AIT PUNE

Unit-4
Energy Sources

Dr. Sonali Bhosle

• Contents
Introduction (definition, classification of fuel based on chemical reactions
and characteristics of an ideal fuel),
Calorific value (CV): Higher calorific value (HCV) and Lower calorific
value (LCV),
Determination of Calorific value: Principle, construction and working of
Bomb calorimeter and Boy’s gas calorimeter and numerical,
Solid fuel: Coal: Analysis of Coal-Proximate and Ultimate analysis,
numerical,
Gaseous fuel: Hydrogen gas as a future fuel
Alternative fuels: Power alcohol and biodiesel.
Lithium Ion Battery: construction , working, advantages and applications.
Definition:

Fuel: Fuel is a combustible substance, containing carbon

as main constituent, which on proper burning gives large
amount of heat, which can be used economically for
domestic and industrial purposes.
Fuel + O2 (air) high Products + heat
temperature

high
C + O2 (air) temperature C O2 + heat

high
H2 + ½ O2 (air) temperature
H2O + heat
 Classification of Chemical Fuels

Primary or Natural Secondory or derived

solid liquid Gaseous solid liquid Gaseous

Wood Crude oil Natural Gas coke Tar Coal gas
Peat Charcoal Kerosene Water gas
Lignite Petroleum Diesel Oil gas
Coal briquette Petrol Biogas
dung Gasoline Blast furn.
LPG gas
Coke oven
gas
Easy, riskfree transportation, no volatile matter, no ash, it should be
harmless to the environment ,,etc
Calorific value
It is defined as the total quantity of heat liberated when
a unit mass of a fuel is burnt completely

litre

S.I. J / kg J/ m3

1 kcal/kg = 4.187 kJ/kg

1 kcal/m3 = 4.187 kJ/m3

Latent heat of water is 587 kcal/kg or 2450 kJ / kg

 Types of Calorific Value
1 ) Gross Calorific Value (GCV)
Or
Higher Calorific Value (HCV)

H2 + ½ O2 (air) high H2O + heat

temperature
steam

Steam Condensation/ coolwater + latent heat of water vapour

2. Net Calorific Value (NCV)
or
Lower Calorific Value (LCV)
Relation between HCV and LCV

Cal / gm or kcal / kg

- 2450 kj / kg
100
1) Calculate NCV of a Fuel which has 12 % Hydrogen and its
GCV as 6500 cal /gm.
(Given: the latent heat of steam is 580 cal / gm)

2) Calculate NCV of a Fuel which has 20 % Hydrogen and 12 %

moisture and its GCV is 3370 cal /gm.
(Given: the latent heat of steam is 580 cal / gm)

3) Calculate LCV of a Fuel which has 8 % Hydrogen and 20 %

moisture and its GCV is 2550 cal /gm.
(Given: the latent heat of steam is 587 cal / gm)
4) Calculate NCV of a Fuel which has 12 % Hydrogen and its
GCV as 45 MJ/kg.
(Given: the latent heat of steam is 2450 kj / kg)

5) Calculate NCV of a Fuel which has 5 % Hydrogen and 10 %

moisture its GCV as 33.5 MJ/kg.
(Given: the latent heat of water vapour is 2.450 MJ / kg)

6) Calculate difference between GCV and NCV of a Fuel

which has 9 % Hydrogen.
(Given: the latent heat of water vapour is 587 cal/gm)
BOMB CALORIMETER

Bomb
Bomb calorimeter

construction: write
two sentences abou
diagram
Working:
 A Known mass (about 0.5 to 1.0 g) of the given fuel is taken in clean crucible.
 The Crucible is then supported over the ring.
 A Fine magnesium wire, touching the fuel sample, is then stretched across
the electrodes.
 The bomb lid is tightly screwed and bomb filled with Oxygen to 25
atmospheric pressure.
 The bomb is then lowered into copper calorimeter, containing a known mass
of water.
 The stirrer is worked and initial temperature of the water is noted.
 The electrodes are then connected to 6 V battery and circuit completed.
 The sample burns and heat is liberated Uniform stirring of water is continued
and the maximum temperature attained is recorded.
Calculations:

Let, mass of fuel in gm = x gm

Mass of water in calorimeter = W gms
Water equivalent of calorimeter set = w gms
Gross calorific value of fuel = L calories/gm
Rise in temperature of water = t2 - t1
Heat liberated by burning fuel = Heat absorbed by water
and calorimeter
(W+w) (t2-t1)
GCV = L = cal/gm
x

W, w, x is expressed in units of grams

 (W+w) (t2-t1)
1) GCV or HCV (L) = cal/gm
x

2)
= HCV - Cal / gm or kcal / kgm
100
Corrections in formula: for accurate calculations of GCV,
following corrections are made

1) Fuse wire correction (f)

