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FUEL TECHNOLOGY

Definition

Fuels are organic combustible substances used solely or mainly for the production of

useful heat. They may be divided into the three normal classes, solid, liquid and

gaseous fuel.

Solid fuels

The more important of these one:

Natural

• Wood

• Peat

• Lignites

• Hard coals – ranging from bituminous coals to anthracites.

Artificial

• Wood charcoal

• Peat charcoal

• lignite briquettes

• lignites coke

• coal briquettes – uncarbonized.

• coal briquettes – carbonized.

• Low temperature coke

• Medium temperature coke

• High temperature coke

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Liquid Fuels

May be divided into main classes, based on their utilization viz:

1. Light oils or spirit; suitable for use with internal combustion engines and jet

engines.

2. Heavy oils; suitable mainly or exclusively for burning in furnaces.

Oils for Spirits

Suitable for use in engines oils,

(a) The higher, more volatile, factors obtained by distilling or cracking natural

petroleum oils, shale oils and related natural deposits.

(b) The light fractions obtained by the hydrogenation of coal or coal tar , or heavy oil

residues.

(c) The light fractions obtained by the synthesis of hydrogenation by the Fischer

– Tropsch process.

(d) Alcohol, particularly methyl and ethyl alcohol, obtained by synthesis or by

fermentation process.

(e) Benzole, obtained by the distillation of coal or by extraction from coal gas.

Uses

Gasoline or petrol (natural/synthetic)

- benzole

- alcohol

{spark ignition engine}

Higher Fractions

• shale oil

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• synthetic oil

• (under kerosine / naptha)

- jet turbines

Lighter Fractions

• Petroleum oil

• Shale oil

• Synthetic oil

- diesel engines and small furnaces.

Furnaces Oils

Which are heaviest grades of petroleum oils/cracked oil whereby no more valuable

use can be found.

Gaseous Fuels

Classification:

1. Normal Gas – associated with petroleum oil deposits with coal seams or with

the decay of waste organic matter.

2. Manufactured Gases:

(a) From wood by distillation or carbonization.

(b) From peat by distillation or carbonization peat gas.

(c) From coal by carbonization – coal gas or by gasification.

i. In any producer gas

ii. In any and steam – water gas

iii In oxygen and steam – Lurgi Gas

Or by hydrogenation or as a by product of the reduction of ores blast furnace gas.

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(d) from petroleum and oil shale

- By cracking – refinery gas

- By hydrogenation – oil gas

- By partial oxidation – oil gas

(e) from carbides with water – acetylene.

Definitions of Units

Calorific value – the heat evolved by the combustion of unit quantity of the fuel.

For solid and liquid fuels, kcal/kilogram – the number of kilo calories evolved by the

combustion of 1 kg of fuel.

For gaseous fuels – kg-cal/m3 whereby temperature and pressure and immediately

of the gas should be specified to prevent ambiguity.

Gross calorific value (G C.V.)- The heat evolved when all the product of combustion

are cooled to atmosphere temperature, as in a bomb calorimeter. This includes the

sensible and latent heat of evaporation of water in the products of combustion.

Net calorific value (net C.V.) – is the gross calorific value, less the sensible and

latent heats of the water in the products of combustion when cooled to 15.50 C. the

value of this deduction is 586 cal/g of water condensed.

Thermal capacity / specific heat – the quantity of heat required to produce unit

change of temperature in unit mass of a substance.

Alternatively, specific heat is the ratio of water at 15.50 C. (since the thermal

capacity of water at 15.50Cis 1.000)

For gases Cv and Cp are defined accordingly.

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For coal other properties used are:

- Moisture content : Free – the % lost when moist, ground, coal is

allowed to reach equilibrium with the atmosphere at 15.50 C.

- Moisture content : fixed – the % of moisture present in the air- dried

coal. (called interest/equilibrium moisture)

- Ash content - % of residue obtained when coal is burned in air at 8000

C in a muffle furnace under standard conditions.

- Volatile matter content - % of products evolved when coal is heated in

a covered crucible to a temperature of 9250 C under standard

conditions.

- Fixed carbon = 100 – (the sum of % ash, volatile matter and moisture.)

PEAT

Peat is a brown fibrous mass of partially decayed plant material that has

accumulated in silt under water logged conditions. Aerobic bacteria near the surface

(oxidizing conditions) and increasingly anaerobic (reducing) with increasing depth

are the responsible agent for decay.

Types and composition

Composition depends upon type, depth in the deposit, and age. Bog peat consists

mainly of mosses. Forest peat consists of the decayed products of massive trees.

Moisture

Wet peat > 95% H2O

Cut peat 80% - 90% H2O

Air-dried peat 25% H2O

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Ash

Usually above 3.0%, > 10% in areas subject to loading.

Calorific value, average and dry basis 4500 – 5000 cal/g.

