Solved Problem On Flue Gas Analysis & Orsat Analysis

                                                                                 

    Flue Gas Analysis & Orsat Analysis

Natural gas containing 80% CH₄, 15% C₂H₆ and 5% C₃H₈ is burnt with 50% excess air. Assuming that 90% of the carbon in the hydrocarbons are converted to CO₂ and the rest to CO, determine

1. Flue gas analysis
2. Orsat analysis

Calculations:

Basis: 1 mole of natural gas

Component	Mole	Atoms of C	Atoms of H
CH4	0.80	0.8 x 1 = 0.8	0.8 x 4 = 3.2
C2H6	0.15	0.15 x 2 = 0.3	0.15 x 6 = 0.9
C3H8	0.05	0.05 x 3 = 0.15	0.05 x 8 = 0.4
Total	1.0	1.25	4.5

Reactions:

C + O₂ à CO₂ -- I

C + 1/2 O₂ à CO -- II

H + 1/4 O₂ à 1/2 H₂O -- III

90% of Carbon is converted by reaction I, and 10% of carbon is converted by II.

Amount of CO₂ produced = 1.25 x 0.9 = 1.125 mole

Amount of CO produced = 1.25 x 0.1 = 0.125 mole

Amount of H₂O produced = 4.5 / 2 = 2.25 mole

Amount of O₂ used by hydrocarbon = O₂ used by reactions I, II and III.

= 1.125 + 0.125 x (1/2) + 4.5 x (1/4) = 2.3125 mole

Theoretical O₂ needed = Oxygen for complete conversion of C to CO₂ and H to H₂O.

= 1.25 + 4.5 x (1/4) = 2.375 mole

Oxygen entering = 150% of theoretical = 1.5 x 2.375 = 3.5625 mole

Therefore, nitrogen entering = 3.5625 x 79/21 = 13.4018 mole = N₂ in the flue gas

O₂ in the flue gases = O₂ entering - O₂ used = 3.5625 - 2.3125 = 1.25 mole

Flue gas analysis:

Component	Moles	Mole %
CO2	1.125	6.20
CO	0.125	0.69
H2O	2.25	12.39
O2	1.25	6.89
N2	13.4018	73.83
Total	18.1518	100

Orsat analysis (Water free):

Component	Moles	Mole %
CO2	1.125	7.07
CO	0.125	0.79
O2	1.25	7.86
N2	13.4018	84.28
Total	15.9018	100

---

Balance of Individual Reactions:

Basis: 1 mole of Natural gas

CH4	0.80
C2H6	0.15
C3H8	0.05
Total	1.0

CH₄ + 2 O₂ à CO₂ + 2 H₂O -- 1

CH₄ + 3/2 O₂ à CO + 2 H₂O -- 2

C₂H₆ + 7/2 O₂ à 2 CO₂ + 3 H₂O -- 3

C₂H₆ + 5/2 O₂ à 2 CO + 3 H₂O -- 4

C₃H₈ + 5 O₂ à 3 CO₂ + 4 H₂O -- 5

C₃H₈ + 7/2 O₂ à 3 CO + 4 H₂O -- 6

In the above reactions CO₂ is produced from reactions 1, 3 and 5.

Since 90% of Carbon is converted to CO₂ and 10% to CO,

CO₂ produced = (1 x 0.8 + 2 x 0.15 + 3 x 0.05) x 0.9 = 1.125 mole

Similarly CO is obtained from reactions 2, 4 and 6.

CO produced = (1 x 0.8 + 2 x 0.15 + 3 x 0.05) x 0.1 = 0.125 mole

H₂O produced = (2 x 0.8 + 3 x 0.15 + 4 x 0.05) x 0.9 + (2 x 0.8 + 3 x 0.15 + 4 x 0.05) x 0.1 = 2.25 mole

O₂ used up in these reactions = (2 x 0.8 + 3.5 x 0.15 + 5 x 0.05) x 0.9 + (1.5 x 0.8 + 2.5 x 0.15 + 3.5 x 0.05) x 0.1

= 2.3125 mole

Theoretical O₂ needed = moles of O₂ needed for Conversion of C to CO₂ and H to H₂O.

= 2 x 0.8 + 3.5 x 0.15 + 5 x 0.05 = 2.375 mole

O₂ entering = 50 % excess = 150% of theoretical = 2.375 x 1.5 = 3.5625 mole

N₂ entering along with O₂ in the air = 3.5625 x 79/21 = 13.4018 mole (sine air is 21% O₂ and 79% N₂ by volume).

O₂ in the flue gas = O₂ entering - O₂ used up = 3.5625 - 2.3125 = 1.25 mole

N₂ in the flue gas = N₂ entering = 13.4018 mole

Therefore, for 1 mole of Natural gas entering, the flue gas coming out are:

Component	Moles
CO2	1.125
CO	0.125
H2O	2.25
O2	1.25
N2	13.4018
Total	18.1518

On comparing with the data obtained from atomic balance and balance of individual reactions, it can be seen that the results are same from either method. But instead of lengthy calculations for individual reactions, we can very well make use of atomic balances.