Completing Combustion - Episode 19 (Air Pollution Engineering ) | 52 PE Exam Problems in 52 Weeks

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  1. SIGMA Environmental, LLC
  2. Environmental Engineering
  3. Environmental Science
  4. Combustion
  5. Chemical Reactions
  6. PE Exam
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Completing Combustion - Episode 19 (Air Pollution Engineering ) | 52 PE Exam Problems in 52 Weeks

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Since the question asks for volume of air required for combustion, why wouldn't we use the equation for the combustion in air which includes nitrogen rather than the equation for combustion in oxygen?

CharlotteHarrell
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At 2:47, you show that 1 scf C3H8 = 5 scf O2 due to the mole ratio. My understanding is that you also have to multiply by molecular weight ratio (32MW of O2 divided by 44MW of C3H8) before you can use the mole ratio. Why do we not have to factor in molecular weights in this problem?

christiancarron
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I keep arriving at 2, 947.4 scf.

My methodology can be seen below:
1 mole of propane reacts with 5 moles of Oxygen
44 lb/mole of propane reacts with 160 lb/mole of Oxygen
44 lb/mole of propane reacts with (160 * (100/21) ) = 761.9 lb/mole of air

20 lbs of propane / Vapor density, 0.1175 lbs/scf = 170.21 scf of propane

170.21 scf of propane * (761 lb/mole of air / 44 lb/mole of propane gas)
= 2947 scf of air - answer.

with the utmost respect, I think it is incorrect to assume that 1SCF of propane = 5SCF of oxygen. Just like how it is wrong to say 1lb of propane = 5 lb of oxygen.

shaangoo