THERMODYNAMICS ENGINEERING II

(Jaime Postigo - JuanF.Cruz)

COMBUSTION

Introduction.- The thermodynamics of chemical processes plays an important role in

the practice of engineering in a large number of extractive industries as well as the
chemical energy released or absorbed

Combustion.- Combustion is understood as a rapid oxidation of a substance.

accompanied by the transformation of chemical energy into molecular energy and a
substantial increase in the temperature of the substances in the reaction.
Combustion is also accompanied by luminous emission, which depends on the
temperature reached in the reaction.
Its application is in the M.C.I., steam thermal plants, furnaces, etc.

Combustible - It is a substance capable of being burned in the presence of oxygen.

Examples: hydrogen, coal, oil, wood, bagasse, natural gas, etc.

Oxidizer.- It is the substance that causes or activates combustion. This is called

oxygen is present in the atmospheric air

FUELS USED IN THE INDUSTRY (Classification)

Analysis of fuels.

a) Gravimetric analysis or elemental analysis: Specifies the mass percentages

of the chemical elements of the fuel.
b) Volumetric analysis: Specify the percentages by volume or in moles of the
compounds that make up the fuel are generally used for fuels
gaseous.

Example:
A mixture of gases produced in a petroleum distillation plant has the
next composition in volume:

10% CO2, 22%H2, 5% CO, 28% CH4, 12%C2H 4, 6%C2H 6, 7%C3 H 6

The gravimetric analysis of the same gas is:

68.58%

Combustion processes

When a chemical reaction occurs, the bonds of the molecules of the reactants are
bonds break and atoms and electrons regroup to form products.

Basic definitions:

1. Reaction equation - is the quantitative expression of the substances involved.

in the reaction.

Example:
a) C O2 CO2
b) 10CH 23O
4
79N2 8CO2 2CO 20H O2 79N 40 2 2 2

2. Reactants.- They are the substances that enter the combustion like the fuel and the
oxidizer.

3. Products - They are the substances that result from the combustion process.

Combustion scheme

4.- Complete combustion.- It is one in which all the oxidizable elements of

combustible are completely oxidized, this is the C is oxidized to CO2, and the H up to H2O,
etc.

5.- Ideal combustion.- It is the complete combustion in which oxygen

supplied is the minimum necessary.
This equation that is defined by a simple balanced chemical equation is referred to
it is also called theoretical combustion.

6.- Incomplete combustion.- It is one in which the products contain 'CO'.

in these processes, carbon reacts forming "CO" and " CO2in proportions
unknown, depending on the amount of oxygen used and the procedure
employee to mix the oxidizer with the air. Temperature also influences, as if
it is very high there may be dissociation. (Separate, decompose, disunite).

7.- Stoichiometric air.- It is also referred to as theoretical air (at). It is the amount of
air that provides the strictly necessary oxygen for complete combustion of
the oxidizable elements of the fuel.

8.- Stoichiometric mixture.- It is one that contains the corresponding proportions.

stoichiometric air.

Note: that a mixture is stoichiometric does not indicate that the combustion is
necessarily complete.

9.- Air-fuel ratio.- It is the quotient between the mass of air and the mass of
fuel used in combustion. It is expressed in kg of air per kg of fuel.

ma
ra/ c
mc

10.- Fuel-air ratio.- It is the quotient between the mass of fuel and the mass
of air used in combustion. It is expressed in kg of fuel per kg of air.

mc 1
rc/ a
mara/ c

Ideal combustion with oxygen: Let's consider the combustion of the main
combustible elements.

a) Combustion of carbon:

C + O2 CO2
1 kmol + 1 kmol 1 kmol
12 kg + 32 kg 44 kg
1 kg + (32/12) kg (44/12) kg
1 kg + 1.87 m3 1.87m3

b) Combustion of hydrogen:

2 H2 + O2 2 H2O
2 kmol + 1 kmol 2 kmol
4 kg + 32 kg 36 kg
1 kg + 8 kg 9 kg
2 m3 + 1 m3 2 m3

c) Combustion of sulfur:
 S + O2 SO2
1 kmol + 1 kmol 1 kmol
32 kg + 32 kg 64 kg
1 kg + 1 kg 2 kg
1 kg + 0.7 m3 0.7 m3

In the reaction equations for C, H2and we present a balance of

materials, expressing in moles, mass, and volume. In general, the number of moles
the number of products is not equal to the number of moles of the reactants, since this is
proportional to the number of molecules, which varies with the reaction.

The volume of the reactants, measured at normal pressure and temperature conditions,
it is not the same as the volume of the products, under the same conditions.