Resin and Waxes

Montan wax – varies from 3-12%

Resin content of wax – varies from 10 – 45%

COMBUSTION CHARACTERISTICS OF PEAT

1. Low C.V and high proportion of moisture reduces furnace temperature and

efficiency.

2. Low bulk density, reduces capacity of furnace and increases storage and

transport capacity.

3. Friable nature causes appreciable losses on handling.

COMBUSTION PROPERTIES AND CALCULATIONS

a. Calorific value

The most important property of fuel gas possesses is the energy liberated as

heat when it is burned. This may be expressed as the Heat of Formation of its

combustion products on a modern basis

In the fuel industry however this property is much more commonly expressed

as the calorific value which is the quantity of heat released by combustion in a

calorimeter of unit quantity of fuel under given conditions at a constant pressure of

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one atmosphere and at a specified reference temperature of unit quantity of a fuel.

The water formed during the combustion being in the liquid state. Any sulphur in the

fuel being converted to sulphur dioxide and any nitrogen remaining as such.

Complete combustion is possible with gaseous fuels at atmospheric pressure and

the C.V is measured at constant pressure.

Solid and liquid fuels on the other hand, require higher pressure and determination

of C.V are ,made under constant volume conditions in a BOMB calorimeter. If the

C.V at constant pressure for these fuels is required then the heat equivalent of the

work which would be done by the atmosphere on the products of combustion of the

fuel were burned at constant pressure would need to be added to the constant

volume figure. This is not normally done, the C.V being determined and quoted at

constant volume.

Fuel gases which contain H2 or hydrocarbon posses two calorific values; the

gross value Cg and the net value Cn, depending upon whether the water formed in

the combustion is in the liquid or vapour phase.

The gross calorific value provides the basis on which charges are made by

gas industries on their consumers. The calorific value of pure gases is normally

measured using the BOMB CALORIMETER or its recorded version of FAIR

WEATHER.

Calorific values are used in the British system of measurement to standard

temperature and pressure conditions which is always designated as S.T.P (standard

temperature and pressure), which is 600 F and 30.0 is of mercury (saturated).

Under the international system of units S.T.P is expressed by standard reference

condition S.R.C or metric standard condition, M.S.C with temperatures of 150C and

101325 Nm-2 or Pa (dry).

At 150C and 101325 Pa, one metre-cube of dry gas will occupy 1.017 m3 when

saturated with water vapour. The conversion from standard cubic- feet at S.T.P

referred to as (s ft3 )to standard cubic metre (m st ) is given by:

1 s ft3 = 0.02778 s m 3 (st).

ft is a not a straight forward conversion however from s ft3 to m3 (st) because of the

hydration state (saturation)

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Thus for calorific value

1 Btu s ft –3 = 0.03796 MJ m-3 (st)

1 MJ m-3 (st) = 26.34 Btu s ft-3

The simplest method of calculation is calorific value from the gross value is to deduct

the latent heat (assumed) to be 2.455 KJ kg-1 or 1055 Btu lb-1) of the water

condensed from the products of combustion by cooling to 150C or 600F.

The simplest method of calculating net calorific value of multi component fuel gas

may be calculated by a simple additive method from the C.V its constituents.

(Assuming ideal gas behaviour).

Gas Formula Ideal gas C.V’s Btu/

cu ft at 600F & 30

inHg

MJm-3 st

Ideal gas relative

density,

ρgas / ρair

Satd Dry Dry

Carbon dioxide

CO2

-

-

-

1.519

Oxygen O2 - - - 1.104

Nitrogen N2 - - - 0.967

Hydrogen H2 318.7 224.4 12.10 0.069 6

Carbon monoxide CO 315.3 320.8 11.97 0.967

Methane CH4 992.9 1010.5 37.69 0.554

Ethane C2H6 1739.6 1770.2 66.03 1.038

Propane C3H8 2475.3 2579.2 93.97 1.552

Butane C4H10 3206.7 3263.6 121.74 2.006

Ethylene C2H4 1773.1 1601.0 59.72 0.968

Propylene C3H6 2294.3 2334.7 87.09 1.452

Relative density is given by d gas = ρ gas/ ρair.

Where ρgas and ρair are the respective densities at the same conditions of

temperature and pressure.

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Example 1

Calculate the gross C.V of a substitute natural gas produced by the gas recycle

Hydrogenation Process and of the following compositions.

Component % by volume Volume fr. Component gross C.V

CO 3.0 0.030 0.03*11.97 = 0.36

CH4 34.1 0.341 0.341*37.69 =12.85

C2H6 12.9 0.129 0.129*66.03 =8.52

H2 38.4 0.384 0.384*12.10 =4.65

C3H8 11.6 0.116 0.116*93.97 =10.90

100.0 1.000 37.28

The gross C.V is 37.28 MJm-3 (st)

Fuel Gas

Three components:

1. Lean gas (CO + H2)

2. Rich gas (CH4, C2 H6, C3 H8, C4 H10)

3. Ballast gas (inert gases, N2, CO2)

Rather than picture a fuel gas as a complex mixture of individual gases, look on it as

three component mixture as listed above.

