Question

Natural gas containing a mixture of methane, ethane, propane, and butane is burned in a furnace with excess air. (a) One hundred kmol/h of a gas containing 94.4mole% methane, 3.40% ethane, 0.60% propane, and 0.50% butane is to be burned with 17% excess air. Calculate the required molar flow rate of the air.

Answer 1

The required flow rate of air is 1153.01kmole/h

Step-by-step explanation:

we can obtain the number of moles of each gas contained in the mixture

Given a basis of 100kmole/h

94.4% methane CH₄ = (94.4/100)x100 = 94.4kmole/h

3.4% ethane C₂H₆ = (3.4/100)x100 = 3.4kmole/h

0.6% Propane C₃H₈ = (0.6/100)x100 = 0.6kmole/h

0.5% butane C₄H₁₀ = (0.5/100)x100 = 0.5kmole/h

the equation for combustion for each of the gases above is

methane: CH₄ + 2O₂ ⇒ 2H₂O + CO₂

1mole CH₄ ⇒ 2moles O₂

94.4kmole CH₄ ⇒ 2x94.4 = 188.8kmole

ethane: C₂H₆ + 3.5O₂ ⇒ 3H₂O + 2CO₂

1mole C₂H₆ ⇒ 3.5moles O₂

3.4kmole C₂H₆ ⇒ 3.5x3.4 = 11.9kmole

propane: C₃H₈ + 5O₂ ⇒ 4H₂O + 3CO₂

1mole C₃H₈ ⇒ 5moles O₂

0.6kmole C₃H₈ ⇒ 5x0.6 = 3kmole

butane: C₄H₁₀ + 6.5O₂ ⇒ 5H₂O + 4CO₂

1mole C₃H₈ ⇒6.5moles O₂

0.5kmole C₃H₈ ⇒ 6.5x0.5 = 3.25kmole

Total amount of oxygen required will be the sum of the all amount required by individual gases for combustion

188.8 + 11.9 + 3 + 3.25 = 206.95kmole/h

Recall that oxygen is 21% of air,

therefore 21% of what amount of air is 206.95kmole of oxygen, let that amount be x

0.21x = 206.95

x = 985.48kmole/h

and excess of 17% of air is added

Required flow rate of air = 985.48 + 0.17(985.48) = 1153.01kmole/h

The required flow rate is 1153.01kmole/h