However, the total mass of the products must be equal to the total mass of the
reactives.

COMBUSTION WITH AIR.

Combustion processes in practice are not carried out with pure oxygen, but rather with
it uses atmospheric air that contains oxygen.
Combustion with pure oxygen is reserved for laboratory tests. Example:
determination of the calorific value of a fuel or in cutting exercises in
welding.

Atmospheric air composition.-is a mixture of oxygen, nitrogen, argon, carbon dioxide

carbon, neon, hydrogen, helium, and water vapor. (The latter in proportions that vary
with the atmospheric conditions)

Average composition:

78%
Oxygen 21%
Other gases: 1%

In practical applications, it is sufficient to consider air as composed of: 79% of

nitrogen and 21% oxygen by volume.

The gravimetric composition can be considered as 77% nitrogen and 23% oxygen.

In calculations related to combustion with air, it is very useful to consider what

next:

a) 100 Kmoles of air contain 21 Kmoles of oxygen and 79 Kmoles of nitrogen, therefore
so, in combustion:
For each kmol of O2enter79/21 = 3.76kmol of N2.

And for each kmol of O2100/21 = 4.76 mol of air.

As they are ideal gases, the previous relationships are equivalent if we use
volumes, then:
For each m3of O2they enter 3.76 m3of N2
And for each m3of O2they enter 4.76 m3of air.
b) 100 Kg of air contains 23.3 Kg of oxygen and 76.7 Kg of nitrogen, therefore, in the
combustion
 For each kg of O2They enter 76.7/23.3 = 3.30 kg of N2.

And for each kg of O2100/23.3 = 4.29 kg of air.

Note: the composition of air remains constant up to an average height of 20 km above

sea level. From that height, the proportion of oxygen decreases by 0.3% per
every km of height.

IDEAL COMBUSTION WITH AIR.

For ideal combustion with air, we consider that the H2it oxidizes to form H2Oh and the
C is oxidized to form CO2If sulfur existed, it would oxidize to form SO.2(can
react also until forming SO3, which in the presence of H2The liquid will give acid
sulfuric, a highly corrosive compound).

The combustion equation of a hydrocarbon is of the form:

C xH y b(O2 3.76N2) dCO 2 eH2 O fN 2

Example: combustion of propane (C3H8) with stoichiometric air.

C 3 H 8 b(O2 3.76N2) dCO2 eH 2 The fN 2

Carbon balance : d = 3
Hydrogen balance: 2e = 8 => e = 4
Oxygen balance 2b = 2d + e => b = (2x3 + 4)/2 = 5 => b = 5
Nitrogen balance: 2x3.76b = 2f
(2x3.76x5)/2 = f f = 18.8

C 3 H 8 5(O2 3.76N2) 3CO2 4H 2 The 18.8N2

We determine the:

ma5(O2 3.76N2) 5(32 3.76x28

ra/ c
mc C 3H 8 12x3 1x8
5 (32 105.28) 686.4 sick
ra/ c 15.6
36 8 44 kgcomb

Stoichiometric air-fuel ratio.

REAGENTS PRODUCTS
Combustible O2 N2 Total Components total
Moles 1 5 18.8 24.8 3+4+18.8 25.8
Time C 12x3= 36 16x2=32 14x2x3.76=105.28 132+72+526.4 730.4
(kg) H 1 times 8 equals 8

36+8= 44 730.24

Example: from the calculation of the table:

3CO2: 3 (12 + 16x2) = 132

4H2O:4 (1x2 +16) = 72
18.8N2:18.8 (14x2) = 526.4

REAL COMBUSTION.
In practice, we will encounter combustion processes that are neither ideal nor
complete; that is to say, they will be incomplete despite having excessive air.

Factors that influence combustion:

Combustible
Air-fuel ratio
Geometry of the combustion chamber
Combustion temperature
Fuel feeding method
Turbulence in the chamber
Gas outlet velocity

Actual combustion is incomplete, both when theoretical air is used as

when excess air is used.

If ex>0: excess air

If ex=0: stoichiometric air
If ex<0: air defect

Then:
Combustion with stoichiometric air.

Complete combustion.
Incomplete combustion.

Combustion with excess air.

Complete combustion.
Incomplete combustion.

Combustion with air deficiency.

Incomplete combustion.

Note: with excess air we manage to form the least possible amount of CO in the
combustion.

IDEAL COMBUSTION WITH AIR EXCESS

If combustion occurs with only the theoretical amount of air, the fuel does not
it burns completely, since some molecules of the fuel would not be present
with oxygen. For this to happen, more air (excess) will be needed.

Real air (ar) - is the amount of air that enters a combustion process.

