Composition of flue gases by volume on a wet basis and dry basis

[PCal]

Homework Statement

A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume. It is to be fed to the combustion chamber in 10% excess air at 25ºC, where it is completely burnt to carbon dioxide and water. The flue gases produced are to be used to generate 5 bar steam from water at 90ºC. Data: Net calorific value (MJ m–3) at 25ºC of: Butane (C4H10) = 111.7 MJ m–3 Butene (C4H8) = 105.2 MJ m–3 Propane (C3H8) = 85.8 MJ m–3 Air is 21% oxygen, 79% nitrogen by volume and 23.3% oxygen and 76.7% nitrogen by mass. Atomic mass of C = 12, O = 16, N=14 and H = 1.
Determine the composition of the flue gases by volume (assuming the inlet air is dry)
(i) on a wet basis
(ii) on a dry basis
I know the main part of the question has been posted before but I cant find any threads about the flue gas analysis. Any help would be greatly appreciated as my lessons don't seem to give much help!

Relevant Equations

Balanced reactions
Butane C4H10 + 6.5O2 = 4CO2 + 5H20
Propane C3H8 + 5O2 = 3C02 + 4H2O
Butene C4H8 + 6O2 = 4CO2 + 4H20

Wet basis

0.75mol C4H10

Requires 4.875 mols O2

Produces 3 mols of CO2 and 3.75 mols of H2O

0.1mol C3H8

Requires 0.5 mols O2

Produces 0.3 mols of CO2 and 0.4 mols of H2O

0.15mol C4H8

Requires 0.9 mols O2

Produces 0.6 mols of CO2 and 0.6 mols of H2O

Theoretical oxygen= 6.3mol +10% excess = 6.93mol

N2= 26.07

Total mol in flue gas = 35.35mol

Flue gas composition by volume (wet basis)

CO2 3+0.3-0.6=3.9mol

H2O 3.75+0.4+0.6=4.75mol

N2=26.017mol

O2=0.63mol

I’m not sure if this is correct and I also don’t have a clue how to work out the composition on a dry basis.

 

[PCal]

Is the dry basis as simple as saying...

CO2 3+0.3-0.6=3.9mol

N2=26.017mol

O2=0.63mol

 

[PCal]

Any help would be massively appreciated please!

 

[Chestermiller]

Wet basis

0.75mol C4H10

Requires 4.875 mols O2

Produces 3 mols of CO2 and 3.75 mols of H2O

0.1mol C3H8

Requires 0.5 mols O2

Produces 0.3 mols of CO2 and 0.4 mols of H2O

0.15mol C4H8

Requires 0.9 mols O2

Produces 0.6 mols of CO2 and 0.6 mols of H2O

Theoretical oxygen= 6.3mol +10% excess = 6.93mol

N2= 26.07

Total mol in flue gas = 35.35mol

Flue gas composition by volume (wet basis)

CO2 3+0.3-0.6=3.9mol

H2O 3.75+0.4+0.6=4.75mol

N2=26.017mol

O2=0.63mol

I’m not sure if this is correct and I also don’t have a clue how to work out the composition on a dry basis.

They want the mole fractions of the various species (a) for the flue gas as-is and (b) with the water vapor removed. (In common terminology, volume fraction is the same as mole fraction)

 

[PCal]

Fab got it! Thank you!