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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ClassificationsFuels 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 fuelsVisit www.gyandharaonline.com

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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


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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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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 :


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i) C + O2 → CO2ii) 2H2 + O2 → 2H2O OR H2 + (O) → H2Oiii) S + O2 → SO2iv) 2CO + O2 → 2CO2 OR CO + (O) →

CO2v) CH4 + 2O2 → CO2 + 2H2Ovi) 2C2H6 + 7O2 → 4CO2 + 6H2Ovii) C2H4 +3O2 → 2CO2 + 2H2Oviii) 2C2H2 + 5O2 → 4CO2 + 2H2O


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Hint to Solve Problems on Calculation of Quantity of AirRequired 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 → 18Visit www.gyandharaonline.com


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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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SrNo

Types 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. Bituminous

4 83 5 15 7 8000 For metallurgical coke, coal gas, boiler, domestic purpose

4. Anthracite

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 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 airVisit www.gyandharaonline.com


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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


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

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fuels ppts 674816

Engineering