I copied this.
The general consensus is ethanol burns “cooler” than*gas. Technically that is correct but what does that really mean? There’s a lot more to it and I’m throwing out my analysis (btw, this is not copied from any textbook or external resource--just my own back-of-the-envelope calcs so I make no assurances).*
The burning question in my mind has been how does*cooler*burning correlate to*cylinder*temps and power? That is, how can a fuel that burns “cooler” generate more power? How much “cooler” does it burn? Are the EGTs*cooler? To explain this, I did some calcs--the results seem reasonable so I thought I would post them here with explanation.*
If you take one mole (6.02 x 10^23 molecules) of octane (gasoline) and one mole of ethanol and combust them in air, octane would produce 5460 KJ of heat compared to 1368 KJ for ethanol (eqs 1 and 2). The heat given off from the combustion reaction can be experimentally measured to a very precise number. In this case, more heat is evolved from combusting one mole of octane than one mole of ethanol, in fact, about 4 times as much.*
C8H18 + 12.5O2 → 8CO2 + 9H2O DH (heat released) = -5460 KJ/mol [eq 1]
C2H5OH + 3O2 → 2CO2 + 3H2O DH = -1368 KJ/mol [eq 2]
(note: negative sign just means heat is released; more negative = more heat)
So in this sense, ethanol burns*cooler.*
Now, let’s take a look at what happens inside the*cylinder*when the oxygen content is the limiting factor. For this, we need to look at the combustion equations where the oxygen content is set as the limiting reagent (i.e. 12.5 moles). For every 12.5 moles of O2, you can*burn 1*mole of octane and 4.16 moles of ethanol. So you can see in this case, you actually get more heat from ethanol per unit oxygen (approximately 4.4% more) [eq 3].*
C8H18 + 12.5O2 → 8CO2 + 9H2O DH (heat released) = -5460 KJ/mol
4.16C2H5OH + 12.5O2 → 8.3 CO2 + 12.5 H2O DH = -5700 KJ/mol [eq 3]
If we scale down the last equation by 4.4% such that the thermal energy produced from burning ethanol equals that of octane, the equation becomes*
3.98 C2H5OH + 11.97 O2 → 7.95CO2 + 11.97 H2O DH = -5700/1.044 = -5460 KJ/mole [eq 4]
So how does ethanol burn*cooler, yet produce more power? Well, power is a result of*cylinderpressure. For that, we need to take into consideration the total number of moles of combustion products from octane and ethanol. For octane, you get a total of 17 moles of combustion products (eq. 1). For ethanol, you get a total of approximately 20 moles of combustion products (eq 4). That corresponds to approximately 18% more moles of*exhaustproducts from burning ethanol at the same thermal energy level as octane.*
Since P=nRT/V, pressure is proportional to not only T (temperature) but also n (# of moles of total*exhaust*products) at constant V. Well, if EtOH produces 18% more moles of*gas*at the same thermal energy level as octane, then the temp can drop by 18% to produce the same*cylinder*pressure. Hence,*ethanol can burn*cooler*to give the same pressure as burning octane because it produces a greater amount of combustion products.*
Here’s the rub--everyone wants max power, so you end up burning as much ethanol as there is O2 in the*cylinder*which thereby produces 4.4% more heat than octane with even greater*cylinder*pressures. So while ethanol can burn*cooler*and produce more power, in reality, we end up burning as much ethanol as possible to get max power. This results in higher*cylindertemps than*gas*(but more power too*[
https://www]).
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