How Does Compression Ratio Affect Heat Generation in ICEs?

[Wagon Master]

I'm new here, and don't know much. This question may not even be in the right forum, and if so I apologize; please direct me otherwise.

My question concerns internal combustion engines. I'll put out a few notions that I have and base my question on them. Please correct me if I'm wrong.

Running on pump gas (87 octaine), and ICE can only hit about 10:1 compression before predetonation kicks in. That is to say, for example, that a 1 liter single cylinder engine will take in 1 liter of air/fuel and squish it to 1/10 of a liter, thus making a 10:1 compression. In the compression phase, the air/fuel heats up and will detonate if it becomes too hot. Much above 10:1, the act of compression heats the air/fuel too much.

Assuming that the above is more or less correct, what is the formula/table/rule for an all-other-things-being-equal-enviroment rate of compression=heat? In other words, how do I figure out how much more heat 10:1 makes than say 8:1 or 5:1?

If I'm not being clear, I can try to ask another way.

Thanks

 

[brewnog]

Welcome, Wagon Master. Your post is in the right place!

Your assumption regarding compression ratios is broadly, more-or-less correct.

The fundamental law linking pressure, volume and temperature in a system such as this is the combined gas law, which states that the ratio (PV) / T is constant; where P is pressure, V is volume, and T temperature.

If you're messing with numbers, don't forget that temperature must be absolute (ie in Kelvin).
There are some other factors which affect detonation but that should get you started at an "all other things being equal" level.

 

[Wagon Master]

Thanks for the welcome brewnog .

Math is not my strong suit, so I'll bounce the numbers off of you, if you don't mind.

I use 1 liter and 10:1 compression as easy benchmarks, and will use 80.33*F as well.

T= 300*K (K is Kelvin?)
V= 1000ml (ml is milliliters?)
P= 147psi (14.7 psi atmosphere at sea level times 10:1 compression?)

So 147*1000/300 = 490*K (about 422*F).

Am I close?

 

[brewnog]

Not quite. Firstly, bear in mind that this is all at a basic level, so the heat introduced by the combustion event is irrelevant, and that we've made quite a few assumptions about your engine here. I'm going to change your temperature of 300 Kelvins to something much higher, (let's say 800 Kelvins), because this is much lower than what you'd see in a real engine just before combustion (ie after compression), rather than ambient air temperature.

For your given example, (147psi *1000ml)/800 Kelvins = a constant. Let's call it K, so K=184.

Now, if your compression ratio was only 8:1, your pressure in the equation would be just under 118psi.

So given the same conditions as above (keeping our constant K and the volume the same; again another assumption!):

K = (PV)/T

We know that K is 184, and we know that the new pressure is only 118psi, so we'll rearrange the equation to show us what T will be:

T = (PV) / K

If you do the maths with the new numbers, you'll see that this gives a lower temperature, and therefore less likely to cause engine knock:

(118psi x 1000ml) / 184 = 641 Kelvins

Which is a lower temperature than your higher compression engine, and less likely to cause knock.

You might also be interested to read up on the Ideal Gas Law; the Wikipedia page is very good.

 

[Wagon Master]

Not quite. Firstly, bear in mind that this is all at a basic level, so the heat introduced by the combustion event is irrelevant, and that we've made quite a few assumptions about your engine here. I'm going to change your temperature of 300 Kelvins to something much higher, (let's say 800 Kelvins), because this is much lower than what you'd see in a real engine just before combustion (ie after compression), rather than ambient air temperature.

For your given example, (147psi *1000ml)/800 Kelvins = a constant. Let's call it K, so K=184.

Now, if your compression ratio was only 8:1, your pressure in the equation would be just under 118psi.

So given the same conditions as above (keeping our constant K and the volume the same; again another assumption!):

K = (PV)/T

We know that K is 184, and we know that the new pressure is only 118psi, so we'll rearrange the equation to show us what T will be:

T = (PV) / K

If you do the maths with the new numbers, you'll see that this gives a lower temperature, and therefore less likely to cause engine knock:

(118psi x 1000ml) / 184 = 641 Kelvins

Which is a lower temperature than your higher compression engine, and less likely to cause knock.

You might also be interested to read up on the Ideal Gas Law; the Wikipedia page is very good.

You've lost me and I need some bread crumbs...

You mention the Kelvins just before combustion, which is exactly what I'm looking for: I need to know what the temp rise will be on the very first compression stroke, with no residual heat from any prior combustions.

You use 800 Kelvins as the sample temp, stating that it's closer to the actual temp in the engine at that point. You also point out that the math with an 8:1 drops the temp to 641 Kelvins. The question I have is: Is 800 the actual temp at 10:1 just before combustion?

I know that 8:1 is lower than 10:1, I need to know the actual temp that the air/fuel will reach at 10:1.

