Subject Contents

1. Fuel and Combustion

2. Steam Generator/ Boiler

3. Nozzles

4. Steam Turbines

5. Gas Turbines

6. Air Compressors

Recommended Books

i. Applied Thermodynamics for Engineering Technologists

By: T.D. Eastop & A. McConkey

ii. Basic Thermodynamics

By: Reynar Jeol

iii. Thermodynamics An Engineering Approach

By: Yunus A. Cengel & M.A. Boles

iv. Thermodynamics for Engineers

BY: Bhalchandra V. Karlekar

Reference Books

i Text book of Thermal Engineering

By: R.S. Khurmi

ii Thermal Engineering

By: Sarao

What is a Fuel ?

Any material that can be burned to release thermal energy is called a Fuel.

Most familiar fuels consist primarily of hydrogen and carbon.

They are called hydrocarbon fuels and are denoted by the general formula CnHm.

Type of Fuels

i. Liquid (gasoline)

ii. Solid (coal)

iii. Gas (natural gas)

Liquid Fuels

Furnace oil

Light diesel oil

Petrol / Gasoline

Kerosene

Ethanol

LSHS (low sulphur heavy stock)

Liquid Fuels and Calorific value

Fuel Oil Gross Calorific Value (kCal/kg)

Kerosene 11,100

Diesel Oil 10,800

L.D.O 10,700

Furnace Oil 10,500

LSHS 10,600

Contamination of Liquid Fuel

1. Sulphur content:

1. Depends on source of crude oil and less on the refining process

2. Furnace oil: 2-4 % sulphur

Effects: Sulphuric acid causes corrosion

2. Ash content

Inorganic material in fuel

Typically 0.03 - 0.07%

Effects: Corrosion of burner tips and damage to materials /

equipments at high temperature

3. Carbon residue

Tendency of oil to deposit a carbonaceous solid residue on a hot

surface

Residual oil: >1% carbon residue

4. Water content

Normally low in furnace oil supplied (

Solid Fuels

Coal classification

Anthracite: hard and geologically the oldest

Bituminous

Lignite: soft coal

Further classification:

semi- anthracite,

semi-bituminous, and

sub-bituminous

Physical properties

Heating or calorific value (GCV)

Moisture content

Volatile matter

Ash

Chemical properties

Chemical constituents: carbon, hydrogen, oxygen, sulphur

Solid Fuel Contents

1. Moisture content

0.5 to 10% of moisture in fuel

Effects:

Reduces heating value of fuel

Weight loss from heated and then cooled powdered raw coal

2. Volatile matter

Methane, hydrocarbons, hydrogen, CO, other

Typically 25-35%

Easy ignition with high volatile matter

Weight loss from heated then cooled crushed coal

3. Ash

Impurity that will not burn (5-40%)

Important for design of furnace

Ash = residue after combustion

4. Fixed carbon

Fixed carbon = 100 (moisture + volatile matter + ash)

Carbon + hydrogen, oxygen, sulphur, nitrogen residues

Heat generator during combustion

Analysis of Coal

1. Proximate analysis of coal

Determines only fixed carbon, volatile matter, moisture and ash

Useful to find out heating value (GCV)

Simple analysis equipment

2. Ultimate analysis of coal

Determines all coal component elements: carbon, hydrogen,

oxygen, sulphur, other

Useful for furnace design

(e.g flame temperature, flue duct design)

Laboratory analysis

Gaseous Fuels

Classification of gaseous fuels

1. Fuels naturally found in natureNatural gasMethane from coal mines

2. Fuel gases made from solid fuelGases derived from coalGases derived from waste and biomassFrom other industrial processes

3. Gases made from petroleumLiquefied Petroleum gas (LPG)Refinery gasesGases from oil gasification

4. Gases from some fermentation

Comparing of Fuels

Constituents Fuel Oil Coal Natural Gas

Carbon 84 41.11 74

Hydrogen 12 2.76 25

Sulphur 3 0.41 -

Oxygen 1 9.89 Trace

Nitrogen Trace 1.22 0.75

Ash Trace 38.63 -

Water Trace 5.98 -

Advantages of Gaseous Fuel

They are free from all solid and liquid impurities.

Simplest burners systems

Burner systems require least maintenance

Environmental benefits: lowest GHG and other emissions

They do not produce smoke and ash.

They undergo complete combustion with minimum air supply.

A good fuel should have:

Low ignition point

High calorific value

Freely burn with high efficiency, once it is ignited

Not produce harmful gases

Produce least quantity of smoke and gases

Economical , easy to store and convenient for transportation

Calorific Value of FuelsIt is amount of heat given out by complete combustion of one kg of

fuel, solid, liquid or gas.

