Fuel and Combustion

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

 FUELS
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o A fuel is simply a combustible substance.

o Most familiar fuels are hydrocarbon fuels and they fall

into one of the three categories- coal, liquid hydrocarbons,
or gaseous hydrocarbons based on their phases.

o Liquid hydrocarbon fuels are commonly derived from

crude oil through distillation and cracking processes.

o Examples are gasoline, diesel fuel, kerosene, and other

types of fuel oils.

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

 FUELS: Fractional Distillation
Fractional distillation separates a mixture into a number of different
parts, called fractions. Crude oil is a mixture of hydrocarbons.

The crude oil is evaporated and its vapours condense at different

temperatures in the fractionating column.

During the fractional distillation of crude oil:

• heated crude oil enters a tall fractionating column , which is
hot at the bottom and gets cooler towards the top.
• vapours from the oil rise through the column.
• vapours condense when they become cool enough.
• liquids are led out of the column at different heights.

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST 3

 FUELS: Fractional Distillation

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST 4

 FUELS
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o The most volatile hydrocarbons vaporize first, forming

what we know as gasoline.

o The less volatile fuels obtained during distillation are

kerosene, diesel fuel, and fuel oil.

o The composition of a particular fuel depends on the source

of the crude oil as well as on the refinery.

o Although liquid hydrocarbon fuels are mixtures of

many different hydrocarbons, they are usually considered to
be a single hydrocarbon for convenience in
combustion calculation.
Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST
 FUELS
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Fuels Chemical Name Formula

Gasoline Octane C8H18

Diesel Dodecane C12H26
CH3OH
Methanol Methyl Alcohol

LNG Methane
CH4
(Liquefied natural gas)

LPG Propane C3H8

(Liquefied petroleum gas)

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

 Combustion
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A chemical reaction during which a fuel is

oxidized and a large quantity of energy is released
is called combustion.

Fuel Type Combustible Incombustible

Components Components

Solid Fuels C, S, H N, O, ash, moisture

Liquid Fuels C, H, S N, O

Gaseous Fuels H2, CO, hydrocarbons, N2, CO2, O2, SO2

NH3, H2S etc.

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

 Modeling Combustion Air
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 Reactants Products

 For Combustion, chemical equation will be

reformed as: Fuel + Oxidizer Products

 1 mole of dry air = 20.9% O2 + 78.1% N2 + 0.9% Ar +

other gases (CO2, He, Ne & H2)

 In the analysis of combustion processes, the argon

in the air is treated as nitrogen.

 The gases that exist in trace amounts are disregarded.

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

 Modeling Combustion Air
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 Hence, 1 mole of dry air = 21% O2 + 79% N2

 The molar ratio of the Nitrogen to the Oxygen is

0.79/0.21 = 3.76.
 Therefore, when air supplies the oxygen in a combustion
reaction, every mole of oxygen is accompanied by 3.76
moles of nitrogen.
 That is, 1 kmol O2 + 3.76 kmol N2 = 4.76 kmol air

 For Every moles of Oxygen there are 3.76 moles of

Nitrogen out of a total of 4.76 moles of air.

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

 Modeling Combustion Air
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 We also assume that the nitrogen present in the

combustion air does not undergo chemical reaction. That
is, nitrogen is regarded as inert.

 To account for the use of air, instead of pure oxygen,

equation for the combustion may be revealed as follows:

 C + (O2 + 3.76 N2) = CO2 + 3.76 N2

 S + (O2 + 3.76 N2) = SO2 + 3.76 N2
1
 H2 + (O2 + 3.76 N2) = H2O + 1.88 N2
2
 CH4 + 2(O2 + 3.76 N2) = CO2 + 2H2O + 7.52 N2

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

Stoichiometric or theoretical air: The minimum amount of air needed for the
complete combustion of a fuel. Also referred to as the chemically correct
amount of air, or 100% theoretical air.
Stoichiometric or theoretical combustion: The ideal combustion process
during which a fuel is burned completely with theoretical air.
Excess air: The amount of air in excess of the stoichiometric amount. Usually
expressed in terms of the stoichiometric air as percent excess air or percent
theoretical air.
Deficiency of air: Amounts of air less than the stoichiometric amount. Often
expressed as percent deficiency of air.
Equivalence ratio: The ratio of the actual fuel–air ratio to the stoichiometric
fuel–air ratio.

