture
(iii) Percentage of volatile matter
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For gaseous fuels :
(i) Percentage of combustible gases e.g. CO, H2,
CH4, C2H4, C2H6, C4H10, H2S etc.
(ii) Percentage of non-combustible gases e.g. N2,
CO2 etc.
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Calorific Value
number of units of heat evolved during complete
combustion of unit weight of the fuel.
A British Thermal Unit: the heat required to raisethe temperature of one pound of water from 60r F to
61r F.
The Calorie: the heat required to raise the temperatureof one kg of water from 15rC to 16rC.
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High and Low Calorific Values
Calorific values are of two types as,
(i)High or Gross Calorific Value (H.C.V. or G.C.V.)
(ii)Low or Net Calorific Value (L.C.V. or N.C.V.)
High calorific value may be defined as, the total
amount of heat produce when one unit of the fuel has
been burnt completely and the combustion productshave been cooled to 16rC or 60rF.
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LCV: is the net heatproduced when unit mass or
volume of fuel is completely burnt and products are
allowed to escape.
Net or lower C.V.= Gross C.V. Latent heat of
water formed
Or Gross C.V Mass of hydrogen v 9v Latent heat
of steam (587 cal/g)
(Because 1 part by weight of hydrogen produces 9
parts (1 + 8) by mass of water)
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The calorific value of fuels (e.g. Coal) is determined
theoretically by Dulong formula, or I.A. Daviesformula.
Dulong formula can be expressed as,
HCV = 1/100 [8,080 C+ 34,500(H- O/8)+ 2240 S]
Where C = % Carbon, H = % Hydrogen, O = %
Oxygen, S = % Sulphur
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Oxygen in fuel (coal) is in combined state as
water and hence it does not contribute to
heating value of fuel.
LCV = [HCV 0.09 H(%) 587] cal/g
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Sr.
No.
Property Solid Fuels Liquid Fuels Gaseous Fuels
1. Calorific value Low Higher Highest
2. Specific gravity Highest Medium Lowest
3. Ignition point High Low Lowest
4. Efficiency Poor Good Best
5. Air required for
combustion
Large and excess
of air
Less excess of air Slight excess of air
6. Use in I.C. engine Cannot be used Already in use Can be used
7. Mode of supply Cannot be piped Can be piped Can be piped
8. Space for storage Large 50% less than
solid fuel
Very high space
9. Relative cost Cheaper Costly More costly than
other two
10. Care in storage and
transport
Less care required Care is necessary Great care required
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Sr
No
Types
of Coal
Moisture
of Air
Dried At
40rC
C
%
H
%
O
%
Ash
%
Calorific
Value
(kcal/kg)
Uses
1. Peat 25 57 6 35 2 5400 Power generation and
domestic purpose.
2. Lignite 20 67 5 20 8 6500 Manufacture of
producer gas, thermal
power plants.
3. Bitumi
nous
4 83 5 15 7 8000 For metallurgical
coke, coal gas, boiler,domestic purpose
4. Anthra
cite
2 92 3 2 3 8600 Boilers, metallurgical
fuel, domestic
Classification of Coal
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Analysis of Coal
The proximate analysis is easy and quicker and it gives a fair
idea of the quality of coal.
The ultimate analysis is essential for calculating heat
balances in any process for which coal is employed as a fuel.
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Moisture
It is determined by heating about one gm. of finely
powdered coal at 105rC to 110rC for an hour in
electric oven. The loss in weight is reported as due to
moisture.
% Moisture = [loss in wt of sample 100]/wt of coal
taken
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Ash
A known weight of sample is taken in a crucible and
the coal is burnt completely at 700rC 750rC in
muffle furnace until a constant weight is obtained.The residue left in the crucible is ash content in coal
which is calculated as
% of Ash = [wt of residue left in cruciblev 100]/ wt
of coal taken
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Volatile matter
For determining volatile matter content, a known
weight of dried sample is taken in a crucible with
properly fitting lid. It is then heated at9
50rCs
20rC
for exactly seven minutes in previously heated muffle
furnace. The loss in weight is due to volatile matter
which is calculated as
Volatile matter = [loss in wt at 9500C 100]/wt of
coal sample
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iv) Fixed carbon :
% of Fixed carbon = 100 (% of moisture + % of ash +
% of volatile matter)
In any good sample of coal, the percentages ofmoisture, ash, volatile matter should be as low as
possible and thus the percentage of fixed carbon
should be as high as possible.