It is expressed in terms of kJ/kg of fuel.

Types of calorific value

i. Gross/higher calorific value (H.C.V)

ii. Net/lower calorific value (L.C.V)

i. Gross/higher calorific value:

Total amount of heat produce by one kg of fuel

ii. Net/lower calorific value:

Net amount of heat used to produce steam per kg of heat

L.C.V = H.C.V Heat of steam formed during combustion

What is Combustion ?

A chemical reaction during which a fuel is oxidized and a large quantityof energy/ heat is released is called combustion .

Necessary Constituents of Combustion:

i. Fuel

ii. Supporter of combustion(Air/oxidizer)

iii. Ignition / kindling temperature

Ignition temperature for different fuels

Fuel Ignition Temperature (oC)

Gasoline / Petrol 260

Carbon 400

Hydrogen 580

Carbon monoxide 610

Methane 630

The oxidizer most often used in combustion processes is air, which is freeand readily available.

Pure oxygen O2 is used as an oxidizer only in some specialized applications, such as cutting and welding.

Oxygen is the key to combustion

Principle of Combustion

Composition of Air

Types of combustion Process

1. Complete combustion

2. Incomplete combustion

1. Complete Combustion Process

A combustion process is complete if all the carbon & hydrogen in

the fuel burns to form, CO2,& H2O.

All the sulfur (if any) burns to SO2.

It means, all the combustible components of a fuel are burned tocompletion during a complete combustion process.

2. Incomplete Combustion Process

The combustion process is incomplete if the combustion products

contain any unburned fuel or components such as, C, H2, CO, or OH.

Insufficient oxygen is an obvious reason for incomplete combustion,but it is not the only one.

Incomplete combustion occurs even when more oxygen is present inthe combustion chamber than is needed for complete combustion.

Another cause of incomplete combustion is dissociation, which becomes important at high temperatures.(During adiabatic combustion max. Temp. Lower than the calculatedone)

Chemical equations are balanced on the basis of the conservation ofmass principle (or the mass balance),

Stated as follows: The total mass of each element is conservedduring a chemical reaction.

That is, the total mass of each element on the right-hand side of thereaction equation (the products) must be equal to the total mass ofthat element on the left-hand side (the reactants).

Even though the elements exist in different chemical compounds in the reactants and products.

Example:

2H2 + O2 2H2OReactants Product

Theoretical / Stoichiometric Combustion Process

The ideal combustion process during which a fuel is burnedcompletely with theoretical air(stoichiometric air) is called thestoichiometric or theoretical combustion process.

For example,

the theoretical combustion of methane is notice that theproducts of the theoretical combustion contain no unburned

methane and no C, H2, CO, OH, or free O2.

Stoichiometric air:

Minimum mass / volume of air required for completedcombustion of 1kg of fuel is known as Stoichiometric air.

Actual combustion process

In actual combustion processes, it is common practice to use

more air than the stoichiometric amount to increase the chances

of complete combustion or to control the temperature of the

combustion chamber.

Excess Air

The amount of air in excess of the stoichiometric amount is called

excess air.

The amount of excess air is usually expressed in terms of thestoichiometric air as percent excess air or percent theoretical air.

For example,50 percent excess air is equivalent to 150 percent theoretical air,

200 percent excess air is equivalent to 300 percent theoretical air.

Deficiency of Air

Amounts of air less than the stoichiometric amount are called

deficiency of air and are often expressed as percent deficiency of air. For example, 90 percent theoretical air is equivalent to 10 percent deficiency of

air.

Air-Fuel RatioA frequently used quantity in the analysis of combustion processes

to quantify the amounts of fuel and air is the airfuel ratio (AF).

AF represents the amount of air used per unit mass of fuel during acombustion process

It is usually expressed on a mass basis and is defined as,

The mass m of a substance is related to the number of moles N through the relation m= NM, where M is the molar mass.

Example:

The airfuel ratio can also be expressed on a mole basis as the ratioof the mole numbers of air to the mole numbers of fuel. But we willuse the former definition.

The reciprocal of airfuel ratio is called the fuelair ratio.

Example:One k mol of octane (C8H18) is burned with air that contains 20 kmol of O2.Assuming the products contain only CO2, H2O, O2, and N2, determine themole number of each gas in the products and the airfuel ratio for thiscombustion process. Assume the molar mass of air is 29.0 kg/kmol.

The chemical equation for this combustion process,

X, y,z and w represent the unknown mole numbers of the product, theseunknowns are determined by applying the mass balance to each of theelements.

