PETROLEUM ENGINEERING - QUESTION 25.1 : Fact 1 : Dry air contains 20.95 % oxygen

PETROLEUM ENGINEERING - QUESTION 25.1 : Fact 1 : Dry air contains 20.95 % oxygen, 78.09 % nitrogen, 0.93 % argon, 0.039 % carbon dioxide, and small amounts of other gases by volume. Fact 2 : Volume occupied is directly proportional to the number of moles for ideal gases at constant temperature and pressure. Fact 3 : 12.5 moles of pure oxygen is required to completely burn 1 mole of pure octane. Fact 4 : Air-fuel ratio (AFR) is the mass ratio of dry air to fuel present in a combustion process such as in an internal combustion engine or industrial furnace. Fact 5 : Molecular weight of oxygen gas is 31.998 g / mole and molecular weight of nitrogen gas is 28.014 g / mole. (a) Find the molar ratio of nitrogen and oxygen, or (moles of nitrogen) / (moles of oxygen) in dry air, by assuming ideal features of nitrogen and oxygen gases. (b) How many moles of nitrogen are available if dry air is used to completely burn the 1 mole pure octane? (c) Find the mass of fuel of 1 mole of octane with molecular weight of 114.232 g / mole. (d) Find the mass of dry air with 12.5 moles of pure oxygen by assuming only oxygen and nitrogen gases exist in the air. (e) Find the air-fuel ratio (AFR) when octane is used as fuel. (f) Find the fuel-air ratio (FAR) when octane is used as fuel.

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PETROLEUM ENGINEERING - QUESTION 25.1 : Fact 1 : Dry air contains 20.95 % oxygen, 78.09 % nitrogen, ..

Answer / kangchuentat

PETROLEUM ENGINEERING - ANSWER 25.1 : (a) Ideally (moles of nitrogen) / (moles of oxygen) = (volume of nitrogen) / (volume of oxygen) = 78.09 % V / 20.95 % V = 3.7274 where V is the volume of dry air. (Fact 1, Fact 2). (b) Moles of nitrogen = (moles of oxygen) x (volume of nitrogen) / (volume of oxygen) = 12.5 x 3.7274 = 46.5925 moles. (c) Mass of 1 mole of octane fuel = 1 mole x 114.232 g / mole = 114.232 g. (d) Mass of dry air = mass of oxygen + mass of nitrogen = 12.5 mole x 31.998 g / mole + 46.5925 mole x 28.014 g / mole = 1705.217295 g. (Fact 5) (e) Air-fuel ratio (AFR) = [ Answer in (d) ] / [ Answer in (c) ] = 1705.217295 g / 114.232 g = 14.928. (Fact 3, Fact 4) (f) Fuel-air ratio (FAR) = 1 / [ Answer in (e) ] = 1 / 14.928 = 0.067. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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