Theoretical auto engine thermodynamics question

While thinking about EGR it occurred to me to wonder about the way IC engines work. I assume that fuel and air burn, the heat released heats up the components of air not involved in combustion, namely N2, causing them to expand (Dalton's) and push the piston down. Now here's the question: Would different ratios of O2 and N2 affect efficiency enough to matter?

JazzMan

Reply to
JazzMan
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As I understand it the piston is pushed down by the combination of the combustion products being higher volume than the pre-combustion fuel and pressure caused by temp increase.

Anything that increases combustion efficiency (turbo charging, air intake design, N02 injection) will increase the efficiency of the engine.

Reply to
marks542004

No, because the thermodynamic properties of all gasses are the same. Unlike the chemical properties.

Reply to
Misterbeets

welll...if you're talking about ideal, PV = NRT stuff, okay. But once you are talking about IC engines, you have chem and thermo stuff all in there together, causing heat, work, losses...

I'm not going to go look it up, but I'm sure that the enthalpies of combustion of N and O are different, so yeah, it would make a difference. Air is pretty much straight 21% O2/ 79%N2, what would you do to change that? k wallace

Reply to
k wallace

Even that isn't quite true. That equations describes the ideal gas law, and many gases fit closely, but some deviate a lot.

Reply to
<HLS

that's why I said, "if you're talking about ideal, PV= NRT stuff...." which doesn't take into account Z, the compressibility factor. Z is solved for with an infinite series expansion (which you probably know since you discuss pvnrt as the ideal gas law) so I won't go into details, as likely no one else here is interested :)

k wallace

Reply to
k wallace

Yes. Frequently in analysis of IC engines, one takes the properties of air as the working fluid. It simplifies the situation to use air as the working medium. While in truth the working fluid contains many species of combustion products, the result is not that far off from just considering the compression, heating, and expansion of air. If I remember right, air has reasonably good thermo properties. A diesel, of course, frequently operates with very high amounts of air compared to fuel and combustion products. In spark ignition engines operating in air, the mixture ratio cannot vary that much to really affect things much unless one uses a lot of EGR.

Reply to
Don Stauffer

Cramming more air into the engine may not in itself increase the thermal efficiency. It does affect both the volumetric efficiency at high rpm, and the power to weight ratio. If, however, it increases the actual compression ratio, not the geometric ratio, it can affect thermal efficiency.

Ordinarily the efficiency computed is the full throttle efficiency. Throttling reduces actual CR, and hence efficiency. That is why, if there were no power richening of mixture, engines would be most efficient at a pretty high throttle opening, just below point where volumetric efficiency starts to take a nosedive at higher rpm. So engine is most efficient at low to mid rpm but full throttle (again, not including power enrichment).

Turbocharging helps efficiency not only by boosting actual CR, but by utilizing exhaust enthalpy that is ordinarily discarded, so turbocharged engine ordinarily more efficient than mechanically supercharged ones.

Reply to
Don Stauffer

Not quite true. Thermal properties of actual gases are not quite the same as ideal gas. In cases where the efficiency or other parameter are exponential with a given thermal property, it can make a difference. The differences are ordinarily not extreme (except for helium) but because of the strong dependence, it can make some difference.

Reply to
Don Stauffer

The law applies to actual gases as long as you have the right value of R.

Reply to
Don Stauffer

Sorry, Don, I lost this thread about a month ago. It has been all too convenient over the years to assume that 0.082046 lt-at/(deg mol) fit the wide range of gases. And it does, as long as the pressure is relatively low.

I guess I would normally attempt to use Van der Waals corrections or a different equation of state rather that to try to derive a function which would describe R (hopefully a universal constant) for complicated cases. (Glad Im not a physicist. Not my favorite cuppa)

Reply to
<HLS

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