For Carbon(C): 8 = x x = 8For Hydrogen (H): 18 =2y y = 9For Oxygen (O) 20x2=2x+ y+2z z =7.5For Nitrogen (N2): (20)(3.76) =w w = 75.2

Now, substitute the values and yields

Note that the coefficient 20 in the balanced equation above represents the Number of moles of Oxygen, not the number of moles of air.

The latter is obtained by

20x3.76 =75.2 moles of nitrogen to the 20 moles of oxygen.

Therefore, total moles of air is 95.2

The air-fuel ratio (AF) is determine by

AF = 24.2 kg air / kg fuel

It mean 24.2 kg of air is required to burn each kilogram of fuel during thisCombustion process.

Combustion Equations of Solid Fuels

When carbon burns in sufficient quantity of oxygen and produced carbon dioxide along with large amount of heat.

C + O2 = CO2+Heat

1mol + 1mol =1mol

12kg + 32kg = 44kg

1kg + 8/3kg = 11/3

It means that 1kg of carbon requires 8/3 kg of oxygen for its complete combustion and produces 11/3 kg of carbon dioxide gas.

If sufficient oxygen is not available, then combustion process is not complete and produces carbon monoxide instead of carbon dioxide.

2C + O2 = 2CO

2mol + 1mol =2mol (by volume)

2x12kg + 2x16kg =2x28kg (by mass)

1kg + 4/3kg = 7/3

It means that 1kg of carbon requires 4/3 kg of oxygen and produces 7/3 kg of carbon monoxide.

If carbon monoxide burn further and produces carbon dioxide

2CO + O2 = 2CO2

2mol + 1mol =2mol (by volume)

2x28kg + 2x16kg =2x44kg (by mass)

1kg + 4/7kg = 11/7

It means that 1kg of carbon monoxide requires 4/7 kg of oxygen and

produces 11/7 kg of carbon dioxide.

When sulphur (if any) burns with oxygen and produced sulphur dioxide.

S + O2 = SO2

1mol + 1mol =1mol (by volume)

32kg + 2x16kg = 64kg (by mass)

1kg + 1kg = 2kg

It means that 1kg of sulphur requires 1 kg of oxygen for its complete combustion and produces 2 kg of sulphur dioxide gas.

Combustion Equations of Gaseous Fuels

Gaseous fuels are generally measured by volume (m3) than by mass

When carbon monoxide burns with oxygen and produced carbon dioxide.By Volume:

2CO + O2 = 2CO2

2volumes + 1vol =2vol

2m3 + 1m3 = 2m3

By mass:

2CO + O2 = 2CO2

2x28kg + 1x32kg= 2x44kg

1kg + 4/7 kg = 11/7kg

It means that 1kg of carbon monoxide requires 4/7 kg of oxygen and produces 11/7 kg of carbon dioxide gas.

When hydrogen burns with oxygen produces:By Volume:

2H2 + O2 = 2H2O

2volumes + 1vol =2vol

2m3 + 1m3 = 2m3

By mass:

2H2 + O2 = 2H2O

2x2kg + 1x32kg= 2x18kg

1kg + 8 kg = 9 kg

It means that 1kg hydrogen requires 8 kg of oxygen and produces 9 kg of water or steam.

For burning of CH4 with O2,

CH4 + 2O2 =CO2 + 2H2O

16kg + 2x32kg =44kg + 2x18kg

1kg + 4 kg = 11/4 kg +9/4kg

Now, consider 1kg of a Fuel, which contains C,H and S

Let, Mass of Carbon = C kg

Mass of Hydrogen = H kgMass of Sulphur = S kg

We know that for the complete combustion of fuel requires sufficient Oxygen for constituent element such as,

1kg Carbon (C) 8/3 kg of Oxygen

1kg Hydrogen (H) 8 kg of Oxygen

1kg Sulphur (S) 1 kg of Oxygen

Therefore, total oxygen required for complete combustion of 1 kg of fuel

= 8/3 C +8 H2 +1 S kg

If some quantity of oxygen is already present in the fuel, then

=[8/3 C +8 H2 + S kg]- O2 kg

We know that oxygen is obtained from the atmospheric air for the combustionWhich mainly consists of Nitrogen, Oxygen and negligible quantity ofCarbon dioxide and inert gases (Neon, Argon and Krypton).

For the calculations these negligible quantities are not considered.Therefore,

N2 = 77% and O2 = 23% By massN2 = 79% and O2 = 21% By volume

It is obvious that for obtaining 1kg of oxygen, amount of air req.

=100/23=4.35kg

=100/23[(8/3 C+8H2+S)-O2]

Example:A fuel has the following composition by mass,Carbon 86%, Hydrogen 11.75% and Oxygen 2.25%.Calculate the theoretical air supply per kg of fuel and the mass ofproducts of combustion per kg of fuel.

