Just the first part please<3 Problem 3.32 Methane Dilution A mixture of methane
ID: 700920 • Letter: J
Question
Just the first part please<3
Problem 3.32 Methane Dilution
A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%.
A mixture containing 9.00 mole% methane in air flowing at a rate of 1300.0 kg/h is to be diluted with pure air to reduce the methane concentration to the lower flammability limit.
Calculate the required flow rate of air in mol/h and the percent by mass of oxygen in the product gas.
(Note: Air may be taken to consist of 21.0 mole% O2 and 79.0 mole% N2 and hence to have an average molecular weight
Problem 3.32 Methane Dilution
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A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%.
A mixture containing 9.00 mole% methane in air flowing at a rate of 1300.0 kg/h is to be diluted with pure air to reduce the methane concentration to the lower flammability limit.
Calculate the required flow rate of air in mol/h and the percent by mass of oxygen in the product gas.
(Note: Air may be taken to consist of 21.0 mole% O2 and 79.0 mole% N2 and hence to have an average molecular weight
Explanation / Answer
Average Molecular weight of inlet mixture
= molecular weight of air x mol fraction of air + molecular weight of CH4 x mol fraction of CH4
= (29 x 0.91) + (16 x 0.09)
= 27.83 g/mol = 27.83 x 10^-3 kg/mol
Molar flow of inlet = (1300kg/h)/(27.83 x 10^-3 kg/mol)
= 46712.18 moles/hr
Molar flow of Methane in inlet mixture
= 46712.18 x 0.09 = 4204.09 moles/hr
Molar flow of air in inlet mixture = 42508.08 moles/hr
reduce the methane concentration to the lower flammability limit i.e. 5%
Balance the CH4
Inlet flow = outlet flow
4204.09 = 0.05 x molar flow rate of outlet mixture
molar flow rate of outlet mixture = 84081.8 mol/h
Air balance
Inlet flow + dilution air = outlet flow
42508.08 + dilution air = 84081.8 x 0.95
dilution air = 37369.63 mol/h
= 37.369 kmol/h
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