# 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 0.8 C2H6 0.15 C3H8 0.05 Total 1

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.