2) Acid correction (a)
3) Cooling correction (tc) = t. dt

(W+w) (t2-t1 + tc) – (a + f) cal/gm

GCV or L =
x

- Cal / gm or kcal / kgm

100
 Numerical on Bomb calorimeter
1) Calculate NCV of a fuel containing 7.5 % Hydrogen
with the following data of bomb calorimeter experiment
Mass of fuel = 0.928 gm
Water equivalent of calorimeter set = 850 gm
Mass of water in calorimeter = 2050 gm
Rise in temperature = 2.78 OC
2) Calculate NCV of a fuel containing 3.5 % Hydrogen with
the following data of bomb calorimeter experiment
Mass of fuel = 1.5 gm
Water equivalent of calorimeter set = 660 gm
Mass of water in calorimeter = 1697 gm
Rise in temperature from = 21.5 OC to 23.7 OC
3) Calculate GCV in kj / kg of 0.75 gm of a fuel containing
80 % c when burn in bomb calorimeter increased the
temperature of water from 27.3 to 29.7 oC. The calorimeter
contains 250 gm of water and its water equivalent is 150 gm.
 acid correction

4) 0.60 gm of a fuel on complete combustion in a bomb

calorimeter increased the temperature of water in the
calorimeter from 26.3 to 28.4 oC. The calorimeter contains
500 gm of water and water equivalent of the calorimeter is
1500 gm. The corrections are as follows:
Cooling correction: 0.02 oC
Fuse wire correction: 15 cal
Acid correction : 65 cal
If the fuel contains 5 % Hydrogen , calculate GCV and NCV
of the fuel.
 Boy’s Calorimeter
Principle: The calorific value of liquid, gaseous fuel is determined by
complete combustion of known mass of fuel at constant volume and absorption
of heat by surrounding water.
 Construction: 1)A combustion chamber surrounded by water tube made of
copper with two thermometers T1 and T2 attached.
2) There is a gas burner in which known volume of gas is burnt at known pressure.
3) Pressure rate of 3 – 4 litre / minutes.

Working: 1) Water is allowed to pass through the water tube and

temperature of incoming and outgoing water is recorded.
2) When fuel is burnt, water taken all the heat produced
3) The set up is enclosed in an insulated container

Calculations:
Where: V – volume of
1) HCV or L = W * ( T2 - T1 ) fuel burnt, m – mass of
V
water condensed, W –
2) NCV = HCV – m * 587 kcal / m3
V mass of cooling water
 Numericals

1) Calculate NCV and GCV of the fuel. The following

observations were noted in the Boy’s gas calorimeter
experiment
Volume of the gas burnt at STP = 0.1 m3,
Mass of cooling water used = 28 kg,
Rise in temperature of water = 8.2 oC
Mass of steam condensed = 0.05 kg.

Calculations:
1) HCV or L = W * ( T2 - T1 )
V

2) NCV = HCV – m * 587 kcal / m3

V
 Wood / Classification of Coal

Aerobic decay
Peat

Long time,
alkaline condition
Anaerobic bacteria Anaerobic bacteria
Anthracite Acidic condition

Longer time
Lignite
Bituminous
 Classification of Coal

wood %C, Calorific value, density,

lusture, hardness, Moisture, % of N, H, O,
black colour intensity and S, ash
Peat

Lignite

Bituminous

Increases Decreases

Anthracite
The process of conversion of
wood into coal is coalification.
Coal contains mainly Carbon
atoms with SP 2 hybridization.
Atoms like O, S, N, H are
covalently bonded to carbon
atoms.
Coal also contains water and
mineral particles entrapped.
 Peat (turf)

Peat (turf) is an accumulation of partially decayed vegetation

(brown colour)
 Lignite (Brown coal or rosebud coal)

Lignite brequette

Brownish black in colour

Denser than lignite

bituminous More % of carbon
CV more than lignite
 Anthracite coal
 It is highest ranking grade coal
 Calorific value = 8700 cal/gm (92-98 % carbon)
 It burns with non-smoky blue flame
 It is lustrous black and hard
 It is costly

It is used for
1. Industrial purpose:
2. As metallurgical fuel
3. Making electrodes
4. High temperature heating
Purpose of analysis
1. To decide price of coal
2. To determine quality
3. Specify particular purpose
4. Calculate C.V of coal
5. Design the furnace
Defination: study of analysis of coal sample in which moisture, ash,
volatile matter and fixed carbon are found out
3.
4.
 Significance of proximate analysis

