STOICHIOMETRY OF COMBUSTION


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1 STOICHIOMETRY OF COMBUSTION
2 FUNDAMENTALS: moles and kilomoles Atomic unit mass: 1/ C ~ kg Atoms and molecules mass is defined in atomic unit mass: which is defined in relation to the 1/12 of carbon 126 C. Mole: (Avogadro number) Volume of 1 mole (perfect gas) Kmole: atoms l (T= 0 o C, p = 1 atm) 10 3 moles Mass of one mole (kmole) is a number of grams (kilograms) equal to the relative atomic mass.
3 CHEMICAL REACTIONS Consider a simple chemical reaction of two species A and B giving a single product C A + 2B C 1 mol + 2 moles 1 mol A, B and C are reactants. A and B are substracts. C is a product of reaction. Stoichiometry of chemical reactions means that species react in exact proportions.
4 SPECIE S CONCENTRATIONS AND FRACTIONS Concentration of i specie: (moles of i specie) /volume (mass of i specie)/volume Mole fraction: (moles of i specie)/(total number of moles of species) Mass fraction: (mass of i fraction)/(total mass)
5 STOICHIOMETRY OF HYDROCARBONS OXIDATION Fossil fuels are mainly compounds of carbon and hydrogen (hydrocarbons  C m H n ). The reaction of its oxidation can by written by the equation of stoichiometry: C m H n + (m + n/4)o 2 mco 2 + n/2h 2 O 1 mol + (m + n/4)moles (m +n/2)moles This is an equation of stoichiometry of combustion. It is important that for one mole of fuel C m H n there is necessary exactly: (m + n/4) moles of oxygen for complete combustion.
6 TYPES OF OXIDIZERS In combustion processes the oxidizer could be: 1. Oxygen (O 2 ) 2. Air (21%O %N 2 ) 3. Air enriched with oxygen (O 2 > 21%) 4. Some compounds containing oxygen, like nitrogen oxides: N 2 O
7 COMPOSITION OF FUEL MIXTURE When fuel and oxidizer composition in the mixture (fuel and oxidizer) results from the equation of stoichiometry we say that the mixture is stoichiometric. If combustion of a stoichiometric mixture is complete in flue gas cannot be nor fuel neither oxygen.
8 TYPES OF COMBUSTIBLE MIXTURES FUEL MIXTURE Rich Stoichiometric Lean Excess of fuel Stoichiometric content of fuel and oxygen Excess of oxidizer
9 BURNING OF STOICHIOMETRIC METHANE MIXTURE IN AIR Air: 79% N % O 2 CH 4 + 2O N 2 CO 2 + 2H 2 O N 2 Attention: there is not fuel neither oxygen in flue gas.
10 BURNING OF STOICHIOMETRIC METHANE MIXTURE IN AIR: flue gas composition A number of moles of flue gas N: N = 1 mole CO moles H 2 O mole N 2 = moles According to the wet (water is steam/liquid) analysis of flue gas the concentration of the components is as follows: [CO 2 ] = 1mole CO 2 /10.52 = 9.5% CO 2 vol. [H 2 O] = 2 moles H 2 O/10.52 = 19% H 2 O vol. [N 2 ] = 7.52 mole N 2 /10.52 = 71.5% N 2 vol.
11 BURNING OF LEAN METHANE MIXTURE IN AIR Assumption: AIR EXCESS = 5% CH O N 2 CO 2 + 2H 2 O + 7.9N O 2 Attention: there is oxygen in flue gas.
12 BURNING OF LEAN METHANE MIXTURE IN AIR: flue gas composition A number of moles of flue gas N equals: N = 1 mole CO moles H 2 O mole N mole O 2 = 11 moles According to the wet (water is liquid) analysis of flue gas the concentration of the components is as follows: [CO 2 ] = 1mole CO 2 /11 = 9.09% CO 2 vol. [H 2 O] = 2 moles H 2 O/11 = 18.2% H 2 O vol. [O 2 ] = 0.1 mole O 2 /11 = 0.91% O 2 vol. [N 2 ] = 7.52 moles N 2 /10,52 = 71.82% N 2 vol.
13 STOICHIOMETRIC AIR The theoretical air required to complete combustion of fuel results from the equation of stoichiometry of oxygen/fuel reaction. Stoichiometric air means the minimum air in stoichiometric mixture. The stoichiometric air/fuel ratio (AFR) can be calculated from the reaction equation (g/g). For gas AFR is usually determined in m 3 /m 3. The actual combustion air depends also on the assumed air excess (equivalence ratio or stoichiometric ratio).
14 THE STOICHIOMETRIC RATIO (λ) (WSPÓŁCZYNNIK NADMIARU POWIETRZA) λ = Actual air Stoichiometric air λ [0 2 ]
15 AN EQUIVALENCE RATIO φ (WSPÓŁCZYNNIK EKWIWALENCJI) In English more often than stoichiometric ratio λ the equivalence ratio φ is in use. φ = F/A (F/A) s φ = 1 λ F number of moles of fuel, A number of moles of air. s  stoichiometric
16 AIR/FUEL RATIO (AFR) and FUEL/AIR RATIO (FAR) Form the practical reasons in aviation (aircrafts) mostly in use is the ratio of air massflow to the fuel massflow determined as AFR (Air/Fuel Ratio) AFR = F/A A air massflow (A air), kg/s F fuel massflow (F fuel), kg/s In similar way FAR (Fuel/Air Ratio) is defined.
17 STOICHIOMETRIC RATIO/EQUIVALENCE RATIO AND TYPE OF MIXTURE Type of flame/type of mixture Rich Stoichiometric Lean (weak) λ <1 φ > 1 λ =1 φ = 1 λ >1 φ < 1
18 AIR EXCESS (n) n = actual  stoichiometric air stoichiometric air The relationship between the stoichiometric ratio λ and air excess n n = (λ 1) 100%
19 STOICHIOMETRIC AIR/FUEL RATIO FOR SELECTED GASES Gaz CO H 2 CH 4 C 2 H 2 C 2 H 6 C 3 H 8 Natural gas Stoichiometric air/fuel ratio AFR, m 3 /m (approx. 82% CH 4 )
20 EXAMPLE Stoichiometric air for methane burning (CH 4 ) is m 3 /m 3 Assuming: the rate feeding of methane V = 10 m 3 /h the rate feeding of air V = m 3 /h We have: λ= /104.2 = 1.1 n = ( ) 100% = 10%
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