Solution:

C=0.86 kg, H2 = 0.1175kg, O2 = 0.0225kg

(i) Theoretical air supply per kgwe know that

Airtheo = 100/23[(8/3C+8 H2 +S)- O2] kg

Substitute the values Airtheo = 100/23[(8/3x0.86+8 x0.1175 +0)- 0.0225] kg

Airtheo = 13.96 14kg

(ii) Mass of products of combustion

We know that,

C+O2 =CO22H2 + O2 = 2H2O

We know that 1kg of carbon produces 11/3 kg of carbon dioxide and 1kg of hydrogen produces 9 kg of water.

Therefore, total mass of the products of combustion,

=11/3xC+9H2kg

= 11/3x0.86+9x0.1175= 4.21 kg

Example:Calculate the stoichiometric A/F ratio for the combustion of a sample of dry Anthracite of the following composition by massC90% ;H 3%; O2 2.5% ; N1% ; S 0.5% and ash 3%Determine the A/F ratio and the dry and wet analysis of products of combustionBy volume, when 20% excess air is supplied.

Constituent Mass/kg Combust:Equation

Oxygen (O2) req./ kg

Products / kg

C 0.9 C+O2 CO212kg+32 44 kg

0.9x 32/12= 2.4kg

0.9x 44/12 =3.3kg of CO2

H 0.03 2H2+O2 2H2O1kg + 8kg 9kg

0.03x8=0.24kg

0.03x9 =0.27 kg H2O

O 0.025 0.025kg -0.025kg

N 0.01 0.01kg N2

S 0.005 S+O2 SO232kg+32kg=64kg

0.005x32/32=0.005kg

0.005x64/32=0.01kg SO2

Ash 0.03

Total 1.0 2.62kg

From table O2 required per kg of coal =2.62 kg

Therefore,

(i) Air required per kg of coal = 2.62/0.233= 11.25kg

(Assume 23.3% O2 by mass)

(ii) N2 associated with this air =0.767x11.25 =8.63kg

Total N2 in product =8.63+0.01=8.64kg

The stoichiometric A/F ratio = 11.25/1

For an air supply which is 20% excess

We know that % excess air = A/F(Actual)- A/F(Stioch)/ A/F (Stioch)

A/F(Actual)=11.25+20/100x11.25=1.2x11.25

A/F(Actual)=13.5/1

Therefore,

N2 supplied =0.767x13.5=10.36kg

O2 supplied =0.233x13.5=3.144kg

In the products, we have

N2 =10.36+0.01= 10.37kg

and excess oxygen

O2 =3.144-2.62=0.524kg

Example:The volumetric analysis of a flue gas is CO214% ,CO 1%, O2 5% and N280%Calculate the flue gas composition by mass. Solution:By volumeCO2= 0.14m

3 ,CO=0.01m3 , O2 =0.05m3 and N2= 0.8m

3

The volumetric analysis may be converted into mass analysis by Completing following table

Constituent Volume of gas (m3)

(a)

Molecular mass

(b)

ProportionalMassc=axb

Mass in/kgd=c/c

%by massdx100

CO2 0.14 44 6.16 6.16/30.44=0.202

20.2%

CO 0.01 28 0.28 0.28/30.44=0.009

0.9%

O2 0.05 32 1.60 1.60/30.44=0.053

5.3%

N2 0.80 28 22.40 22.40/30.44=0.736

73.6%

Total 1.00 c=30.44 1.00 100

The flue gas composition by mass is given in the last column

CO2=20.2%CO = 0.9%O2 =5.3%

N2 = 73.6%

Example:A flue gas has the following percentage composition by massCO213.3% ,CO 0.95%, O2 8.35% and N277.4%

Calculate the flue gas composition by volume. Solution:By volumeCO2= 13.3% ,CO=0.95% , O2 = 8.35% and N2= 77.4%The mass analysis may be converted into volumetric analysis by Completingfollowing table

Constituent % mass analysis

(a)

Molecular mass

(b)

Proportionalvolumec=a/b

Volume in (m3)

d=c/c

%volumetricdx100

CO2 13.3 44 13.3/44=0.302

0.302/3.357=0.090

9.0

CO 0.95 28 0.95/28=0.034

0.034/3.357=0.010

1.0

O2 8.35 32 8.35/32=0.261

0.261/3.357=0.078

7.8

N2 77.4 28 77.4/28=2.76

2.76/3.357=0.822

82.2

Total 100.0 c=3.357 1.00 100

The flue gas composition by volume is given in the last column

CO2=9%CO = 1%O2 =7.8%

N2 = 82.2% Ans