Or did you just explain that, and I really don't get it?

 

[brewnog]

I know that 8:1 is lower than 10:1, I need to know the actual temp that the air/fuel will reach at 10:1.

Or did you just explain that, and I really don't get it?

Ahh, no, I didn't explain that, sorry. I thought you wanted a hypothetical comparison between 8:1 and 10:1.

To determine actual temperatures, you need to look at the ideal gas law:

pV = nRT

where p, V and T are pressure, volume and temperature respectively, n the number of moles of the air held within the cylinder, and R the gas constant.

Have a look at:
http://en.wikipedia.org/wiki/Ideal_gas_law

Remember to use SI units for any calculations done with this. And remember that you've got to consider pressure and volume at the condition you want the temperature at; ie under compression.

Even considering all this, there are a few things to remember:
- Some of the heat due to compression will be lost to the combustion chamber walls
- Some heat may be picked up due to heat in the cylinder walls
- It's very difficult to determine true cylinder pressure (P) prior to combustion, because it's extremely difficult to determine when combustion actually begins
- There will always be some residual exhaust gas (and possibly unburnt fuel) in the cylinder from the last cycle; so your gas constant value won't be entirely accurate.
- Possibly some other things...

Can I now ask what you're actually trying to find out?

 

[tharptroy]

its not really safe to assume any particular maximum compression without considering engine design, as the compression ratios where detonation occur vary depending on spark timing and engine design.

you might have an old truck that that detonates on 87 octane with 9:1 compression, or a modern sportbike (more in your engine size range) with 12.3:1 that runs fine on 87 octane.

 

[Wagon Master]

Ahh, no, I didn't explain that, sorry. I thought you wanted a hypothetical comparison between 8:1 and 10:1.

To determine actual temperatures, you need to look at the ideal gas law:

pV = nRT

where p, V and T are pressure, volume and temperature respectively, n the number of moles of the air held within the cylinder, and R the gas constant.

Have a look at:
http://en.wikipedia.org/wiki/Ideal_gas_law

Remember to use SI units for any calculations done with this. And remember that you've got to consider pressure and volume at the condition you want the temperature at; ie under compression.

Even considering all this, there are a few things to remember:
- Some of the heat due to compression will be lost to the combustion chamber walls
- Some heat may be picked up due to heat in the cylinder walls
- It's very difficult to determine true cylinder pressure (P) prior to combustion, because it's extremely difficult to determine when combustion actually begins
- There will always be some residual exhaust gas (and possibly unburnt fuel) in the cylinder from the last cycle; so your gas constant value won't be entirely accurate.
- Possibly some other things...

Can I now ask what you're actually trying to find out?

Yes, and thank you for asking! Others that I've mentioned this to just get a glazed look and say it can't be done, or that only deisels can run at 20:1.

I was actually looking at the Wiki page while you were responding, but there are too many over-3-sylible words, and I didn't get it.

So based on the above parameters, here's what I need to know:

How high can I compress an 87 octaine air/fuel mixture ONCE before the act of compression (no residual heat because it's the first compression) heats the air/fuel mixture to the point where it auto-detonates just from pressure rise?

Each new test stroke will restart at zero (ambient temp).

Following on that, how much explosive pressure will the resulting combustion develop?

If I can get to 20:1 before auto-detonation on pump gas, will the resulting pressure blow my engine apart?

 

[brewnog]

Firstly, please don't be put off by any multisyllabic words; just ask!

How high can I compress an 87 octaine air/fuel mixture ONCE before the act of compression (no residual heat because it's the first compression) heats the air/fuel mixture to the point where it auto-detonates just from pressure rise?

The annoying answer is "it depends". It depends on the temperature and design of the combustion chamber, the rate of compression, the heat lost to the cylinder walls, the air fuel ratio, and potentially some other things.

Your question about compression of mixture in a single cycle is very hard to answer. Practically, in an engine (as you probably know), the detonation limit depends on a lot of things: throttle position, manifold temperature, manifold pressure, air-fuel-ratio, engine speed, load, and in-cylinder temperature.

As for compressing a mixture once, I'm not sure what your purpose is. Even if you could (hypothetically) get to 20:1 by doing this, it would be irrelevant to what happens inside your engine. Given a cold combustion chamber, I suppose you could compress an air-fuel mixture by a huge amount without autoignition. But this isn't what happens in a running engine.

You probably realize that the efficiency of an engine is strongly related to its compression ratio; which is why diesels win here. But I don't know how you'd get to 20:1 compression ratio without severe detonation in a running engine, even if you had managed it on a 'single firing' test rig.

In real terms, most manufacturers sell their engines at the optimum compression ratio they can, for the market and application they're intended for.

If I can get to 20:1 before auto-detonation on pump gas, will the resulting pressure blow my engine apart?

