Annex IV
AIV-1. Adiabatic Flame Temperature At Lower Flammability Limit The AFTLFL is explained as an example in which the fuel gas is methane, CH4. Let’s consider the chemical reaction of methane where its concentration is 5% (value of LFL).
(
0.05CH 4 + (1 − 0.05)·( 0.21O2 + 0.79N2 ) → Pr oducts CO2 , H2O, O2 and N2
)
(AIV-1) Dividing through by 0.05 gives: CH 4 + 3.99O2 + 15.01N2 → CO2 + 2H2O + 1.99O2 + 15.01N2
(AIV-2)
As the original mixture was fuel lean, the excess of oxygen will contribute to the total thermal capacity of the product mixture. For obtaining the AFTLFL, the method of section must be applied. Considering an initial temperature of 25ºC, the AFTLFL is 1200 ºC.
AIV-2. Adiabatic flame temperature at upper flammability limit The AFTUFL is also explained as an example in which the fuel gas is methane, CH4. Let’s consider the chemical reaction of methane where its concentration is 15% (value of UFL). Stoichiometric combustion XCH4 + CH4 + 2O2 + 7.52N2 → CO2 + 2H2O + 7.52 N2 + XCH4 (AIV-3) For obtaining what is the fuel non burnt:
UFL =
X +1 = 0.15 X + 1 + 2 + 7.52
(AIV-4)
Therefore, X is equal to 0.68.
0.68CH4 + CH4 + 2O2 + 7.52N2 → CO2 + 2H2O + 7.52 N2 + 0.68CH4 (AIV-5) As the original mixture was lack of oxygen, the excess of fuel will contribute to the total thermal capacity of the product mixture. For obtaining the AFTUFL, the method of clause must be applied. Considering a initial temperature of 25ºC, and a thermal capacity of methane equal to 81.3 J/molK the AFTUFL is 1565 ºC.
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Annex IV
AIV-3. Adiabatic Concentration
Flame
Temperature
At
Limiting
Oxygen
The AFTLOC is also explained as an example in which the fuel gas is methane, CH4. Let’s consider the chemical reaction of a gaseous mixture as follows:
• • • •
CH4 concentration is: 6.6 % O2 concentration is: 12.0% N2 concentration is: 45.14% Inert concentration is: 35.25%
Therefore,
(
0.066CH 4 + 0.5714 ( 0.21O2 + 0.79N2 ) + 0.3525Inert → Pr oducts CO2 , H2O,O2 and N2 (AIV-6) Dividing through by 0.066 gives: CH4 + 1.818O2 + 6.834N2 + 5.34Inert → CO2 + 2H2O + 6.834N2 + 5.34Inert (AIV-7)
At LOC the mixture is found in stoichiometric relation, therefore no oxygen or fuel is found in the combustion products. Now, the inert will contribute to the total thermal capacity of the product mixture. For obtaining the AFTLOC, the method of clause must be applied. Considering an initial temperature of 25ºC and N2 as inert species, the AFTLOC is 1523 ºC.
AIV-3. Example of diffusive flame limit, Le Chaterier’s rule As an example, let’s suppose a mixture at 700K consisting of 10% of hydrocarbons in the form of CH2, 2% of CO, 1% of H2, 15% of CO2, 2% of O2 and 70% of N2. In order to check if the mixture can be flammable when it is mixed with air fresh at 300K, the following procedure is done: The working equation is Eq. (AIV-8). The first step is to write the balanced chemical equation for stoichiometric burning. 0.1CH2 + 0.02CO + 0.01H2 + 0.02O2 + 0.7N2 + 0.15CO2 + x ( O2 + 3.78N2 ) → 0.27CO2 + 0.11H2O + ( 0.7 + 3.78XN2 )
(AIV-8)
We can find “x” by requiring that both sides of this equation have the same amount of oxygen.
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Annex IV
CO
O2
CO2
air CO2
H 2O
0.02 / 2 + 0.02 + 0.15 + x = 0.27 + 0.11/ 2 → x = 0.145
(AIV-9)
The concentration in the stoichiometric mixture can be determined from the balanced chemical equation Ci = ( n i / n T ) 100%
(AIV-10)
n T 0.1 + 0.02 + 0.01 + 0.02 + 0.7 + 0.15 + 0.145 + 0.145·3.78 = 1.693
(AIV-11)
CHC = ( 0.1/1.693 )·100% = 5.9%
(AIV-12)
CCO = ( 0.02 /1.693 )·100% = 1.2%
(AIV-13)
CH2 = ( 0.01/1.693 )·100% = 0.6%
(AIV-14)
Similarly, the number of moles of products per mole of reactants can be determined from the chemical equation n p = ( 0.27 + 0.11 + 0.7 + 0.145·3.78 ) /1693 = 0.962
(AIV-15)
This is lower than typical values of 1 to 1.1, because the unknown hydrocarbon mixture is taken as CH2. This is not an error, since CH2 has been consistently used for the heat released and heat capacity as well. For convenience, we will use constant average specific heat taken from Drysdale, 1999, see Table 2-2 of Chapter 2.
n pCp = n p ∑ ( Ci /100 ) Cp,i = 0.96 [( 0.162·54.3 ) + ( 0.066·41.2 ) + ( 0.772·32.7 )] (AIV-16) One may note that the average specific heat is near that of nitrogen, since it is the major constituent of mixture. In calculating To, the initial temperature of the mixture, we will ignore variations in Cp between the hot and the cold layers.
To =
( n h Th + nc Tc ) ( nh + nc )
=
(1·700K + 0.69·300K ) 1.69
nT = nc + nh
= 537K
(AIV-17) (AIV-18)
Where nh and nc are the number of moles originating in the hot and cold layers, respectively. Substituting into Eq. (8-38) of Chapter 8 and assuming a combustion efficiency factor of 1.0: n
( Ci /100 ) ∆HC,i
∑ C (T i =1
p
f,SL,i
− To )
=
0.059·620·103 + 35.3·(1700 − 537 ) 0.012·283·103 0.006·242·103 + = 1.07 35.3·(1450 − 537 ) 35.3·(1080 − 537 ) (AIV-19)
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Since the result is greater than 1, the hot layer will ignite and burn.
AIV-4. Correlation of AFTLOC and AFTLFL for Alkenes and Alkanes 2300 2200
ALKANES - C[n]H [2n+2]
2100
ALKENES - C[n]H[2n]
2000
AFT_LOC [K]
1900 1800 1700 1600 1500 1400
AFT_LOC = 1.8345[AFT_LFL] - 796.75
1300
R2 = 0.7768
1200 1100 1000 1000
1050
1100
1150
1200
1250
1300
1350
1400
1450
1500
1550
1600
1650
1700
AFT_LFL [K]
AIV-5. Combustion heat approximation 6000 5500
For C, H, O-containing fuels
Real Combustion heat [kJ/mol]
5000
12.8[kJ/g(O2)]
4500 4000 3500 3000 2500 2000
Average deviation = -4.29%
1500 1000 500 500
1000
1500
2000
2500
3000
3500
4000
4500
5000
5500
6000
Combustion heat calculated [kJ/mol]
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