WOBBE NUMBER

Consider the flow of gas through an orifice. For a constant pressure drop,

V α d–1/2 where V= volume rate of flow

d= relative density

The potential heat contained in the gas is proportional to CV example:

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Potential heat α CV

Hence, if the gas is supplied to the appliance at a fixed pressure, then we can say

that the flow in cubic metres per hour through any particular jet is

CV d–1/2 termed as the WOBBE index or number.

It gives a measure of the relative heat input to a burned fuel at a fixed gas pressure

of any fuel gas.

Manufactured Gases

Gas Group Mean Wobbe Number

U.K units S.I units

G1} not now designated

G2} not now designated

G3 800 (+40) 30.3

G4 730 (+30) 27.6

G5 670 (+30) 25.4

G6 615 (+25) 23.3

G7 560 (+30) 21.2

U.K Gas Groups

Natural Gas

British natural Gases mean Wobbe number 50.7 SI units.

(1335 UK units + 5%)

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Wobbe number is an important criterion in the interchange ability of gases and in

burned design the gas groups fit into an international classification based on gas

families. The international gas classification.

Gas Family Wobbe number

3 subdivisions namely 17.8 – 35.8 S.I units

a) coke oven gas 470 – 943 U.K units

b) manufactured gas

c) hydrocarbon-air mixture

Gas Family 2

Natural Gas

35.8 – 53.7 S.I

943 – 1415 U.K

35.8 – 57.6 (94.3 – 1361)

57.6 – 53.7 (1361 – 1415)

Gas Family 3

Liquified Petroleum Gases

71.5 – 87.2 S.I Units

(1885 – 2300) U.K Units

Malaysian gas groups?

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FUEL COMBUSTION

Definitions of terms used in combustion

To study the combustion of a fuel to form its products there are several terms that

need to be defined. These terms are accordingly defined in the section as follows.

a. Combustion Process

It is usually assumed that the process is complete. This means that if the fuel

contains carbon, hydrogen and sulphur (called the combustible components) then all

the carbon burns to CO2 , all the hydrogen burns to water and any sulphur present

forms SO2.

On the other hand, a combustion proces is incomplete if the combustion products

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

What causes incomplete combustion ?

One would assume from the chemical reaction that proceeds, incomplete

combustion must be due to the insufficiency of oxygen supplied. However this is

undoubtedly one factor but an over supply of oxygen may also cause incomplete

combustion. Although air supply may be more than what is required, but insufficient

mixing in the combustion chamber during the limited time that the fuel and oxygen

are in contact, contributes to the incomplete combustion process of the fuel.

The chemistry of combustion can be complicated. The reaction between carbon and

oxygen can take place as follows;

Single-step

C + O2 → CO2

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Two-step

C + ½ O2 = CO

CO + ½ O2 = CO2

Thus, another factor of incomplete combustion is dissosciation of the product which

becomes important at high temperatures.

Oxygen is more strongly attracted to hydrogen than it is to carbon.. Hence, if there is

less oxygen present than needed, all the hydrogen present will form water vapour

leaving some of the carbon ending up as CO or just plain carbon.

b.Stoichiometric or theoretical air

The minimum amount of air needed for the complete combustion of a fuel is called

the stoichiometric. This means that the air supply contains the chemically correct

amount of oxygen to burn the fuel to form the products. Alternatively, it can be said

that the fuel was burned with 100 percent theoretical air.

Since air contains 21 % by volume of oxygen and 79 % of nitrogen, then by ratio

O2 : N2

21 : 79

1 : 3.76

e.g. CH4 + 2(O2 + 3.76N2) →CO2+2H2O+7.52N2

Hence the stoichiometric amount of air for the combustion of methane is

2 x 4.76 = 9.52 mol per mol of methane.

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c. Excess air

In practice it is usually necessary to use rather more air than the stoichiometric

requirement since to obtain complete and rapid combustion the fuel and air have to

be intimately mixed. Also excess air is required to control the combustion

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

excess air. The amount of excess air is usually expressed as percent excess air or

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

percent theoretical air and thus 200 percent excess air is 300 percent theoretical air.

Excess air = ( air ) actual - ( air ) stoic

( air ) stoic

d. Deficiency of air

This is defined as the amount of air less than the stoichiometric amount and often

expressed as percent deficiency of air. For example, 90 percent theoretical air is

equivalent to 10 percent deficency of air.

