Combustion

 Combustion – a chemical reaction in which an oxidant rapidly

combines with a fuel to liberate thermal energy (i.e. high-temperature
gases).
• oxidant is usually oxygen in the air
• fuels are usually HCs in either elementary or compound form
• products of combustion include CO2, CO, H2O, SOx (SO2 or
SO3), N2 (some N2 may react with O2 at the elevated temp. to
produce NO), particulate matter, and ash.

 Combustion process – liberation of thermal energy - associated with a

luminous gas or flame.
• Flame has a flame front that separates the luminous and non-
luminous gases.
 Combustion
 Burning zone exists on the flame side of the flame front
- Burning region – a reaction zone – where the unburned gases
are ignited and oxidation begins
- Luminous zone – where the oxidation is completed and light
is emitted.

 Experimental efforts have established - in a flame various

intermediate reactants and products exist for a short period of time.
 Thus combustion includes various intermediate steps or reactions
that are understood for only the most elementary fuels.
 Usually the term combustion refers to the end or final products of
the fuel oxidation process.
 Combustion

Air • Bringing a fuel into intimate contact

with O2 is not sufficient to start a
combustion process.
Fuel • Fuel must be brought above its
ignition temperature to start the
combustion.
• Minimum ignition temperature in atmospheric air:
~ 260°C for gasoline,
~ 400°C for C,
~ 580°C for H2,
~ 610°C for CO,
~ 630°C for CH4
• Moreover, the proportions of the fuel and air must be in the proper
range for combustion to begin. For example, NG does not burn in air
in concentrations <5% or >15%.
 Combustion
 During the combustion process, the mass of each component remains
constant.
 Solving the chemical reaction equation involves the conservation of
mass.
 The following reactions involve complete combustion (i.e. all the C,
H and S are completely oxidized) of the most common components
in HC fuels – C, H and S:
Reactants Products
C + O2  CO2
H2 + 0.5O2  H2O
S + O2  SO2
 Ideal combustion – reaction involves the exact proportion of O2 and
fuel – results in complete combustion – known as theoretical/
stoichiometric quantities of air/O2 and fuel.
 Combustion
• On a mole or a volume basis, dry air is composed of 20.9% O2, 78.1%
N2, 0.9% Ar, and small amounts of CO2, He, Ne, and H2.
• Dry air can be approximated as 21% O2 and 79% N2 by mole numbers.
Therefore, each mole of O2 entering a combustion chamber is
accompanied by 0.79/0.21 = 3.76 mol of N2:

1 kmol O2 + 3.76 kmol N2 = 4.76 kmol air

• Assumption - N2 present in the combustion air does not undergo
chemical reaction – regarded as inert.
1 mol C + 1 mol O2 = 1 mol CO2 (total number of moles is not
conserved during a chemical reaction).
12 kg C + 32 kg O2 = 44 kg CO2 (mass is conserved)
1 kg C + (32/12) kg O2 = (44/12) kg CO2
 Important Parameter
• Air–Fuel Ratio (A/F) and its reciprocal,
the fuel–air ratio (F/A):
• A/F is simply the ratio of the amount of
air in a reaction to the amount of fuel.
 can be expressed as molar basis
(moles of air/ moles of fuel) or as
mass basis (mass of air/ mass of fuel).

• Conversion between these values is accomplished using the molecular

weights of the air, Mair, and fuel, Mfuel,

A/F =
 Problem
# One kmol of octane (C8H18) is burned with
air that contains 20 kmol of O2, as shown in
Fig. Assuming the products contain only CO2,
H2O, O2, and N2, determine the mole number
of each gas in the products and the air–fuel
ratio for this combustion process.
Solution: Chemical equation for the combustion process is –
1(C8H8)+ 20(O2 + 3.76 N2) → xCO2 + yH2O + zO2 + wN2
Unknowns x, y, z and w are determined by applying the mass balance to
each of the elements
C: 8 = x → x = 8
H: 18 = 2y → y = 9
O: 20 × 2 = 2x + y + 2z → z = 7. 5
N: (20).(3.76) = w → w = 75.2
Substituting yields
C8H8+ 20(O2 + 3.76 N2) → CO2 + 9H2O + 7.5O2 + 75.2N2
 Problem

C8H8+ 20(O2 + 3.76 N2) → CO2 + 9H2O + 7.5O2 + 75.2N2

• The molar mass of air is Mair = 28.97 kg/kmol ≌ 29.0 kg/kmol

• So, 24.2 kg of air is required to burn each kilogram of fuel

during this combustion process.
 Complete and Incomplete Combustion
 A combustion process is complete if all the carbon in the fuel burns
to CO2, all the hydrogen burns to H2O, and all the sulfur (if any)
burns to SO2.