moisture, : % more it decreases the CV, Lower is the moisture

better is the coal

ash, : Lesser the amount of ash then better is the quality

volatile matter : Lesser the amount of VM then better is the

quality of coal

fixed carbon : Better % of FC better is the quality of coal

1)
2.2 gm of coal sample was heated in a silica crucible in an electric
oven at 110 oC for 1 hour. The weight of the residue was 1.98 gm.
The residue was then heated in a silica crucible covered with a lid
at 950 oC for exactly 7 minutes. The residue weighed 1.2 gm. The
crucible was then heated without any lid to a constant weight of
0.240 gm. Calculate the fixed carbon of the coal sample.
2)
1.50 gm of coal sample was heated in a silica crucible in an
electric oven at 110 oC for 1 hour. The weight of the residue was
1.43 gm. The residue was then heated in a silica crucible covered
with a lid at 950 oC for exactly 7 minutes. The residue weighed
1.03 gm. The crucible was then heated without any lid to a
constant weight of 0.12 gm. Calculate the fixed carbon of the coal
sample.
3)
1.20 gm of coal sample was heated in a silica crucible in an
electric oven at 110 oC for 1 hour. The weight of the residue was
1.15 gm. This residue was then burned completely to a constant
weight of 0.09 gm. In an another experiment 1.20 gm of the same
coal samples was heated in a crucible with vented lid at 950 oC
for 7 minutes. After cooling the residue weighed 0.82 gm.
Calculate the fixed carbon of the coal sample.
4)
1.29 gm of coal sample was heated in a silica crucible in an
electric oven at 110 oC for 1 hour. The weight of the residue was
1.21 gm. This residue was then burned completely to a constant
weight of 0.09 gm. In an another experiment 1.29 gm of the same
coal samples was heated in a crucible with vented lid at 950 oC
for 7 minutes. After cooling the residue weighed 1.10 gm.
Calculate the fixed carbon of the coal sample.
Principle:
Method:
Principle: N in coal gets converted to ammonium sulphate, by action of hot
conc. H2SO4 and then on treatment with alkali solution, equivalent amount of
NH3 is liberated.

Method:

% N = (Blank – Back reading)* normality of NaOH * 1.4

weight of coal sample

1000 ml of 1N H2SO4 = 14 g of nitrogen

% N = volume of acid consumed * normality of NaOH * 1.4

weight of coal sample

3. Estimation of sulphur
O2 ½ O2 H2O BaCl2
S SO2 SO3 H2SO4 BaSO4

Principle:
Sulphur present in coal converts to first SO3 which is soluble in water
forming H2SO4. H2SO4 is then converted to BaSO4 precipitate when
treated with BaCl2.
32 g of sulphur forms 233 g of BaSO4

% S = 32 ₓ Weight of BaSO4 ₓ 100

233 Weight of Coal

Therefore % of Oxygen can be calculated by

% Oxygen = 100 - % [ C+ H + N + S + Ash]

Significance of Ultimate analysis analysis

C % Is more then better is the coal quality, increases CV

H % is in the form of H2O n VM it decreases CV.....is less

better is the coal

N % does not burn during combustion ..negligible

amount N%...no effect CV

S % create pollution ..acid rain..decreases CV..% S is low

then better is the coal

O% is in the form H2O....decreases CV..O is low then

better is the coal
 Numerical

1) Calculate C, H, N and S % present in the coal sample.

1) 4.8 gm of sample was analyzed by sulphur or Eschka method and gave
0.466 gm of BaSO4.
2) 0.3 gm of sample on combustion gave 0.81 gm of CO2 and 0.0318 gm of
H2O.
3) 1.2 gm of coal in KJ method ammonia absorbed in 100 ml 0.1 N H2SO4,
solution required 10 ml of 0.1 N NaOH for complete neutralization. The
blank reading was 50 ml.
2) Calculate C, H, N and S % present in the coal sample.
1) 0.55 gm of sample was analyzed by sulphur or Eschka method and
gave 0.025 gm of BaSO4.
2) 0.25 gm of sample on combustion was found to increase in weight of
CaCl2 U tube by 0.12 gm and KOH tube by 0.57 gm.
3) 1.6 gm of coal in KJ method ammonia absorbed in 50 ml H2SO4,
solution required 14 ml of 0.1 N NaOH for complete neutralization of
acid in back titration. The blank reading was 25 ml.
Calculate Oxygen % from 1.5 gm of coal sample when it contain Carbon
25 %, Hydrogen 12 %, Nitrogen 5 %, Ash 16 % and Sulphur 0.25%.

Calculate Oxygen % from 3.3 gm of coal sample when it contain Carbon

37 %, Hydrogen 16 %, Nitrogen 2.5 %, Ash 13 % and Sulphur 0.68%.

Calculate S % in coal when 1.7 gm of coal is combusted in Bomb

calorimeter. Solution from Bomb on treatment with BaCl2, forms 0.9255 gm
BaSO4. Answer = 7.4 %

Calculate S % in coal when 0.55 gm of coal is combusted in Bomb

calorimeter. Solution from Bomb on treatment with BaCl2, forms 0.025 gm
BaSO4. Answer = 0.62%
Calculate C and H % in coal when 3.0 gm of sample on
combustion gave 1.15 gm of CO2 and 0.033 gm of H2O.

