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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12/13/2010 Dr P B Dwivedi, NMIMSUniversity, Mumbai 7

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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12/13/2010 Dr P B Dwivedi, NMIMS

University, Mumbai

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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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University, Mumbai

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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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University, Mumbai

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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

fuels ppts

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