I'd like to know the physics behind forced induction and compression
ratio to answer the following : Why can an engine run on 15psi of
gauge boost, yielding twice as much air consumption, yet the
compression ratio of a similar engine cannot be doubled without
detonation? My understanding is that if there is 15 psi of gauge
pressure, then twice as many moles of air are entering the engine in a
given amount of time. If the compression ratio is 8:1, then that means
the air is being compressed "seemingly" twice as much since there's
twice as much in there. Yet if the compression ratio is doubled to
16:1 and no boost is used (0 gauge pressure) you get the naturally
aspirated number of moles but squeeze them twice as much. 2 moles in
8:1 compression will work. 1 mole in 16:1 compression will blow up in
your face. Why is this?

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Clarification of Question by juugcatm-ga on 18 Apr 2003 12:29 PDT

Also, I good set of resources about Adiabatic processes would be very helpful.

The reason for the blow-up is not that the pressure reaches a certain
value, or that there's a certain amount of air in the cylinder.  It's
that the temperature rises to a certain value.  When you squeeze air,
its temperature increases, because you are doing work on it.  So if
you start with air at ambient temperature and 1 atm pressure (~15psi)
and squeeze it to 1/16 its original volume, it will be hotter than if
you start with air at ambient temperature and 2 atm (~30 psi) pressure
and squeeze it to 1/8 its original size.  To work out exactly how big
the difference in termperature is for these two cases, you need to use
the fact that for adiabatic processes (no heat transfer) p*V^gamma is
a constant, where p is pressure, V is volume, and gamma is the ratio
of specific heats, about 1.4 for air I think.

Yep, air is like 99% oxygen and nitrogen, both of which are diatomic
gases, so gamma is more or less exactly 7/5.  From the relation in the
previous comment, p*V^gamma = constant, and from pV/T = constant, you
get:

T2 / T1 = (V2 / V1)^(1 - gamma)

Since gamma is 7/5, that means the ratio of final temperature to
initial temperature is the ratio of the final volume to the initial
volume to the -2/5 power.  For compression ratios of 8 and 16
respectively, the temperature ratios would be (1/8)^(-2/5) = 2.30 and
(1/16)^(-2/5) = 3.03.  That means if you start with air at room
temperature (300 K) the temperature will rise to 689 K, or 416 C, or
781 F in the 8:1 compression ratio engine, but to 909 K, or 1182 C, or
2160 F in the 16:1 compression ratio engine.  I imagine if you look up
the flash point of vaporized gasoline, you will find it is between
these two temperatures.

Thanks for the help. I will modify the question to include a request
for more information about adiabatic processes (I never understood
them in the heat section of my basic physics class).