Deficiency air = ( air ) actual - ( air ) stoic

( air ) stoic

The negative sign of the result indicate that air supply is deficient.

e. Equivalence Ratio

The amount of air used in combustion process is also expressed as equivalence

ratio. This is the ratio of actual air-fuel ratio to the stoichiometric air-fuel ratio.

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Equivalence ratio = (air – fuel)actual

(air – fuel)stoic

MEASUREMENT OF FUEL PROPERTIES

The properties of solid and liquid fuels can be divided into the chemical and physical

properties.

Chemical properties

The composition of a fuel can be determined by carrying out by performing ;

i. Ultimate analysis

Involves the determination of the major elements C, H, O, N, S, ash

ii. Proximate analysis

This the measure of the composition based on moisture content, volatile

matter (combustible), fixed carbon (the residue left after removal of volatile

matter) and ash (the final residue left after combustion in an open crucible)

Physical properties

i. Energy content / C.V.

The energy content determined by the use of bomb calorimeter or by

calculation if elemental composition is known.

ii. Fusion point of ash

This the temperature at which the ash melts and form clinker by fusion and

agglomeration typically around 1100 – 1200 C.

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CALCULATION OF AIR REQUIREMENT FOR FUEL COMBUSTION

To release the potential heat contained in a fuel it is necessary to burn it with a

sufficient quantity of air. Insufficient air will lead to loss of potential heat by

incomplete combustion while an excess may give rise to an unduly large loss of

sensible heat (heat which is possible to be used or recover). It is important therefore

to know the theoretical or stoichiometric requirement or air to gas ratio (air:- gas)

which will give complete combustion of fuel.

This cannot be achieved with solid and liquid fuel which means that to ensure

complete combustion, excess air must be used. Gaseous fuels on the other hand,

particularly when the gas and air are premixed before the burning may allow

complete combustion to occur using essentially the theoretical air requirement.

Consider the combustion equation for methane as mentioned earlier,

CH4 + 2O2 → CO2 + 2H2O

From Avogadro’s law which stated that equal volume of gases under the same

condition of temperature and pressure contain the same number of molecules, it

follows that since one molecule of methane plus two molecules of oxygen will equal

one molecule of carbon dioxide and two molecules of water, then

1 volume CH4 + 2 volume O2 = 2 volume H2O + 1 volume CO2 assuming ideal gas

behavior.

Since the composition of fuel gases are expressed by % of volume this leads to

considerable simplification of the combustion calculation. In the case of solid or

liquid fuel where the composition are normally quoted on % by weight basis it is

necessary to convert to a molar basis to carry out the calculation. For gaseous

fuels, the important combustion equation are shown below. Since air contained

20.95% of oxygen (by volume), the air requirement, O2 req * 100/29.5.

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Each volume of O2 is accompanied by 3.76 volumes of N2 to make up 4.76 volumes

of air.

CH4 + 2O2 + 7.52 N2 → CO2 + 2H2O +7.52 N2

CH4 + 9.52 air → CO2 + 2H2O +7.52 N2

1 2 3 4

Gas Equation Stoic O2 /

Gas ratio

Stoic Air / unit.

volume of gas

CO CO + 1/2 O2 = CO2 0.5 2.38

H2 H2 + 1/2 O2 = H2O 0.5 2.38

CH4 CH4 + 2O2 = CO2 + 2H2O 2.0 9.52

C2H6 C2H6 + 7/2O2 = 2CO2 + 3H2O 3.5 16.66

C3H8 C3H8 + 5O2 = 3CO2 + 4H2O 5 23.8

C4H10 C4H10 + 13/2O2 = 4CO2 + 5H2O 6.5 30.94

C2H4 C2H4 + 3O2 = 2CO2 + 2H2O 3 14.28

C3H6 C3H6 + 9/2O2 = 3CO2 + 3H2O 4.5 21.42

CO2 - - -

N2 - - -

O2 -1.0 - 4.76

Example 2

For a town gas manufactured by the ICI 500 process.

(1) (2) (3) (4)

Constituent % By volume O2/gas (2)x (3)/100

CO2 14.8 - -

CO 2.7 0.5 0.0135

H2 48.9 0.5 0.2445

CH4 33.6 2.0 0.6720

100.0 0.9300

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Therefore the stochiometric Air / Gas ratio = 0.93 X 4.76

= 4.43

Excess Air

Excess Air is normally expressed as a % of the theoretical air requirement thus for

the above example if the actual air: gas was 5:1 ,

Then, the excess air = (5.0 - 4.43) X 100

4.43

= 12.9 %

CALCULATION OF THE ANALYSIS AND QUANTITY OF THE COMBUSTION

PRODUCT.

The volume and composition of the products of complete combustion may be

calculated from analysis of flue gas as shown below using the same gas as before.