Excess air (ex).

a r at
%ex x100
at

Theoretical air percentage.

ar
% at x100
at
Rich mixture - is one that contains a lower amount of air than stoichiometric (air
in default).

Rich mixture - it is the one that contains a greater amount of air than the stoichiometric.
(excess air)

If combustion is complete, the excess oxygen will appear in the products and the
the equation of the reaction will have the following form:

C x H y B(O2 3.76N2) dCO2 eH2The fN2 gO2

Example: combustion of propane with 40% excess air.

From the previous example, the equation with stoichiometric air.

C 3 H 8 5(O2 3.76N2) 3CO2 4H2 O 18.8N2

If 40% excess air is used, the combustion will be carried out with 140% of theoretical air.
Then the real equation will be.

Replacing in the theoretical equation we obtain:

3 H 8 1.4x5(O2 3.76N2) 3CO 2 4H 2 The 18.8x1.4N2 0.4x5O2

C 3 H 8 7(O2 3.76N2) 3CO2 4H 2 O 26.32N2 2O2

Theoretical air-fuel ratio:

mat5(O2 3.76N2) 5(32 3.76x28

(ra/ c) t
mc C 3H 8 12x3 1x8
5(32 105.28) 686.4 kgaire
(ra/ c ) t 15.6
36 8 44 kgcomb

Actual air-fuel ratio:

mthe7(O2 3.76N2) 960.96 kgaire

(r a/ c ) r 21.84
c C 3H 8 44 kgcomb

Then:

(ra/ c) (ra/ c) r
1.40 %at x100
(ra/ c ) t (ra/ c) t

Importance of complete combustion.

Combustion is a process that is carried out to harness the chemical energy released.
so much for the reaction of H2towards H2Oh, how the reaction of C towards CO2. The H2for his/her
affinity with O2reacts completely towards H2Oh, on the other hand, C reacts towards CO.2y
CO.
When one kmol of C reacts completely to CO2, it releases 3.5 times more energy than
when the kmol of C reacts completely to CO. This justifies the tendency to reduce
at least the formation of CO to achieve complete combustion.
COMBUSTION REAL WITH EXCESS AIR (INCOMPLETE)

C xH y B(1 (O2 3.76N2) iCO 2 JCO eH2 The f ( 1  e x2 g.ex O2

Example: 1

A fuel whose gravimetric analysis is:

82% Carbon2y 4%O2it burns with 150% of theoretical air. It is known that 80% of C
reacts forming CO2and 20% reacts to form CO. Determine the air-fuel ratio.
fuel used.

Solution:
For the understanding of this problem, we will assume the following:
In 100Kg of fuel, we have: 82 Kg of C, 14 Kg of H.24 Kg of O2

The number of moles of each element will be:

mass (kg)
M molecular weight (kg/kmol)

m 82kg
n C 6.83kmol
M 12 kg/kmol
14
H2 7 kmol
2
4
O2 0.13 kmol
32
Then the equation with stoichiometric air will be:

6.83 degrees Celsius 7H 2 0.13O2)  b(O2 3.76N2) dCO2 eH2 The fN2

Carbon balance: 6.83 = d

Hydrogen balance: 7x2 = 2e => e = 7
-Oxygen balance: 0.13x2+2b = 2d+e = 2x6.83+7; b = 10.2
-Nitrogen balance: 2x3.76x10.2 = 2f => f = 38.35

Replacing:
(6.83C 7H 2 0.13O2)  10.2(O2 3.76N2) 6.83 CO2 7H 2 The 38.35N2

Then: for the actual combustion, we have that carbon is distributed as follows
way
80% form CO2 6.83x0.8 5.46 moles
20% form CO 6.83x0.2 1.37 moles

So the real equation with 150% theoretical air will be:

(6.83C 7H 2 0.13O2)  10.2x1.5(O2 3.76N2) 5.46CO2 1.37CO 7H 2 The 38.35x1.5N2 0.5gO2

(6.83C 7H 2 0.13O2)  15.30(O2 3.76N2) 5.46CO2 1.37CO 7H 2 The 57.53N2 0.5gO2

Oxygen balance to determine g:

0.13 times 2 15.30x2 5.46x2 1.37 7 0.5x2g
g 11.57
For the O2:it has to be: g = 11.57 x 0.5 = 5.79; g = 5.79

The equation is expressed

(6.83C 7H 2 0.13O2)  15.30 (O2 3.76N2) 5.46CO2 1.37CO 7H 2 The 57.53N2 5.79O2

15.30(32 3.76x28 2100.384 Kg.air

(ra/ c) r 20.97
6.83x12 7 times 2 0.13x32 100.12 Kg.comb

Example 2.
Butane burns (C4H10) with 90% theoretical air. Find the reaction equation.

