So if the temperature drops are as much as I estimated the other night, 130 kelvin (ethanol) and 30 kelvin (octane), it would explain why low autoignition temperature ethanol can have a higher compression ratio than high autoignition temperature isooctane. (You have to evaporate about four times as many ethanol molecules.) With that in mind, it comes as no surprise then that evaporating the ethanol required for combustion drops the temperature about 4 times as much as evaporating octane. But the figures per molecule only tell part of the story, as an ethanol molecule has two carbons and octane has eight. If you look at the enthalpy of vaporization of ethanol (around 40 kJ/mol but varies depending on source in the literature!) and that of octane, the numbers are broadly similar. What I am saying is at stoichiometric AFRs ethanol looks like at could support significantly higher compression ratios than isooctane due to the large amounts of evaporative cooling. So air that might reach 760 due to compression (in the absence of fuel) could reach 630 due to ethanol evaporation (a drop of 130) and just start to autoignite. The autoignition temperatures I had were 630K for ethanol and 690K for isooctane. Can it be right? And similar calculations (so similar mistakes if wrong) gave iso-octane removing about 30 kelvin. I did an estimate the other day, which I won't put here as the calculations aren't handy, but from memory, at 9:1 air fuel ratio, ethanol evaporating could take about 130 kelvin out of the air.
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