Calculate C and H % in coal when 2.7 gm of sample on

combustion gave 1.9 gm of CO2 and 0.0279 gm of H2O.
 Alternative fuel 1)Power alcohol
When ethyl alcohol C2H5-OH is blended with petrol (25%) and
is used as fuel in internal combustion engine, it is called as
Power Alcohol.

Manufacture of
ethyl alcohol

Advantages: Disadvantages:
High octane number Lowers the CV because of
Burns clean ethanol
Capable of absorbing moisture High surface tension
Alcohol reduces chances of overheating of engine May cause corrosion
 2) Biodiesel / Biofuel
Chemically biodiesel obtained from vegetable oil (like soyabean or palm oil) or
animal oils by transesterification, it is the mixture of methyl esters of long chain
carboxylic acids.
Its mainly used as fuel for diesel engine

Advantages:
Reaction
• Non toxic, non conventional Limitations:
R= CH3COO17H35
• Higher flash point, safe Highest cloud point
storage
May dissolve gaskets
• Highest cetane number
Becomes gummy
• Burns clean because of oxygen
 Hydrogen Gas as a future fuel

Properties:
Non toxic , clean burning
High CV, most abundant element on earth

Applications:
• Used in Rocket fuel
• Used in ICE / internal combustion
engine
• Electricity generation
• Used in electrodes
 production of Hydrogen gas

Industrial or commercial method

(A) Steam reforming of methane

 Hydrogenation in which methane is mixed with steam and passed

over nickel catalyst at 700 - 1100 oC to yield mixture of CO and H2
CH4 + H2O CO + 3H2
Water gas shift reaction
Further reaction water gas produces more H2
At lower temperature 400 oC and using FeO catalyst
CO + H2O CO2 + H2
 (B) Steam reforming of coke (coal and steam reaction)
 Coke is obtained by carbonization of coal at 1000 oC and Ni catalyst
to yield mixture of CO and H2
C + H2O CO + H2
Water gas shift reaction
Further reaction water gas produces more H2
At lower temperature 400 oC and using FeO catalyst
CO + H2O CO2 + H2
Liquid CO2 removed by following ways
1) Scrubbing with alkali solution
2) Treating with K2CO3
3) Evaporation process
 Storage and transportation of Hydrogen
1) Physical storage: in the form of compressed H2 , liquefied H2 and adsorption on
porous carbon material
2) Chemical storage: in form of metal hydride, sodium alanates (NaAlH4)
3) Difficulties in Transportation: a) H2 is the lightest gas because of its low
molecular weight 2 i.e. it occupy volume of 22.4 lit. at STP
b) High pressure H2 storage in steel cylinders causes decarburization and steel
becomes brittle, may result in explosion
c) Ignition temperature of H2 is lowest of all gases and it is highly inflammable.
d) It is difficult to liquefy, as its boiling point is very low i.e. -253 oC
 Lithium Ion Battery (LIB)
Definition: “The batteries where Lithium ions are used instead of lithium
metal and movement of lithium ion through electrolyte takes place from one
electrode to another electrode”
e.g. Lithium cobalt oxide [LiCoO2]
• It has high volumetric and
energy densities
• It uses metal oxide as
cathode (+ electrode)
• It uses porous carbon
as Anode ( negative electrode)
• And Electrolyte as
conductor
 During discharge Lithium ions flow from anode to cathode through
electrolyte and separator
 Charging reverse the directions and ions flow from cathode to anode
 Cell reaction in the battery is merely the migration of lithium ions between
+ ve and –ve electrode
 No chemical change during process

Construction:
 Anode is made of carbon electrode with thin copper coil

 Cathode is [LiCoO2] with thin Aluminium foil as current collector

 Porous polyethylene or polypropylene film used as separator
 Lithium salt with organic solvent used as electrolyte
 This electrolyte transport Lithium ion to cathode during discharge of
battery
 Working

 During discharge Lithium ions flow from

anode to cathode through electrolyte  During charging Lithium ions flow from
 Electron moves Anode to cathode cathode to anode
 Current cathode to anode  Charge process referred as intercalation
 Discharge is referred as de-intercalation
 During discharging

Reactions are revered during charging

 Applications:
 Advantages:  Disadvantages:  Li-ion batteries used in cameras,
 high energy density  Expensive calculators
 Lighter in weight  Not available in standard  Cardiac pacemakers and
 High voltage 4 V cell types implantable device
 safety, more resistance  Instruments, radios, Tys
 Fast charge and  Laptops, computers, mobile
discharge rate phones and aerospace
applications
Prepare your Mind map

Thank you