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Combustion Equation-Thermodynamics-Lecture Slides, Slides of Thermodynamics

Amity University - BiharThermodynamics

Prof. Jalal Mukul delieved this lecture for Thermodynamics course at Amity University. It includes: Combustion, Equation, Stoichiometric, Proportions, Products, Design, Measurements, Temperature, Efficiency, Coiler, Requirements, Reaction

Typology: Slides

2011/2012

Uploaded on 08/12/2012

sharmila_sehgal
sharmila_sehgal 🇮🇳

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1. Applications of the Combustion
Equation
•(1) Stoichiometric proportions for finding
the correct air supply rate for a fuel
(2) Composition of the combustion products
is useful during the design, commissioning
and routine maintenance of a boiler
installation
•On-site measurements of flue gas composition
and temperature are used as a basis for
calculating the efficiency of the boiler at
routine maintenance intervals.
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1

1. Applications of the Combustion

Equation

  • (1) Stoichiometric proportions for finding

the correct air supply rate for a fuel (2) Composition of the combustion products is useful during the design, commissioning and routine maintenance of a boiler installation

  • On-site measurements of flue gas composition

and temperature are used as a basis for calculating the efficiency of the boiler at routine maintenance intervals.

2

2. Combustion Air Requirements: Gaseous

Fuels

  • Calculating the air required for gaseous fuels combustion is most convenient to work on a volumetric basis.
  • The stoichiometric combustion reaction of methane is :

CH 4 + 2O 2 → CO 2 + 2H 2 O

which shows that each volume (normally 1 m^3 ) of methane requires 2 volumes of oxygen to complete its combustion.

4

  • The complete relationship for stoichiometric combustion:

CH 4 + 2O 2 + 7.52N 2 → CO 2 + 2H 2 O +7.52N 2

as the volume of nitrogen will be 2× 79 ÷21=7.52.

  • A very small amount of nitrogen is oxidized but the resulting oxides of nitrogen (NOX) are not formed in sufficient quantities to concern us here. However, they are highly significant in terms of air pollution.

5

  • It can be seen that the complete combustion of one volume of methane will require (2+7.52=9.52) volumes of air, so the stoichiometric air-to-fuel (A/F) ratio for methane is 9.52.
  • In practice it is impossible to obtain complete combustion under stoichiometric conditions. Incomplete combustion is a waste of energy and it leads to the formation of carbon monoxide, an extremely toxic gas, in the products.

7

  • It is sometimes convenient to use term

excess air ratio, defined as:

  • Where sub-stoichiometric (fuel-rich) air-

to-fuel ratios may be encountered, for instance, in the primary combustion zone of a low-NOX burner, the equivalence ratio is often quoted. This is given by:

/ /

actual A F ratio stoichiometric A F ratio

/ /

stoichiometric A F ratio actual A F ratio

8

3. Flue Gas Composition-Gaseous

Fuels

  • The composition of the stoichiometric combustion products of methane is: 1 volume CO 2 7.52 volumes N 2 2 volumes H 2 O
  • Given a total product volume, per volume of fuel burned, of 10.52 if water is in the vapor phase, or 8.52 if the water is condensed to a liquid. The two cases are usually abbreviated to “wet” and “dry”.

10

  • Considering the combustion of methane with 20% excess air, the excess air (0.2×9.52) of 1.9 volumes will appear in the flue gases as (0.21×1.9)=0.4 volumes of oxygen and (1.9-0.4)=1.5 volumes of nitrogen.
  • The complete composition will be: constituent vol/vol methane CO 2 1 O 2 0. N 2 9. H 2 O 2 giving a total product volume of 12.42 (wet) or 10. (dry).

11

  • The resulting composition of the flue

gases, expressed as percentage by volume, is: Constituent % vol (dry) % vol (wet) CO 2 9.6 8. O 2 3.8 3. N 2 86.6 72. H 2 O – 16.

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  • (a) Stoichiometric Air Requirement

On the basis of 1 volume of the fuel gas, the propane content requires 0.7 × (5 + 18.8) = 16.7 vols air and the butane requires 0.3 × (6.5 + 24.5) = 6.3 vols air Hence the stoichiometric air-to-fuel ratio is 23:1.

14

  • (b) Excess Air The combustion products (dry) will contain (0.7 × 3) + (0.3 × 4) = 3.3 vols CO 2 (0.7 × 18.8) + (0.3 × 24.5) = 20.5 vols N 2 plus υ volumes excess air, giving a total volume of products of (23.8 + υ ).
  • Given that the measured CO 2 in the products is 9%, we can write:

hence υ = 12.87 vols

  • The stoichiometric air requirement is 23 vols so the percentage excess air is:

9 3. 100 (23.8 )

 

12.87 (^) 100% 55.9% 23

 

16

  • Each constituent is considered separately via its own combustion equation. For the carbon: C + O 2 → CO 2 12kg 32kg 44kg or for 1 kg of fuel
  • So each kg of oil requires 2.29 kg oxygen for combustion of its carbon and produces 3.15 kg CO 2 as product.

32 44 0.86 0.86 0.86 (kg) 12 12

   

17

  • Similarly

H 2 + ½ O 2 → H 2 O 2kg 16kg 18kg or per kg of fuel

  • In order to burn the hydrogen content of the

oil 1.12 kg oxygen are needed and 1.26 kg water is formed.

16 18 0.14 0.14 0.14 (kg) 2 2

   

19

  • We can now establish that 3.41 kg oxygen,

which is the stoichiometric requirement, will be associated with:

  • The stoichiometric air-to-fuel ratio is thus

3.41 + 11.23 = 14.6 : 1

3.41 11.23 kg nitrogen

 

20

5. Combustion Products-Solid and Liquid

Fuels

  • The stoichiometric combustion products from

combustion of the oil are: CO 2 3.15 kg H 2 O 1.26 kg N 2 11.23 kg

  • The combustion products would normally be

needed as a volume percentage, so the reverse operation to that which was performed for air above is required.