Methane CH4 is fed into a furnace with a flow rate of 100 mo

Methane (CH4) is fed into a furnace with a flow rate of 100 mole/s, where it is burned with excess air. Complete combustion occurs. The stream leaving the furnace is at 600 degree C, 2 atm. and has a dew point of 40 degree C. Calculate the flow rate of the air stream into the furnace. Feed of air into furnace mole/s CH_2 + 2O_2 rightarrow CO_2 + 2H_2O

Solution

Ans
Given The reactionfor combustion of methane:

CH₄ + 2O₂ ===> CO₂ + 2 H₂O ---(1)

CO₂ is the product of complete combustion of carbon and H₂O is the

product of complete combustion of hydrogen.

This reaction is completely balanced. The sum of reactant masses is 80 { CH₄ = 12 +4 , 2 O₂= 2X16X2}

& is equal to sum of masses of products{12+ 2X16 for CO₂ & 2 H₂0 = 2X 18}..

mole fraction of H₂0 = 2/3 =0.67 & corresponding mass fraction of water= 2X18/80=0.45

& mole fraction of CO₂ = 1/3=0.33 & corresponding mass fraction = 44/80=0.55

For Complete Combustion: Additional reactant is nitrogen. As air is a mixture of 21% oxygen and about 79% nitrogen ,

Mass of are = 0.79 X 28 + 0.21 X 32 = 28.84

Hence mass fraction of O₂ = [0.21 X 32]/ 28.84 =0.233

And mass fraction of nitrogen is = [0.79 X 28]/ 28.84=0.767

Now in air , for every one mole of O₂, (79/21) =3.76 mole of N₂ will be present.

hence for complete combustion, equ (1) becomes;

CH₄ + 2O₂ + 2 x(3.76) N₂==> CO₂ + 2 H₂O + 2 x(3.76) N₂ ---(2)

Now the product steam is at 2atms pressure and 600K temperature . As the corresponding

due point is 40 deg C, the steam is completely in vapour form.

From equ (2), we find that the Air to Fuel ratio for methane is:

= [2 X 32 + 2X (3.76) X 28] / (12 +4) =17.16   ----(3)

And the corresponding oxygen to fuel ratio , from equ (2) , is

= [2X32] / (12 +4) =4   ---(4)

From equations (3) & (4) we find, for each 1 Kg of methane, at least 4 kg of oxygen or 17.16 Kg of

air must be supplied to have complete combustion.

No, in the given situation, the input rate is 100 mole/sec for methane.

As 1 mole of methane means 16.043 gm

100 mole = 1.604 kg ---(5)

Hence, from eqns (3) & (5), theoretically the required rate of air = 1.604 X 17.16 kg/sec=27.525 Kg/sec

As molar mass of air=28. 9645 g/mol= 0. 0289645 Kg/mol , we get the reuired rate in terms of mole

= 27.525 kg/sec / 0.0289645 Kg/mol = 950.28 mol/sec

However, for practical purpsore excess air is always needed to have complete the reaction.

The fraction of excess air can also be calculated using gas law.

The mole fraction of the reactant air per sec = n_air / (n_air+ 100)

as the rate of input methane is 100 moles/sec.