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28 August, 16:39

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.

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  1. 28 August, 17:41
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    The required flow rate of air is 1153.01kmole/h

    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
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