Chemical Engineering - Chemical Engineering Basics - Discussion
Discussion Forum : Chemical Engineering Basics - Section 7 (Q.No. 8)
8.
Air/fuel ratio on molar (volume) basis for combustion of methane with theoretical quantity of air with be
Discussion:
2 comments Page 1 of 1.
Faycz said:
7 years ago
Molar Ratio:
CH4 + 2 O2 -> C02 + H2O.
Molar ratio.
= (2mole of O2/1 mole of CH4 ) x (100 moles of air/29 moles of O2).
=> 9.5238 mol of air/1 mole fuel.
CH4 + 2 O2 -> C02 + H2O.
Molar ratio.
= (2mole of O2/1 mole of CH4 ) x (100 moles of air/29 moles of O2).
=> 9.5238 mol of air/1 mole fuel.
Sushil said:
8 years ago
It should be 17.2:1.
Here's how
CH4 + 2(O2) --> CO2 + 2(H20).
If we look up the atomic weights of the atoms that make up octane and oxygen, we get the following numbers:
Carbon (C) = 12.01.
Oxygen (O) = 16.
Hydrogen (H) = 1.008.
So 1 molecule of methane has a molecular weight of: 1 * 12.01 + 4 * 1.008 = 16.042.
One oxygen molecule weighs: 2 * 16 = 32.
The oxygen-fuel mass ratio is then: 2 * 32 / 1 * 16.042 = 64 / 16.042.
So we need 3.99 kg of oxygen for every 1 kg of fuel.
Since 23.2 mass-percent of air is actually oxygen, we need : 3.99 * 100/23.2 = 17.2 kg air for every 1 kg of methane.
So the stoichiometric air-fuel ratio of methane is 17.2.
Here's how
CH4 + 2(O2) --> CO2 + 2(H20).
If we look up the atomic weights of the atoms that make up octane and oxygen, we get the following numbers:
Carbon (C) = 12.01.
Oxygen (O) = 16.
Hydrogen (H) = 1.008.
So 1 molecule of methane has a molecular weight of: 1 * 12.01 + 4 * 1.008 = 16.042.
One oxygen molecule weighs: 2 * 16 = 32.
The oxygen-fuel mass ratio is then: 2 * 32 / 1 * 16.042 = 64 / 16.042.
So we need 3.99 kg of oxygen for every 1 kg of fuel.
Since 23.2 mass-percent of air is actually oxygen, we need : 3.99 * 100/23.2 = 17.2 kg air for every 1 kg of methane.
So the stoichiometric air-fuel ratio of methane is 17.2.
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