Ethanol

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Ethanol as an alternative fuel has been a thing for quite some time. Only recently are cars in the States coming from the manufacturer with a flex fuel sensor. This sensor determines the ethanol content in the fuel of the vehicle. I’ve talked with quite a few people that have tried to use ethanol in their brand new vehicles, and their only response is, “I got terrible gas mileage, I wont be filling up with that again.” Yes, gas mileage is worse. That’s all because of the stoichiometric fuel/air mixture for ethanol which is 9:1. Octane gas has a stoichiometric ratio of 14.7:1. So for the fuel to fully burn and have no traces of it left after the reaction occurred, it needs almost 40% more fuel for an ethanol burn.
Often when ethanol is being discussed, someone mentions how it burns colder than octane gas. Let’s talk about how a colder burn correlates to temperatures in the cylinder and power. Some questions can be brought up right off the bat. How can a fuel that burns “colder” generate more power? How much “colder” does it burn? To help explain and understand these questions, here are equations to help you see just what ethanol does in comparison to octane gas.

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 Ethanol Vs Gasoline


When one mole  (6.02 x 10^23 molecules) of octane gas and a mole of ethanol are combusted, gas creates 5460 KJ of heat and ethanol creates 1368 KJ. From these numbers you see the colder burn right away. Octane gas is 4 times hotter then ethanol but takes more oxygen to complete a burn.
Gasoline – C₈H₁₈ + 12.5 O₂ → 8 CO₂ + 9 H₂O DH (heat released) = -5460 KJ/mol
Ethanol – C₂H₆O + 3 O₂ → 2 CO₂ + 3 H₂O DH = -1368 KJ/mol

Now when we apply this where oxygen is the limiting factor, things get interesting! When air is the limiting factor, you can burn 4.16 moles of ethanol vs one mole of gasoline. If we were to assume that the exhaust products were ideal gasses, the temperature of the burned fuel would be 4.4% higher. “Huh? But didn’t we say ethanol burns colder… and… now you are saying you get more heat from it?”

Gasoline – C₈H₁₈ + 12.5 O₂ → 8 CO₂ + 9 H₂O DH (heat released) = -5460 KJ/moL
Ethanol – 4.16 C₂H₆O + 12.5 O₂ → 8.3 CO₂ + 12.5 H₂O DH = -5700 KJ/mol

Simply by scaling down the ethanol content in the equation by 4.4%, we can reach the same thermal energy you get from octane gas.

Ethanol – 3.98 C₂H₆O + 11.97 O₂ → 7.95 CO₂ + 11.97 H₂O DH = -5700/1.044 = -5460 KJ/mole

Based off what you have read so far, mathematics and chemistry have shown that ethanol burns colder. But, the amount of molecules in 4.16 moles of ethanol are actually more than in a single mole of gasoline. Avogadros law (or as I like to call it Avocado’s Law) states that “equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.” Based on this, we also know that the pressure is already going to be higher, and it turns out when we factor in the temperature and the fact that there are more combustion products, we actually see an increase in cylinder pressures of around 18%!

Now let’s raise our nerdy level even higher, and talk about the ideal gas law a little more! P=nRT/V. We know that pressure is proportional to temperature and the number of moles on the exhaust at the constant V. So if ethanol produces 18% more moles of exhaust gas at the same thermal energy as octane, then you could also drop the temperature by that 18% to produce the same cylinder pressure as gasoline. #MindBlown.

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