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Annex VI

AVI-1. Molecular Weight Of A Gaseous Mixture Let’s be ΨI, the molar fraction of the gas species “i”, and MWi its molecular weight. The molecular weight of a gaseous mixture is defined as follows: MWmixt =

∑ MW ·ψ i

(AVI-1)

i

i

AVI-2. Limiting Reactant Let us consider the combustion reaction between methane and oxygen. The reaction can be represented by the equation: CH 4 + 2O2 → CO2 + 2H2O

(AVI-2)

This balanced reaction equation indicates that one mol of CH4 would react with two moles of O2. Thus, if you have 2 moles of CH4, 4 moles of O2 will be required. If there is an excess of moles of O2 (more than 4), they will remain unreacted, some will remain as an excess reactant, and the CH4 is a limiting reactant. It limits the amount of the product that can be formed.

AVI-3 Specific Heat of O2, H2, N2, CO, H2O,

CO2

cp,O2 = -4·106 T2 + 0.0128T + 25.802

for O2

(AVI-3)

cp,H2 = 2·10-6 T2 - 0.0004T + 28.913

for H2

(AVI-4)

cp,N2 = 0.0051T + 27.403

for N2

(AVI-5)

cc,CO = 0.0053T + 27.5

for CO

(AVI-6)

cp,H2O = 0.0111T + 29.959

for H2O

(AVI-7)

cp,CO2 = −10−5 T2 + 0.042T + 26.475

for CO2

(AVI-8)

Where T is give in [K] and cp is obtained in [J/molK].

AVI-4 Percentage of Unburnt Species Reacting in a Deflagration In a combustion reaction with lack of oxygen, not all the unburnt products react until exhaustion. Let’s explain how the oxygen is shared among the unburnt products with an example. Let’s suppose a mixture formed of methane, dioxide carbon, monoxide

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Annex VI carbon, hydrogen, water, oxygen and nitrogen. The molar fractions in the compartment of the species are indicated below: ψi

Molar fraction

CH4 [%] 13.6

CO2 [%] 1.65

CO [%] 0.15

H20 [%] 7

O2 [%] 13.5

H2 [%] 0

N2 [%] 64

First of all, the stoichiometric reaction of each unburnt gas must be established, where the value before the O2 is represented in this section as Xi: CH 4 + 2O2 → products ( H2O,CO2 )

(AVI-9)

CO + 1 2 O2 → products ( H2O, CO2 )

(AVI-10)

H2 + 1 2 O2 → products ( H2O,CO2 )

(AVI-11)

Then, the ratio specie-mixture is calculated according to Eq. (AVI-12) and Eq. (AVI-13): mol total = molO2 + mol i ϕi =

(AVI-12)

mol i mol total

(AVI-13)

Resolving for CH4: mol total = molO2 + molCH4 = 2 + 1 = 3

ϕi = 100

molCH4 mol total

(AVI-14)

= 33.33

(AVI-15)

The following Table shows the results for every species. FROM Eq. (AVI-9) Eq. (AVI-10) Eq. (AVI-11)

MolesTotal [mol] 3.0 1.5 1.5

Moles of i 1 1 1

Moles of O2 2 0.5 0.5

ϕunburnt [%] 33.3 66.6 66.6

ϕoxygen [%] 66.6 33.3 33.3

molOi 2 required 27.20 0.075 0.000

Table AVI-1: Ratio specie-mixture Assuming 100 mol of mixture, the required oxygen for a stoichiometric combustion is obtained as:

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Annex VI molO2 required =

∑ mol

i O2 required

= 100∑ ψ i ·X i

i

(AVI-16)

i

Column 7 of Table AVI-1 shows the required oxygen for the species present in the mixture. This data is needed when trying to share in equality the mol of oxygen present in the initial mixture. It is done according to: mol

i O2 → unburnt

=

molO2 _mixture ·molOi 2 required

(AVI-17)

molO2 required

Once molOi 2 →unburnt has been obtained, the moles of unburnt species that react is given by:

mol

i_ react deflagration

=

molOi 2 →unburnt

(AVI-18)

Xi

Table AVI-18 show the result obtained using Eq. (AVI-18): Xi [--]

Eq. (AVI-9) Eq. (AVI-10) Eq. (AVI-11)

2 0.5 0.5

Moles of i in mixture 13.45 0.05 0.00

molOi 2 →unburnt [mol] 13.46 0.037 0.000

i_ react

mol deflagration [mol] 6.730 0.074 0.000

Table AVI-2: Moles of unburnt species that react Eq. (AVI-19) is used for obtaining the excess of oxygen, 2 _ excess 2 2 react molOtignition = molOtignition − molOdeflagration

(AVI-19)

2 _ excess molOtignition = 13.5 − (13.45 + 0.05 ) = 0.0

(AVI-20)

Eq. (AVI-21) is used for obtaining the excess of oxygen, excess _react mol i_ = mol itignition − mol ideflagration tignition

(AVI-21)

4 _excess molCH = 13.6 − 13.45 = 0.15 tignition

(AVI-22)

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Annex VI

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Annexvi