No, not necessarily. It would depend on how the resulting pressure compares to that produced in the same engine at its peak load. But this is hypothetical.

What is your aim in asking this question?


Finally, you might be interested to read this:
http://en.wikipedia.org/wiki/Octane_rating

 

[Wagon Master]

Well, I can't tell you everything, but here's an example: I'm magic. I have the power to lower the combustion chamber to about 422*K before every compression stroke. It seems to me that with this ability, I should be able to use a much higher than current compression ratio, since there will be far less residual heat.

My magic will work on any design of engine, so the specific combustion chamber design can be taken out of the equation. Since the temp is lowered to about 422*K every time, some of the other variables (like in-cylinder temp) could possibly also be ignored.

Engine speed, A/F ratio, load, etc. can be added to the equation later. Manifold pressure is assumed to be constant and to always fill the chamber with a full 1 liter of air/fuel with each stroke.

I used the term "compressing the mixture once" to simulate that there will be no residual rise in temps, because it is magically brought back down before each combustion.

And don't get me wrong; the "magic" I'm referring to is recently existing technology, it just hasn't been used in this way before.

 

[Wagon Master]

I'm not an engineer (in case you hadn't noticed that part ), I'm just a mechanical type who sees some tech that looks compatible and thinks "Why not?". If I'm right, and I can get pump gas to run at 20:1, I'd think that it would put out a lot more power, and thus be able to get much better fuel economy.

 

[tharptroy]

Since detonation isn't really all that simple of a phenomenon, my guess is that if you don't want to talk to anyone about it, you're going to have to man up and start testing. Without talking to a combustion engineer, I doubt you're going to get a concrete answer on what the compression ratio limit is.

it sounds an awful lot like you're after a crower six cycle to me...

http://en.wikipedia.org/wiki/Crower_six_stroke

 

[Wagon Master]

Since detonation isn't really all that simple of a phenomenon, my guess is that if you don't want to talk to anyone about it, you're going to have to man up and start testing. Without talking to a combustion engineer, I doubt you're going to get a concrete answer on what the compression ratio limit is.

it sounds an awful lot like you're after a crower six cycle to me...

http://en.wikipedia.org/wiki/Crower_six_stroke

The Crower engine is an interesting idea, and I wish him the best. My idea is different and far more complex.

As for "maning up and start testing", I'd love to: But since I don't know where to start, maybe you'd like to fund the idea? No? Then I'm probably better off trying to get some of the math out of the way first.

I wasn't aware that there was a such thing as "combustion engineer". I'll do a google search and see what comes up.

 

[tharptroy]

complexities often mean that the system won't be able to be used in practice, especially in something like a street car.

Mechanical engineering covers such a broad array of topics that no man could master them all. The higher up guys are often quite specialized in what they research. The nature of your question is so complex that unless there is experimental data that already exists, you will need to consult with someone who specializes in ICEs.

for example: one materials engineer might specialize in solar cells while another specializes in fracture mechanics.

the SAE might have some tech papers on the necessary conditions for detonation, but they charge for them, so choose wisely.

the engineering tips forum has both an engine section and a thermodynamics section, so perhaps they will be of some use.

you could also contact your local university and see if you can get some council there.

I'm pretty sure that the hottest thing in the chamber is usually the exhaust valves, so make sure that your cooling strategy can take care of those.

 

[Wagon Master]

It can deal with the exhaust valves, and can lower the compression ratio in real time on a stroke by stroke basis. This can go a long way toward dealing with a too hot (for whatever duration) condition.

I will heed your advice and seek out the university types.

Thanks

 

[Old Pokey]

Hi. I'm new here too, and likely can't help much, but would like to try if I may, given the interest I have in your experiment.

Maybe this is an article that has some insight. http://www.streetrodstuff.com/Articles/Engine/Detonation/
I don't have any "rule of thumb" for the compression incremental scale you'd like to have, but you should be able to find pieces of it here and there with articles like the one in the link and make your own graph. However, reading your original question, I think "time" is missing. I could be wrong, but I think the speed that you compress the fuel/air is critical to finding out the amount of heat generated. IMO anyway. Find info about "piston speed" and see if it helps you with your question. A shorter connecting rod will give you a faster piston speed than a longer one. Faster compression of the fuel /air will give you a different reaction in heat development than compressing it slower. IMO, just my 2 pennies, etc. If I'm wrong, I'd like to learn so I can correct it.

 

[dwain]

Looks like my previous reply didn't go thru. Too bad. Here's the short version. Get Paul Cooper's book, "Explosives Engineering". Good examples of contained detonation, calculations, explanations. A cannon and a cylinder of an engine are very much alike. The field of balistics started as an application of thermodynamics in France before the civil war. There has been a lot of experimenation and data collected on the state of matter of explosives in containers over the years.