Respuesta :
Answer:
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.