 The combustion process is incomplete if the combustion products

contain any unburned fuel or components such as C, H2, CO, or OH.
 Reasons for Incomplete combustion: Soot
- insufficient oxygen/ air supply
- insufficient/ improper mixing of fuel with air in the combustion
chamber during the limited time
- dissociation, which becomes important at high temperatures.
 Stoichiometry
 The minimum amount of air needed for the complete combustion of a
fuel is called the stoichiometric or theoretical air or chemically
correct amount of air, or 100 percent theoretical air.
 Fuel completely burned with theoretical air - no free O2 in the product
gases.
 A combustion process with less than the theoretical air must be
incomplete.
 Ideal combustion process –- fuel is burned completely with theoretical
air – known as the stoichiometric or theoretical combustion of that
fuel.
 For example, the theoretical combustion of methane is
CH4 + 2(O2 + 3.76 N2 ) → CO2 + 2 H2O + 7.52 N2

 Notice that the products of the theoretical/ stoichiometric combustion

contain no unburned CH4 and no C, H2, CO, OH, or free O2.
 Stoichiometry (Excess Air)
 Actual combustion - common practice - using more air than the
stoichiometric amount - enhances the chances of complete
combustion or controls the temperature of the combustion chamber.
 The amount of air in excess of the stoichiometric amount is called
excess air.
 excess air is usually expressed in terms of the stoichiometric air as
percent excess air or percent theoretical air.
 For example, 50% excess air = 150% theoretical air, and 200 percent
excess air = 300% theoretical air. Again, 90% theoretical air is
equivalent to 10% deficiency of air.
  Equivalence Ratio or Fuel-air equivalence ratio:
It is the ratio of the actual fuel–air ratio to the stoichiometric
fuel–air ratio.
Equivalence Ratio, Φ =
 Stoichiometry (Excess Air)

 Actual combustion processes are hardly ever complete, even in the

presence of excess air.
 So, it is impossible to predict the exhaust gas composition on mass
balance basis.
 Only alternative - measuring the amount of each component in the
products directly.
 Commonly used devices: Orsat apparatus and gas analyzers.
 Combustion of Coal with Theoretical Air
# Coal from Pennsylvania which has an ultimate analysis (by mass) of 84.36%
C, 1.89% H2, 4.40% O2, 0.63% N2, 0.89% S, and 7.83% ash (noncombustibles)
is burned with a theoretical amount of air. Disregarding the ash content,
determine the mole fractions of the products and the apparent molar mass of
the product gases. Also determine the air–fuel ratio required for this
combustion process.
• Consider 100 kg of coal for simplicity in analysis
• Conversion from mass fraction to mole fraction of the
constituents

• The combustion equation may be written as

7.03C + 0 .945H2 + 0 .1375O2 + 0 .0225N2 + 0.0278S + ath*(O2 + 3.76N2)
→ xCO2 + yH2O + zSO2 + wN2
• Performing mass balances for the constituents gives
C balance: x = 7.03
H2 balance: y = 0.945
S balance: z = 0.0278
O2 balance: 0.1375 + ath = x + 0.5y + z → ath = 7.393
N2 balance: w = 0.0225 + 3.76 ath = 0.0225 + 3.76 × 7.393 = 27.82
• Substituting, the balanced combustion equation without the ash becomes
7.03C + 0.945H2 + 0.1375O2 + 0.0225N2 + 0.0278S + 7.393(O2 + 3.76N2)
→ 7.03CO2 + 0.945H2O + 0 .0278SO2 + 27.82N2
• The mole fractions of the product gases are determined as follows:
Nprod = 7.03 + 0.945 + 0.0278 + 27.82 = 35.82 kmol
 Then, the apparent molar mass of product gases becomes

Finally, the air–fuel mass ratio is determined from its definition to be

That is, 10.2 kg of air is supplied for each kg of coal in the furnace.
 Burning Natural Gas with Excess Air
# A natural gas has the following molar analysis: CH4, 80.62%; C2H6,
5.41%; C3H8, 1.87%; C4H10, 1.60%; N2, 10.50%. The gas is burned with
dry air, giving products having a molar analysis on a dry basis: CO 2, 7.8%;
CO, 0.2%; O2, 7%; N2, 85%. (a) Determine the air–fuel ratio on a molar
basis. (b) Assuming ideal gas behavior for the fuel mixture, determine the
amount of products in kmol that would be formed from 100 m3 of fuel
mixture at 300 K and 1 bar. (c) Determine the percent of theoretical air.
• The chemical equation then takes the form

• Applying conservation of mass to carbon

b(0.078 + 0.002) = 0.8062 + 2(0.0541) + 3(0.0187) + 4(0.0160) ⇒ b = 12.931
• Conservation of mass for hydrogen results in
2c = 4(0.8062) + 6(0.0541) + 8(0.0187) + 10(0.0160) ⇒ c = 1.93
 • Applying conservation of mass to oxygen
12.931[2(0.078) + 0.002 + 2(0.07)] + 1.93 = 2a ⇒ a = 2.892
• The balanced chemical equation is then
(0.8062CH4+ 0.0541C2H6+ 0.0187C3H8+ 0.0160C4H10+0.1050N2) + 2.892(O2+
3.76N2) →12.931(0.078CO2+0.002CO+ 0.07O2+ 0.85N2) +1.93H2O
• The air–fuel ratio on a molar basis is

• The amount of fuel in kmol, nF, present in 100 m3 of fuel mixture at 300K
and 1 bar can be determined from the ideal gas equation of state as
• Total amount of products is b + c = 12.931 + 1.93 = 14.861 kmol of
products per kmol of fuel.
• Accordingly, the amount of product mixture that would be formed from
100 m3 of fuel mixture is (14.861)(4.01) = 59.59 kmol of product gas.
• The balanced chemical equation for the complete combustion of the fuel
mixture with the theoretical amount of air is

• The theoretical air–fuel ratio on a molar basis is

• The percent theoretical air is then