C 4 H10 b(O2 3.76N2) dCO2 H2 The fN2theoretical ec.

Carbon balance: 4 = d
Hydrogen balance: 10 = 2e => e=5
Oxygen balance: 2b = 2d + e => b = (2x4 + 5) / 2 = 6.5
Nitrogen balance: 6.5 x 2 x 3.76 = 2f => f = 24.44

C 4 H 106.5(O 3.76N
2
4CO
2
5H O 24.44N
2
theoretical
2
ec. 2 Balanced.

Then the combustion equation with 90% theoretical air will be:

C 4 H10 6.5(0.9)(O2 3.76N2) I CO2 j CO eH2 The fN2

Carbon balance: i + j = 4 ; i=4-j

Hydrogen balance: 10 = 2e => e=5
Oxygen balance: 2x0.9x6.5 = 2i + j + e 11.7 = 2(4-j) + j + 5
j = 1.3
i = 2.7
Nitrogen balance: 6.5x 0.9x2 x 3.76 = 2f => f = 22

C 4 H10 5.85(0.9)(O2 3.76N2) 2.7CO2 1.3CO 5H2 The 22N2

5.85(O2 3.76N2) 5.85(32 105.28

(ra/ c ) r 13.85
C 4H 10 12 times 4 1x10 48 10
kgaire
(ra/ c) r 13.83
kgcomb
6.5(O2 3.76N2) 6.5(32 1051.68
(a/ c) translatedText 15.38
C 4H 10 12 times 4 1x10 48 10
kgaire
(ra/ c) t 15.38
kgcomb

a a
r r
a r a t c r c t 13.85 15.39
ex x100% x100% 10%
at a 15.39
r
c t

ANALYSIS OF COMBUSTION PRODUCTS

The analysis of combustion products is carried out for the purpose of determining the
composition of the products of the process. With this analysis, information can be obtained
about the process and this can be:
a) Air-fuel ratio.
b) Approximate composition of the fuel.
c) Losses due to incomplete combustion (due to the presence of CO).
d) Excess or deficiency of air.

Among the different devices for the experimental determination of the composition of the
the combustion products are:
The Orsat analyzer, gas chromatograph, the infrared analyzer, the detector of
flame ionization and others.
The data from these instruments are used to determine the molar fractions of the
gaseous products of combustion.
The analyses are usually performed on a dry basis. This means that the fractions
Molar amounts are given for all gaseous products except water vapor.

Calculations based on the analysis of the products.

The following cases can be presented:

a) given the fuel formula and the analysis of the products.

b) Given the composition of the fuel and the analysis of the products.
c) Given the analysis of the products and unknown fuel.

Example: In a combustion process of flow and steady state, at constant pressure of

1 bar., butane burns (C4H10) and the analysis of dry products reveals the following:
7.8% CO28.2% O21.1% CO and 82.9% N2; determine the % of O2 used in the combustion.

Solution:
Taking 100 moles of dry products, then the equation will be of the form:

aC4 H10 b (O2 3.76N2) 7.8CO2 1.1CO 8.2O2 dH2 The 82.9N2
Balancing:
Carbon: 4a = 7.8 + 1.1 => a = 2.23
Hydrogen: ax10 = 2d => d= (22.30)/2 = 11.5 d = 11.5
Oxygen: 2b = 7.8x2 + 1.1 + 8.2x2 + d
b = 44.25 / 2 = 22.3 b = 22.3
Nitrogen: 2x3.76xb = 82.9x2 b = 22.04

Note: the oxygen balance is more reliable.

Replacing the data, the equation becomes:

2.23C4 H10 22.3(O2 3.76N2) 7.8 CO2 1.1CO 8.2O2 11.5H2 The 82.9N2

23.13(32 105.28 Kg.air

(ra/ c) r 23.48 23.48
2.23(48 10) Kg.comb

Determine the equation with stoichiometric air:

C 4 H 10 b(O2 3.76N2) dCO2 eH2 The fN2

Carbon: => 4=d
Hydrogen: 10 = 2e=> e=5
Oxygen: 2b = 2d + =>e b = 6.5
Nitrogen: 2x3.76x6.5 = 2f => f = 24.44

C 4 H 10 6.5(O2 3.76N2) 4CO 2 5H2 The 24.44N2

6.5(32 105.28 Kg.air

(ra/ c) 15.38 23.48
(48 10) Kg.comb
(ra/ c ) r 23.48
%at x100 x100 152.6%
(ra/ c ) t 15.38