Constituent % Equation Product of Combustion(/100

vol.Gas)

CO2 H2O

CO2 14.8 14.8 -

CO 2.7 CO→CO2 2.7 -

H2 48.9 H2→ H2O - 48.9

CH4 33.6 CH4→ CO2 + 2H2O 33.6 67.2

51.1 116.1

In the above example, the products of combustion are presented by 1 vol of fuel

burned , i.e. :

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0.511 vols CO2 + 1.161 vols H2O (vapour) per vol. of fuel gas.

In addition to CO2 and H2O, there is also present any N2 in the original fuel gas and

the nitrogen in the air used for combustion. This amounts to 3.76 X O2.

Example:

3.76 X 0.93 = 3.5 vol of N2 .

The volume of gases resulting from the stoichiometric combustion of 1 volume of the

town gas is

CO2 0.51

H2O 1.16

N2 3.50

5.17

This then is the theoretical volume of combustion products measured made under

the same conditions of pressure and temperature of the fuel gas. Normally m3

standard of s ft3. Their actual volume will of course be much greater than this due to

their high temperature and this volume may be calculated from gas law.

Combustion product analysis is normally done on dry basis.

Hence the composition is presented as :

CO2 0.51 0.51/4.01= 12.7%

N2 3.50 3.5/4.01= 87.3%

4.01

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Example 3

Any water vapour in the gas (in excess of that involved in saturated gas at 15°C) is

assumed to have been condensed before analysis. Thus on this dry basis the

volume of combustion product would be

CO2 0.51

N2 3.50

4.01

*Percentage analysis:

[CO2]0 = 0.51 = 12.7%

4.01

[N2 ] 0 = 3.50 = 87.3%

4.01

0 = stoichiometric condition.

DEW POINT TEMPERATURE OF PRODUCTS

Example 4

Ethane is burned with 20% excess air during a combustion process. Assuming

complete combustion and a total pressure of 100 kPa, determine

a) the air-fuel ratio (kg air/kg fuel), and

b) the dew-point temperature of the products.

Solution

Since ethane is burned completely, therefore the combustion products contain only

CO2, H2O, N2 and the unused O2.

Air supplied is 20% excess = 120% theoretical air or 1.2 ath

Hence the chemical reaction can be represented as

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C2H6 + 1.2 ath(O2 + 3.76 N2) → 2CO2 + 3H2O + 0.2 ath O2 + (1.2 x 3.76)ath N2

To solve for ath, we need to carry out O2 balance :

1.2 ath = 2 + 1.5 + 0.2 ath

on solving ath = 3.5 so that the chemical equation can be rewritten as :

C2H6 + 4.2(O2 + 3.76N2) → 2CO2 + 3H2O + 0.7O2 + 15.79N2

Therefore the air-fuel ratio (AFR) is calculated as

AF = mair = (4.2 x 4.76 kmol) (29kg/kmol)

mfuel (2kmol)(12kg/kmol) +

(3kmol)(2kg/kmol)

= 19.3 kg air / kg fuel

b) The dew-point temperature of the products is the temperature at which the

water vapour in the products starts to condense as the products are cooled.

The dew-point temperature of a gas-vapour mixture is the saturation

temperature of the water vapour corresponding to its partial pressure, p .

pv = nv = 3kmol (100kPa)

pprod nprod 21.49 kmol

pv = 13.96 kPa .

From steam tables, this pressure corresponds with 52.3 °C .

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INTERPRETATION OF COMBUSTION PRODUCT ANALYSIS

Analysis of the flue gases from a fuel appliance yield valuable information regarding

the efficient operation of the appliance. If incomplete combustion or combustion with

an undue excess of air is occurring, this is shown up in flue gas analysis. Flue gas

analysis in conjunction with temperature measurement allows fuel losses to be

calculated. A number of numerical formula have been developed which although not

precise are sufficiently accurate for routine purpose.

FORMULA BASED ON O2 CONTENT.

Let:

A. actual volume of air per unit volume = A

B. actual volume of dry combustion products per unit volume or weight of fuel =

V

C. theoretical volume of air per unit volume or weight of fuel = A0

D. theoretical volume of dry combustion products per unit volume or weight of

fuel = V0

E. % of CO2 actually present in dry combustion products = [CO2]

F. excess air factor:

air actually used = A = n

theoretical air A0

then volume of excess air = (n-1) A0

so % of excess air = (n-1) A0 = 100 (n-1)

A0

Volume of dry flue gas, V = V0 + (n-1) A0

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FORMULA BASED ON O2 CONTENT.

Vol. of dry flue gas/unit quantity of fuel = V0 - (n-1) A0

Volume of air in the flue gas = (n-1) A0

Since O2 content of air = 21%

% of O2 in flue gas [O2] = 21 (n-1) A0 100

100 V0 + (n-1) A0

[O2] = 21 (n-1) A0

V0 + (n-1) A0

n-1 = [O2]

(21- [O2])

or % excess air = [O2] x 100

(21- [O2])

CORRECTION FOR PRESENCE OF CO

Where part of the carbon content of the fuel is burned to CO only, parts of the

theoretical O2 requirement is unused and appears as free O2 in the flue gases.

