Descripción: Combustión y ecuaciones sobre carburación.
Good Document for Combustion Analysis and Testing
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Basic combustion theory and explanation
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05 CombustionFull description
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Combustion Combustion is a chemical process in which a substance reacts rapidly with oxygen and gives off heat. The original substance is called FUEL, and the source of oxygen is called the OXYDISER. For combustion FUEL, OXYDISER and HEAT must be present. As a result of combustion EXHAUST are created and HEAT are produced. We can control or stop the combustion by controlling the amount of fuel or the amount of oxygen or source of heat. Reactants are the components that exist before the reaction of a of a combustion process. Products are the components that exist after the reaction. Why study Combustion?
1. Production and reduction of hostile of hostile fires which annually kill over thousands of people and leave many thousands more with life long physical and psychological scars. 2. Reduction of harmful of harmful liquid waste matter which are slowly but surely damaging the civilization that grew to depend on combustion. 3. It promotes to design of efficient of efficient and safe furnace to dispose wastes. 4. It lets us to improve efficiency of chemical of chemical energy resources. 5. Innovation methods to extract energy from sources which are uneconomical before.
Stoichiometry Stoichiometry air is the quantity of air of air required to burn a unit quantity of fuel of fuel completely with no oxygen appearing in the product of combustion. of combustion. Air contains oxygen by mass 23.2% and by volume 21%. Also air contains Nitrogen by mass 76.8% and by volume 79%. 1
Stoichiometric Combustion Equations: Take an example of combustion of combustion of carbon of carbon with oxygen Lecturer: Ram Chandra Sapkota | Note prepared by Surya Narayan Mandal
C + O2
=
CO2.
This is balanced equation. Here carbon and oxygen are called reactants and CO2 is product. This is called Stoichiometric Combustion Equations. Next Example: H2 + O2
= H2O.
2 H2 + O2
=
This is unbalanced equation.
2 H2O. This is balanced equation, which is called Stoichiometric
Combustion Equations. Third Example: C6H14 2 C6H14
+
O2 +
19 O2
= CO2
+
H2O.
= 12 CO2 + 14 H2O.
Which is unbalanced equation. Which is balanced equation which is called
Stoichiometric Combustion Equations. Analysis on mass basis:
Let us take an example. 2 H2 + O2
= 2 H2O
This equation says that 4 mass of H2 of H2 combined with 32 mass of O2 of O2 to produce 36 mass of H2O. i.e. 1 mass of H of H2 + 8 mass of O of O2
= 9 mass of H2O.
1 kg H2
= 9 kg H2O if unit if unit is kg.
+ 8 kg O2
Oxygen is combined in air and is 23.2% by mass. So the 8 kg O2 = 8kg/0.232 = 34.5 kg air. It means N2 contains 34.5‐8 = 26.5 kg. Analysis on volume basis:
It is based on Avogadro’s Law which states that “ equal volume of different of different gases at same temperature and pressure contain the same number of molecules”. of molecules”. 2 H2O + O2 = 2 H2O That is,
2
1 volume of H of H2 + 0.5 volume of O of O2= 1 volume of H of H2O.
It is minimum amount of air of air which supplies the required amount of oxygen of oxygen for complete combustion. Molar Mass:‐
The mass of one mole of a of a substance is called molar mass. It is assumed that the atmospheric air is the mixture of N of N2 and O2 in the mole ratio of 79:21. of 79:21. One mole of air of air represents 0.79 mole of N of N2 and 0.21 mole of oxygen. of oxygen. Now, Molar mass of N of N2
= 28 X 10‐3 kg.
Molar mass of O of O2
= 32 X 10‐3 kg.
Molar mass of air of air
= 0.79X28X10‐3 + 0.21X32X10‐3 = 29X10‐3 kg.
1 Kmol O2 + 3.76 Kmol N2 = 4.76 Kmol air
Air‐ Fuel Ratio:‐ It is the ratio of the of the mass of air of air to the mass of fuel of fuel for a combustion process. That is
Where m = N X M = Number of moles of moles X Molar Mass. 3
Numerical Problems : 1. Calculate the air‐fuel ratio required for burning of of propane propane (C3H8) with 150% theoretical air. Solution:
The chemical equation for this chemical process can be written as C3H8 + 5 O2 = 3CO2 + 4 H2O with 150% theoretical air, combustion equation becomes C3H8 +1.5 [5O2 + 5*3.76 N2] = 3CO2 + 4 H2O + 2.5 O2 + 1.5*5*3.76 N2 We have,
2.Calculate stoichiometric air fuel ratio on mass basis and on molar basis for combustion of of butane butane (C4H10). Solution:
The chemical equation for this chemical process can be written as C4H10 + 6.5O2 + 6.5*3.76 N2 = 4 CO2 + 5 H2O + 6.5*3.76 N2 4
Mass Basis
We see here that for complete combustion of butane of butane 6.5 mole of O of O2 and 24.44 mole of N2 is to be supplied. Lecturer: Ram Chandra Sapkota | Note prepared by Surya Narayan Mandal
Molar mass of air of air
= Molar mass of O of O2 + Molar mass of N of N2 = 6.5 + 24.44 = 30.94 mole
Thus stoichiometric AF = 30.94 mole air / mole of fuel of fuel Molar Basis Molar Basis
/ /
= [30.94*29]kg/ (4*12+10)kg = 15.45 Problem:
3.During combustion process , C3H15 is burnt with 130% theoretical air. Assuming the complete combustion determine the a) air fuel ratio b) equivalence ratio c) composition of of the the product of of combustion combustion
Solution:
The chemical equation for this chemical process can be written as C8H15 + 11.75 O2 = 8 CO2 + 7.5 H2O with 130% theoretical air, combustion equation becomes C8H15 + 1.3 [11.75 O2 + 11.75*3.76 N2] = 8 CO2 + 7.5 H2O +(1.3*11.75‐11.75)O2 + 1.3*11.75*3.76 N2 We have,
= 19 kg air/kg fuel b) Equivalence Ratio: It is ratio when C8H15 react with 100% air. Then, C8H15 + 11.75 O2 + 11.75*3.76 N2 = 8 CO2 + 7.5 H2O + 11.75*3.76 N2
/ /
[11.75 + 11.75*3.76]Kmol*29 kg/Kmol =‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ [8*12 + 1*15]kg = 14.61 kg air/kg fuel c) Total number of moles of moles = 8+7.5+3.525+57.434
= 76.459
Percentage of CO of CO2
= [8/76.459]*100
= 10.46%
Percentage of H of H2O
= [7.5/76.459]*100
= 9.80%
Percentage of O of O2
= [3.525/76.459]*100
= 4.61%
Percentage of N of N2
= [57.434/76.459]*100
= 75.11%
Problem:
4.An unknown hydrocarbon is burnt with dry air. The volumetric analysis of of the the products on a dry basis is 12.5% CO2, 0.5% CO, 3% O2 and 84% N2. Determine (a) Air‐Fuel ratio (b) Percentage of of the the theoretical air used. Solution:
By considering 100 kmol of dry of dry products the combustion equation can be written as CxHy + a(O2 + 3.76N2) =
12.5CO2+0.5CO+3O2+84N2+bH2O
For finding unknown values of x,y,a of x,y,a & b For N2