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C + O2 p CO2
12 parts p 44 parts
2H2 + O2 p 2H2O
4 parts p 36 parts
% of Carbon = [increase in wt of KOH tube 12
100]/wt of coal taken 44
% of Hydrogen =[increase in wt of CaCl2 tube 4
v100]/wt of coal taken 36
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Nitrogen
Nitrogen is calculated by Kjehldals Method.
% N = [vol of acid consumed in neutralizing NH3 N 1.4]/wt
of coal taken
% Sulphur: [wt of BaSO4 obtained 32 100]/wt of coal taken
233
(v) Oxygen:
The oxygen is determined indirectly by calculation as
% of Oxygen = 100 (% of C + % of H + % of N + % of S + %
of Ash)
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Bomb Calorimeter
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Let x = mass in g of fuel taken in crucible
W = mass of water in calorimeter
w = water equivalent in g of calorimeter, stirrer, thermometer,
bomb etc.
t1 & t2 are initial & final temperatures of water in calorimeter
L = higher calorific value of fuel in cal/g
Then heat liberated by buring of fuel = xL
Heat absorbed by water & apparatus = (W+w)(t2-t1)
But heat liberated = heat absorbed
so, xL = (W+w)(t2-t1)
L = (W+w)(t2-t1)/x cal/g or kcal/kg
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If H = % of hydrogen in fuel
9H/100 g = mass of water from 1 g of fuel= 0.09H g
So heat taken by water in forming steam = 0.09 H
587 cal
LCV = HCV - 0.09 H 587 cal/g
By considering fuse wire correction, acid correction
& cooling corection
L = [{(W+w)(t2-t1+ cooling correction)}- {acid +fuse
correction}]/x cal/g or kcal/kg
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Combustion
Combustion is a process in which oxygen from the
air reacts with the elements or compounds to give
heat. As the elements or compounds combine in indefinite
proportions with oxygen, we need to calculate what is
minimum oxygen or air requiredfor the completecombustion of compounds. The commonly involved
combustion reactions are :
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i) C + O2p
CO2ii) 2H2 + O2p 2H2O OR H2 + (O)p H2O
iii) S + O2p SO2
iv) 2CO + O2p 2CO2 OR CO + (O)p
CO2v) CH4 + 2O2p CO2 + 2H2O
vi) 2C2H6 + 7O2p 4CO2 + 6H2O
vii) C2H4 +3O2-p 2CO2 + 2H2O
viii) 2C2H2 + 5O2p 4CO2 + 2H2O
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Hint to Solve Problems on Calculation of Quantity of Air
Required for Combustion of Fuel :
First write the appropriate chemical reaction with
oxygen and find their relation between the element or
compound on weight or volume basis.
e.g C + O2 p CO2
(12 gm) + (32 gm)
2H2 + O2 p 2H2O
(4 gm) + (32 gm)S + O2 p SO2
(32 gm) + (32 gm)
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2) Calculate the oxygen required on the basis of unit quantity of
fuel.
3) Calculate the total oxygen required for the combustion and
subtract the oxygen which is present in the fuel.
4) The oxygen calculated should be converted into air byknowing that air contains 23 parts by weight of oxygen OR 21
parts by volume of oxygen.
5) The average molecular weight of air is 28.94 gm.
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Weight of
elements per kg.
of coal
Weight of O2 required
for complete combustion
in kg.
C = 0.85 0.85 v32/12 = 2.26 kg.
H = 0.1 0.1 v 8 = 0.8 kg.
O = 0.05
Total oxygen = 3.06 kg.
Weight of oxygen required
= Weight of oxygen needed weight of oxygen present
= 3.06 0.05 = 3.01
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@ Air required for complete combustion
= 3.01 v 100/23
= 13.08 kg. per 1 kg. coal.
@ Air required for 5 kg. of coal
= 13.08 v 5 = 65.40 kg.
Volume of Air@ 28.94 kg. of air = 22,400 ml volume at NTP
@65.4 kg. of air =22400 65.4/ 28.94
=50815.8 ml. Air
=50.8158 litres of air