Since,

CO2 ⇋CO + 1 O2

2

The amount of this oxygen is equal to half the amount present. Thus if CO is

present, the O2 figure example: [O2] in the above equation is given by

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[O2] = observed [O2] – 1 observed [CO]

2

Excess air calculation based on CO2 content :

% Excess air = 3.76 x %O2 x 100

(100 - %CO2 – (4.76 x %O2))

Where correction for [O2] to account for [CO] presence is not necessary.

If we were given the data as shown below:

CO2 10.5

CO 0.5

O2 3.5

N2 85.5

Calculate the % excess air based on O2 content and compare with that based on

CO2 content.

1. To calculate based on O2 content , first need to make corrections for the

presence of CO i.e.

[O2] = observed [O2] – 1 observed [CO]

2

[O2] = 3.5 - 1 x 0.5 = 3.25

2

25

Thus % excess air = [O2] x 100

(21- [O2])

= 3.25 x 100

21 – 3.25

= 18.3 %

2. To calculate based on CO2 content , the following formula is used

% excess air = 3.76 x %O2 x 100

(100 - %CO2 - 4.76 x % O2

= 3.76 x 3.5 x 100 = 18.07%

100 – 10.5 – 4.76 x 3.5

3.To calculate based on O2 , first correction based on [CO] must be made [O2] =

3.5 – 0.5 [0.5]

= 3.25

% excess air = [O2]

21 - [O2]

= 3.25 = 18.3%

21 - 3.25

If the presence of CO was not made and the calculation is based purely on the

amount of O2 measured, then

3.5 x 100 = 20 %

21 – 3.5

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COMBUSTION CALCULATIONS OF COALS AND OILS

Principal reactions

C + O2 → CO2

C + ½ O2 → CO (incomplete combustion)

CO + ½ O2 → CO2

S + O2 → SO2

H2 + ½ O2 → H2O

CH4 + 2 O2 → CO2 + 2H2O

C2H4 + 3 O2 → 2CO2 + 2H2O

C3H8 + 5 O2 → 3CO2 + 4H2O

Composition of air

% O2 % N2 N2 / O2 Air / O2

By vol. 21 79 3.76 4.76

By wt. 23 77 3.35 4.35

Gas molecular volume

Mol. Wt. of any gas in g occupies 22400 cm3 at 760 mmHg and 0°C

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Mol. Wt. of any gas in lb. occupies 359 ft3 at 14.7 p.s.i and 32°F

Mol. Wt. of any gas in kg occupies 22.4 m3 at 101.3 kNm-2 and 273K

e.g. at S.T.P

44 kg of CO2 occupies 22.4 m3

∴ density of CO2 at STP = 44 = 1.96 kgm-3

22.4

From the above reactions

12 kg C requires 32 kg of O2 for complete combustion forming 44 kg CO2

OR 12 kg C requires 22.4 m3 of O2 giving 22.4 m3 CO2

This simple equation gives a lot of information but not the complete picture. It

neglects the nitrogen and heat in bulk for carbon combusted in air, the full situation

is

12kg C + 32 kg O2 + 106 kg N2 → 44 kg CO2 + 106 kg N2

+ 406620 kJ

Nitrogen can be ignored in some mass balance but in energy balance it is very

important.

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Example:

Assume the specific heat of the flame gases is independent of T, and the

dissociation of the combustion products can be ignored, then

Mean flame temperature = heat released

Heat content of flame gas

= 406620

[(44*0.905) + (106*1.02)]

= 2751 K

If N2 is ignored, flame temperature = 406620

44*0.905

= 10, 217 K

In actual fact, dissociation of CO2 at 3000 K is 29% and N2 is 11% which gives a

flame temperature of 2600K

In combustion calculation involving solid and liquid fuels, the weight of the material

can be used as a basis or the “mole” the later commits easy conversion between

mass and volumetric quantities.

(1 kmol of any substance is simply the mol. wt in kg and this occupies 22.4 m3 at

s.t.p if it is gaseous.

