Fuels ppts 674816

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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 fuelsV
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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
Crude oil,
Vegetable oils
Petrol, kerosene,
gas oil, coal tar,
alcohol
Liquid
fuels
Natural gas Producer gas, coke-oven
gas, water
gas, blast furnace
gas, compressed
butane gas, LPG
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.
- efficiency of fuel: how much heat it produces
(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 raise
the temperature of one pound of water from 60° F to
61° F.
• The Calorie: the heat required to raise the temperature
of one kg of water from 15°C to 16°C.

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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 produced when one unit of the fuel
has been burnt completely and the combustion
products have been cooled to 16°C or 60°F.

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• LCV: is the net heat produced when unit mass or
volume of fuel is completely burnt and products are
allowed to escape.
• Net or Low C.V.= Gross C.V. – loss due to water
formed
• Or Gross C.V – Mass of hydrogen ´ 9 ´ 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. Davies
formula.
• 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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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 by fuel = heat absorbed by water, apparatus
• so, xL = (W+w)(t2-t1)
• L = (W+w)(t2-t1)/x cal/g or kcal/kg

• 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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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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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 required for the complete
combustion of compounds. The commonly involved
combustion reactions are :

i) C + O2 ® CO2
ii) 2H2 + O2 ® 2H2O OR H2 + (O) ® H2O
iii) S + O2 ® SO2
iv) 2CO + O2 ® 2CO2 OR CO + (O) ®
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CO2
v) CH4 + 2O2 ® CO2 + 2H2O
vi) 2C2H6 + 7O2 ® 4CO2 + 6H2O
vii) C2H4 +3O2®
2CO2 + 2H2O
viii) 2C2H2 + 5O2 ® 4CO2 + 2H2O

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Hint to Solve Problems on Calculation of Quantity of Air
Required for Combustion of Fuel :
• 1. 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 ® CO2
by weight (12 gm) + (32 gm) (44gm)
by volume 1 1 1
2H2 + O2 ® 2H2O
by volume 2 1 2
S + O2 ® SO2
by volume 1 1 1

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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 by
knowing 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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Calculate the weight and volume of air required
for complete combustion of 5 kg. coal with
following compositions, C = 85%; H = 10%;
O = 5%
Soln. :
Combustion reactions :
C + O2 ® CO2
12 + 32 ® 44
H2 + O2 ® H2O
2 + 16 ® 18

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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 ´32/12 = 2.26 kg.
H = 0.1 0.1 ´ 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 ´ 100/23
= 13.08 kg. per 1 kg. coal.
Air required for 5 kg. of coal
= 13.08 ´ 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

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Sr
No
Types
of Coal
Classification of Coal
Moisture
of Air
Dried At
40°C
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

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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 105°C to 110°C 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
• Decreases calorific value of coal
• Takes away heat in the form of latent heat

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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 at 950°C ± 20°C
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
• Decreases calorific value
• Forms smoke and pollutes air

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Ash (non combustible matter)
• A known weight of sample is taken in a crucible and
the coal is burnt completely at 700°C – 750°C 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 crucible´ 100]/ wt
of coal taken
• Reduces calorific value as it is non burning part
• Ash disposal is a problem

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Fixed carbon
% of Fixed carbon = 100 – (% of moisture + % of ash +
% of volatile matter)
• In any good sample of coal, the percentages of
moisture, 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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Determination of C & H
• Accurately weighed coal sample is burnt in a current
of oxygen in a combustion apparatus, which is heated
to about 350°C.
• Carbon and hydrogen of coal are converted into water
vapour and carbon-dioxide. The products of
combustion are absorbed in anhydrous CaCl2 and
KOH tubes respectively of known weights.
• After complete absorption of H2O and CO2, the tubes
are again weighted.

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C + O2 ® CO2
12 parts ® 44 parts
2H2 + O2 ® 2H2O
4 parts ® 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
´100]/wt of coal taken × 36

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Determination of Nitrogen
Nitrogen is calculated by Kjehldals Method.
The nitrogen is converted to NH3 and passed through a known volume
of standard acid. On neutralization, the excess acid is back titrated
with a base.
1000 ml of x N acid 14 gm of Nitrogen
% N = [volume of acid consumed in neutralizing NH3 × N x 14 x 100
wt of coal taken x 1000
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Determination of Sulphur
% Sulphur: [wt of BaSO4 obtained × 32×
100]/wt of coal taken × 233
Determination of 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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Importance of Ultimate analysis:
• Carbon: Greater the % carbon, better is the quality and
calorific value of coal
• Hydrogen: most of hydrogen is in form of moisture and
volatile matter. Only a small % is combustible, hence it
decreases C.V. Smaller the H% better is quality of coal
• Nitrogen: does not burn, hence it has no C.V. Negligible
N% is good coal
• Sulphur: it increases C.V, but causes Sox pollution.
Hence lower S% is better
• Oxygen: most of oxygen is in form of moisture, hence it
decreases C.V. Smaller the H% better is quality of coal *

Fuels ppts 674816

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