50% excess air = 150% theoretical air

200% excess air = 300% theoretical air.
90% theoretical air = 10% deficiency of air

The complete combustion process with no free oxygen

in the products is called theoretical combustion.
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 Modeling Combustion Air
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 Eqns, 1-4 are stoichiometric relations. They give

chemically correct, i.e., minimum amount of air
required for the complete combustion. (when all the
carbon present in the fuel is burned to carbon dioxide, all the
hydrogen is burned to water, all the sulfur is burned to sulfur dioxide,
and all other combustible elements are fully oxidized)

 This air is referred to as stoichiometric air or

theoretical air.
 If complete combustion is realized with theoretical air,
there is no oxygen remaining unutilized and no free
oxygen appears in the products.

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

 Modeling Combustion Air
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 The theoretical air is also referred to as the chemically

correct amount of air, or 100 % theoretical air.
 A combustion process with less than the theoretical
air is bound to be incomplete.
 Normally the amount of air supplied is either greater or
less than the theoretical amount.
 The amount of air actually supplied is commonly
expressed in terms of the percent of theoretical air.
For example, 150% of theoretical air means that the air
actually supplied is 1.5 times the theoretical amount of
air.

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

 Modeling Combustion Air
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 The amount of air supplied can be expressed alternatively

as a percent excess air or a percent deficiency of air.
Thus, 150% of theoretical air is equivalent to 50% excess
air, and 80% of theoretical air is the same as a 20%
deficiency of air.

 An air-fuel mixture with more than stoichiometric air is

referred as a weak or lean mixture, and the one with less
than stoichiometric air is referred as rich mixture.

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

 Air-Fuel Ratio (A/F ratio)
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 The air–fuel ratio is simply the ratio of the amount

of air in a reaction to the amount of fuel.

 It is usually expressed on a mass basis.

𝑚𝑎 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑖𝑟 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑖𝑟 x (Mair)

 AF = 𝝀 = = =
𝑚𝑓 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑓𝑢𝑒𝑙 x (Mfuel)
𝑀𝑎𝑖𝑟
• = 𝐴𝐹 X
𝑀𝑓𝑢𝑒𝑙

 Where, 𝐴𝐹= Air-Fuel ratio on molar basis.

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

 Combustion characteristics (temperature and velocity)
are depends on equivalence ratio

1) Equivalence Ratio V25

vol fuel 
 vol air  actual Tf Tf
  = vol
 fuel 
 vol air  stoichiometry
V25
 < 1 : fuel lean
 = 1 : stoichiometry
 > 1 : fuel rich
Φ<1 Φ=1 Φ>1

2) Excess air
air % 120% excess air +20% fuel-lean (O2-rich)
100% 0% stoichiometry
80% -20% fuel-rich
 Equivalence Ratio
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 The amount of air used in combustion processes is also

expressed in terms of the equivalence ratio, which is the
ratio of the actual air–fuel ratio to the stoichiometric
air-fuel ratio.
(𝐴/𝐹)𝑆𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐
𝜑=
(𝐴/𝐹)𝑎𝑐𝑡𝑢𝑎𝑙

 𝜑 = 1; Stoichiometric mixture
 𝜑 > 1; Rich mixture
 𝜑 < 1; Lean mixture

Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

Combustion

composition of dry air

1) volume-basis
O2 20.99% N 78
 2 = = 3.76
N2 78.03% O2 21
air 4.76
C + O2 + 3.76N2 = CO2 + 3.76N2 = = 4.76
fuel 1

2) wt.-basis
C + O2 + 3.76N2 → CO2 + 3.76N2
12 32 3.76*28
136
air 136
= = 11.37
fuel 12
Determining the A/F Ratio for Complete Combustion
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Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

Determining the A/F Ratio for Complete Combustion
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Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

Determining the A/F Ratio for Complete Combustion
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Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST

Determining the A/F Ratio for Complete Combustion
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Dr. Md. Kharshiduzzaman/Assoc. Prof./AUST