29

THEORETICAL AIR REQUIRED TO BURN A FUEL OF KNOWN COMPOSITION

Example:

A coal with the following ultimate analysis is used as a fuel :

C, 74.1%; H2, 5.1%, N2, 1.35%; O2, 9.5%, S, 0.95% , moisture, 5.0%; ash, 4.0%

‘Mole’ Method

reactions :

1 kmol C + 1 kmol O2 → 1 kmol CO2

1 kmol H2 + 1/2 kmol O2 → 1 kmol H2O

1 kmol S + 1 kmol O2 → 1 kmol SO2

Basis of 100 kg coal

Component % wt ÷÷÷÷ MW (kg) Kmol Kmol O2 required

C 74.1 12 6.17 6.17

H 5.1 2 2.55 1.28

N 1.35 28 0.05 -

S 0.95 32 0.03 0.03

O 9.5 32 0.3 -0.3

Moisture 5.0 18 0.28 -

Ash 4.0 - - -

100.0 7.18

Air to be supplied = 7.18 * 100 = 34.18 k mol/100 kg coal

21

=34.18 * 22.4 = 7.66 m3/kg coal

100

Mol wt of air=28.9; = 34.18 * 28.9 = 9.87 kg/kg coal

100

30

Composition Of Dry Flue Gas

If the coal used in the previous example is burned with 50% excess air what is the

volumetric composition of the dry flue gas.

Theoretical air = 34.18 kmol /100 kg coal

∴Actual air = 34.18 * 150 = 51.27 kmol/100 kg coal

100

This consists of 79 * 51.27 = 40.50 kmol N2

100

COMPOSITION OF DRY FLUE GAS

If the coal used in the previous example is burned with 50% excess air what is the

volumetric composition of the dry flue gas.

This consists of 79 * 51.27 = 40.50 kmol N2

100

And 51.27 – 40.50 = 10.77 kmol O2

k mol k mol O2 required k mol flue product

C 6.17 6.17 6.17

H 2.55 1.28 2.55

+ 0.28 (fuel)

S 0.03 0.03 0.03

N 0.05 (fuel)

40.50 (air)

-

40.55

O From xss supply 3.59

50.34 kmol

(dry)

31

CO2 + SO2 = 12.32%

O2 = 7.13%

N2 = 80.55%

100.00%

WT OF FLUE GAS AND EXCESS AIR USED

Again for the same coal if it is burned to give a dry flue gas of the following

composition by volume.

CO2, 8.25%; O2 , 11.65%; N2, 80.1%

What is the wt. of the flue gas (including water vapour) produced for 100 kg of coal

burnt and what is the % of excess air used in the combustion.

WT OF FLUE GAS

This is obtained by carrying out a carbon balance.

Example:

100 kg coal contains 6.20 k mol C ( + S)

100 k mol dry flue gas contains 8.25 CO2

hence amount of dry flue gas produced

= 6.20 * 100 = 75.15 k mol/100 kg

8.25

thus for 100 kg of coal dry flue gas product contains:

32

CO2 : 8.25 * 75.15 = 6.20 k mol or 6.20 * 44 = 272.8 kg

100

O2 : 11.65 * 75.15 = 8.76 k mol or 8.76 * 32 = 280.3 kg

100

N2 : 80.1 * 75.15 = 60.19 k mol or 60.19 * 28 = 1685.3 kg

100

Hence, Total weight of dry flue gas = 2238.4 kg

The water originally in the coal and that produced by combustion also appears in the

flue gases in practice while 100kg coal contains 2.55 k mol H2 which forms 2.55 k

mol of water on combustion, the original coal contains 5% i.e. 0.28

k mol, therefore, the total wt of water = (2.55 + 0.28) * 18

= 2.83 * 18

= 50.9 kg

Total wt flue gas = 2289.3 kg/100 kg coal.

ACTUAL AIR USED

This is obtained by carrying out a nitrogen balance.

If x kmol of air per 100 kg of coal

N2 from air + Nitrogen from coal = Nitrogen in flue gas

i.e : 79 x + 0.05 = 80.1 * 75.15

100 100

33

⇒ x = 76.1 k-mol/100 kg coal

∴ wt of air = 76.1 * 28.9 = 22.0 kg/kg coal

100

the value can be obtained from an overall mass balance but not when the

combustion is incomplete.

% excess air = (actual - theoretical) x 100

( theoretical )

= (22.0 – 9.87) * 100 = 123%

9.87

FUEL COMPOSITION

If only the flue gas analysis is known the fuel composition can be calculated.

Example:

An oil fuel containing only carbon and hydrogen is burnt to produce a flue gas

of the following composition.

CO2 10.7%

O2 7.4%

N2 81.9%

Calculate

a) Fuel composition by weight

34

b) Excess air used

SOLUTION

a) Taking100 k moles of dry flue gas

k mol O2 = 10.7 (from CO2 ) + 7.4 (from O2)

= 18.1 k mol

Using an N2 balance actual air = 81.9 * 100

79

=103.7 k mol / 100 k mol

∴ O2 in air supplied = 21 * 103.7

100

= 21.8 k mol.

∴ 21.8 – 18.1= 3.7 k mol O2 must be in the water produced by the combustion of

H2

Also;

2 x 3.7 = 7.4 k mol of H2 , have been burnt.

