A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water.

(a) Calculate the molar composition of this gas on both a wet and a dry basis and the ratio (mol H2O/ mol dry gas).

(b) If 100 kg/h of this fuel is to be burned with 30% excess air, what is the required air feed rate (kmol/h)? How would the answer change if the combustion were only 75% complete.

(c) Determine the composition of wet flue gas in both cases.

a) Molar composition on a wet basis

Methane's mole percent = 81.4 mol%

Ethane's mole percent = 5.79 mol%

Ethane's mole percent = 3.1 mol%

Water's mole percent = 9.65 mol%

On a dry basis,

Methane's mole percent = 90.2 mol%

Ethane's mole percent = 6.41 mol%

Ethane's mole percent = 3.44 mol%

The ratio of mol H₂O to mol of dry gas = 0.107

b) The molar feed rate of air = 68.58 kmol/h.

The amount of air required for 75% combustion is exactly 75% of that required for the 100% combustion, 51.43 kmol/h

c) Molar composition of flue gas

CO₂ molar flowrate = 5.7114 kmol/h

H₂O molar flowrate = 11.288 kmol/h

Air molar flowrate = 57.5 kmol/h

Mole percent without including air

CO₂ molar percent = 33.6%

H₂O molar percent = 66.4%

Mole percent while including air

CO₂ molar percent = 7.67%

H₂O molar percent = 15.2%

Air molar percent = 77.2%

Explanation:

The gas contains:

75wt% of CH₄

10wt% of C₂H₆

5wt% of C₂H₄

The rest, 10wt% is H₂O

a) Molar composition on a wet and dry basis.

Assuming a basis of 100 g of gas.

CH₄ has a mass of 75 g

C₂H₆ has a mass of 10 g

C₂H₄ has a mass of 5 g

And H₂O has a mass of 10 g

We convert each of this to number of moles.

Number of moles = (mass)/(molar mass)

For CH₄, mass = 75 g, molar mass = 16.0 g/mol

Number of moles = 75/16 = 4.6875 moles

For C₂H₆, mass = 10 g, molar mass = 30.0 g/mol

Number of moles = 10/30 = 0.3333 mole

For C₂H₄, mass = 5 g, molar mass = 28.0 g/mol

Number of moles = 5/28 = 0.1786 mole

And H₂O, mass = 10 g, molar mass = 18.0 g/mol

Number of moles = 10/18 = 0.5556 mole

On a wet basis, we include water (its mass or number of moles) in our calculations.

Total number of moles = 4.6875 + 0.3333 + 0.1786 + 0.5556 = 5.755 moles

Methane's mole percent = 100% × 4.6875/5.755 = 81.4 mol%

Ethane's mole percent = 100% × 0.3333/5.755 = 5.79 mol%

Ethane's mole percent = 100% × 0.1786/5.755 = 3.1 mol%

Water's mole percent = 100% × 0.5556/5.755 = 9.65 mol%

On a dry basis, we don't include the mass or number of moles of water in our calculations.

Total number of moles = 4.6875 + 0.3333 + 0.1786 = 5.1994 moles

Methane's mole percent = 100% × 4.6875/5.1994 = 90.2 mol%

Ethane's mole percent = 100% × 0.3333/5.1994 = 6.41 mol%

Ethene's mole percent = 100% × 0.1786/5.1994 = 3.44 mol%

The ratio of mol H₂O to mol of dry gas = 0.5556/5.1994 = 0.107

b) If 100 kg/h of gas is to be burnt,

From the similar calculation in (a), the molar flowrate of the respective gases is

4.6875 kmol/h of CH₄

0.3333 kmol/h of C₂H₆

0.1786 kmol/h of C₂H₄

0.5556 kmol/h of H₂O

The air available has to burn each and everyone of the combustible gases

Combustion of Methane

CH₄ + 2O₂ > CO₂ + 2H₂O

1 mole of CH₄ requires 2 moles of O₂ and gave 1 mole of CO₂ and 2 moles of H₂O

4.6875 kmol/h will require (2 × 4.6875) = 9.375 kmol/h of O₂ and produce (1 × 4.6875) = 4.6875 kmol/h of CO₂ and (2 × 4.6875) = 9.375 kmol/h of H₂O

From the combustion of methane, we have the following details

9.375 kmol/h of O₂ required

4.6875 kmol/h of CO₂ produced

9.375 kmol/h of H₂O produced

For the combustion of ethane C₂H₆

C₂H₆ + (7/2)O₂ > 2CO₂ + 3H₂O

1 mole of C₂H₆ requires (7/2) moles of O₂ and gave 2 mole of CO₂ and 3 moles of H₂O

0.3333 kmol/h will require ((7/2) × 0.3333) = 1.167 kmol/h of O₂ and produce (2 × 0.3333) = 0.6667 kmol/h of CO₂ and (3 × 4.6875) = 1 kmol/h of H₂O

From the combustion of ethane, we have the following details

1.167 kmol/h of O₂ required

0.6667 kmol/h of CO₂ produced

1 kmol/h of H₂O produced

Combustion of ethene

C₂H₄ + 3O₂ > 2CO₂ + 2H₂O

1 mole of C₂H₄ requires 3 moles of O₂ and gave 2 mole of CO₂ and 2 moles of H₂O

0.1786 kmol/h will require (3 × 0.1786) = 0.5358 kmol/h of O₂ and produce (2 × 0.1786) = 0.3572 kmol/h of CO₂ and (2 × 0.1786) = 0.3572 kmol/h of H₂O

From the combustion of ethene, we have the following details

0.5358 kmol/h of O₂ required

0.3572 kmol/h of CO₂ produced

0.3572 kmol/h of H₂O produced

The amount of O₂ required is the sum total from the 3 combustion reactions = 9.375 + 1.167 + 0.5358 = 11.0778 kmol/h

With a 30% excess of O₂ required, the real amount of O₂ = 1.3 × 11.0778 kmol/h = 14.40 kmol/h

Mol% of oxygen in air is 21%

21% of the entire molar flowrate of air = 14.40 kmol/h

The entire molar flowrate of air = 14.40/0.21 = 68.58 kmol/h.

If the combustion were only 75% complete,

The amount of O₂ required = 0.75 of the theoretical molar flowrate of O₂ required for complete combustion = 0.75 × 11.0778 = 8.308 kmol/h

Adding the 30% excess, 1.3 × 8.308 = 10.8 kmol/h

Molar feed rate of air if combustion is 75% = 10.8/0.21 = 51.43 kmol/h

The amount of air required for 75% combustion is exactly 75% of that required for the 100% combustion.

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