This equals 2 x 7.4 = 14.8 kg

For 100 k mol dry flue gas contains 10.7 k mol C

= 12 x 10.7 = 128.4 kg

∴C/H ratio of fuel = 128.4 / 14.8

35

and composition by weight is

C 89.3%

H 10.7%

b) To produce the 10.7 k mol CO2 need 10.7 k mol O2

To produce the 7.4 k mol H2O need 3.7 k mol O2

∴ total theoretical O2 = 14.4 k mol / 100 k mol.

Actual O2 supplied = 21.8

∴ % excess air = (21.8 - 14.4) * 100 = 51.6%

14.4

INCOMPLETE COMBUSTION

The worked example has assumed that all the carbon has been converted to CO2

and none is lost in the ash. This is the ideal but in practice combustion may be

incomplete.

FORMATION OF CO

A coal contains 80% C with a calorific value of 31400 kJ/kg is burnt to produce a flue

gas with the following composition.

CO2 12.0%

CO 1.2%

O2 5.8%

N2 81.0%

36

Calculate

a) The weight of carbon converted to CO per kg coal

b) % heat loss due to incomplete combustion

equations involved :

C + O2 → CO2 , ∆H = 33960 KJ/ kg C

C + ½ O2 → CO, ∆H = 10230 KJ/ kg C

a) Ratio of C forming CO to total C burned

= % CO = 1.2

% CO+ %CO2 1.2+ 12.0

= 0.091 kg / kg C

in burning 1 kg coal i.e. 0.80 kg C

the wt of C forming CO = 0.091 * 0.80

= 0.073 kg/kg coal

b) for every kg of C converted to CO, 33960 – 10230

= 23, 730 KJ are lost

∴heat lost = 0.073 * 23730

= 1732 KJ/kg

or = 1732 * 100 = 5.52% of the heat input

31400

37

LOST OF C IN ASH

Example:

Coal containing 70% C and 15% ash is burned on a grate and the ash discharged is

found to contain 20% C. Calculate :

a) the weight of C burned

b) the % of C burned

c) the heat loss by the incomplete combustion

assuming that the ratio of C associated with the ash

(e.g as carbonates) is the same in both the coal and final ash discharged.

a) Let x kg C be associated with the ash in 100 kg coal.

x = 20

15 100-20

x = 3.75 kg C / 100 kg coal

i.e. 0.0375 kg C are lost in the ash / kg coal.

b) In a 100 kg coal, there are 70kg C of which 70 - 3.75

=66.25kg is burned

∴ % burned = 66.25 * 100 = 94.6%

70

38

b) 31400 kJ/kg coal burned the (heat content)

33960 =∆H of C → CO2

now heat lost = 0.0375 * 33960

= 1274 kJ/kg coal

or 1274 * 100 = 4.06% of the heat input.

31400

SIGNIFICANCE OF % CO2 IN FLUE GASES

1. The flue gases from any fuel burned with theoretical air for complete

combustion (no excess) contain a fixed and characteristic amount of CO2 .

2. For pure dry carbon this is 21% CO2 by volume.

3. Any air supplied in excess of theoretical reduces the % CO2 the flue

gases proportionally.

4. % CO2 is determined using any suitable type of gas analysis apparatus.

The result being % CO2 on the only dry basis.

5. The theoretical CO2 is also on the dry basis.

6. With fuels containing H2 or hydrocarbon gases the theoretical CO2 is

lower because the air supplied to burn the hydrogen forms water which

condenses leaving N2 which dilutes the flue gases formed.

7. SO2 is often analyzed as CO2 but the error is small in low sulphur fuels.

(So for high sulphur coals may indeed affect calculation).

8. Any CO produced reduces the % CO2 formed and must be allowed for.

39

APPLICATIONS OF FUEL IN PROCESS INDUSTRIES

Process industries are capital intensive as large investment is required not only in

the beginning i.e. at plant design stage, purchasing of the equipments, installation

and such, but a lot of funds is required to ensure the successful operation of the

entire plant and hence the products meeting the specifications as pledged.

Process industries also involve a lot of heat in the manufacturing process and this is

usually supplied by the application of fuel. The fuel is burned in devices such as

heaters and furnaces.

Heaters

Heaters as the name implies are common process heating devices in the industries

and are in different services. The type of heater selected depends on the service

requirement of the process heating to be performed on the fluid.

The characteristics of heaters are in two sections:

1. A radiant section which receives heat directly from the flame

2. A convection section that receives heat from the hot gases travelling to the

stack.

Types of heaters can be divided according to process or construction as

follows :

i. Process wise : Reaction or non-reaction types

ii. Construction

40

Reaction type Non-reaction type

• Visbreaker heater Crude heater

• Thermal cracker Reboiler heater

• Coker heater Feed heater

Construction type

• Box-type

• Vertical heater

• Down convection heater

• A frame heater