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TexasKZ wrote:

zed1015 wrote: The PSI measurement should be taken at as high a cranking speed as possible or while the engine is running (on a multi cylinder) to offset the effects of cam timing.

Clearly I am missing something here. If I take an engine with a very short duration camshaft and then install a camshaft with significantly more duration, will I not get a much lower pressure reading the second time as the volume of gas being compressed will be measurably less?

Yes! the cranking pressure is affected by the cam duration but the running psi will be in direct relation to the compression ratio, hence the sentence my previous reply and although there is no exact formula to calculate the compression ratio from cylinder pressure the psi reading is usually in the range of 15 to 20 times the compression ration for a normal road going motor hence factory manuals quoting service limits between figure A and figure B to determine cylinder/engine condition.

I would like to chime in and admit that I when I was trying to teach myself all about cam duration, overlap, LSA, opening and closing times and they all affect each other, I posted some very erroneous information along these lines in a reply to Zaddicts thread about swapping cams in his bike.
www.kzrider.com/forum/2-engine/607394-gp...50-or-zr550?start=40

In my research into the effects of swapping higher duration cams into an engine with low compression pistons, I latched onto the information of how high duration cams cause compression loss at cranking speeds. But for some reason, something I read lead me to believe that the compression loss from valve overlap was constant throughout the entire rpm range until the optimal RPM range for the scavenging effect (caused by the valve overlap) kicked in. It wasn't until later that more research and reading showed just how incorrect this information was.

Threads like this are full of great information and I always learn something very valuable!
So I wish I could go back to my reply in that past thread and edit out all the incorrect information so it doesn't screw up somebody else who is trying to learn about a already very complex subject.

Compression ratio is defined as a change in volume, not pressure.

In an operating engine there is a direct relationship between the volume compression ratio and the pressure compression ratio but not a linear one.

When I do compression tests I don't run the engine. I pull all plugs and measure one at a time.

When checking cylinder pressure on a running engine, wouldn't the incoming fuel charge inflate the reading since gas online is not compressible?

We're getting into mass flow rate and I can only guess at what would happen. You're correct that fuel is not compressible, but in this case it's likely a vapour and therefore just as compressible as the air.

Howdy! Did some math for fun to approximate the volume of gasoline in a single intake charge for hypothetical motorcycle going 65mph at 5,000rpm getting 50mpg, and here's what I calculated (feel free to correct/alter this if you spot something):

(50 miles / gal) * (1 gal / 3785.41 cc) * (60 min / 65 miles) * (5,000 rotations / min) * (1 intake charge /2 rotations) = 30.48 intake charges per cc, or 0.033 cc per intake charge

Looking at a cylinder with 100cc of capacity and treating gasoline as incomprehensible, I think you would then treat the gasoline (up until the point at which it combusts, where it's unclear to any of us yet how to treat it) as "lost volume" on the compression stroke and subtract it from the "compressed volume" value ( i.e. 100:10-0.033 = 10.033:1)

Anyway, this was a really long-winded way of saying that with the volume of gasoline in the intake charge compared to the total cylinder volume, I think it's effect on the actual compression would be pretty negligible. Cool discussion, btw! Lots of good thinking here

Hi all,

hardrockminer mentioned the Ideal Gas Law - PV/T = C. This is correct - the air/fuel mix is close enough to an ideal gas.

You also mentioned Boyle’s Law - P1*V1 = P2*V2.

And while both of these formula are correct, neither of them really apply in this case. And that's because when compressing a gas, it heats up. Boyle’s Law only applies for a set amount of air molecules (which we have) and a constant temperature (which we don't). And while the Ideal Gas Law does apply, we don't know the temperature. So when increasing compression, we can't simply change PV/T = C, since we don't know what T is changing to.

For cylinder compression, the best method to calculate the final pressure and temperature is assuming adiabatic compression. This assumes that no gas or heat enters or exits the system, and the only work that enters or leaves the system is that of the compression. So instead of the standard ideal gas law that uses temperature (PV/T = C) we use the polytropic process formula - PV^n = C, where n is about 1.4 for air in an adiabatic process.

We know the staring pressure and volume, so our constant is:
P1V1^n = C = 100,000 Pascals x (0.0001 m^3)^1.4 = 0.2512

Now that we know our constant, we can use P2V2^n = C to calculate our final pressure:
0.2512 = P2V2^n = P2 x (0.00001 m^3)^1.4

=> P2 = 0.2512/( (0.00001 m^3)^1.4) = 2511886Pa

That's 25.2Bar or 364 psi.

Of course, you'll never see that kind of number on a real engine. The polytropic process formula assumes that no air or temperature leaves the system - you will get some air escaping through rings and valves, and you will get a small amount of heat leaving the system (though only a very small amount for a fast compression like you will see in an engine).

More importantly - at speed, the starting pressure is lower as the cylinder acts as a vacuum to suck in the air, and even at 10:1 the valves don't open close exactly at bottom dead centre. So your starting pressure is lower, your compression ratio isn't truly 10:1.

Does removing volume from the combustion chamber have a linear effect on measured pressure? Is there any reliable relationship between measured volume and measured pressure?

Does that mean that every cylinder, no matter its volume or design, will have the same volume ratio and pressure ratio? I am pretty sure that this is not the case, as changing the cam profile will change the pressure readings. Is there a dependable relationship between the two?

The bottom line is that the relationship between volume and pressure is *not* linear. It would only be linear if the temperature did not change, which is not the case. For a given compression ratio the final temperature can be calculated regardless of cylinder design - but as you mentioned, changing the cam profile will change the pressure readings. That's because the cam profile will change the valves, and will effectively change the compression ratio.

There is no dependable relationship, as there are too many moving variables. The shape and size of the valves change the starting pressure, and the design of the cam changes the effective compression ratio. Throw in the heat and pressure from combustion, and everything changes again.

Thanks for that Irish. I thought also that the heat generated by detonation would have a powerful effect on the PV/T relationship in an operating engine. But doing a cold compression test would eliminate that variable.

Yes, in a hot engine the air heating up from the excess heat in the cylinders will definitely effect the temperature of the gas. It will change the starting temperature, and it will leach temperature into the gas during compression - so in a hot cylinder, the assumption that there is no temperature transfer gets more shaky. In the end we need to make some assumptions though to get any formula to work though.

For the process I mentioned (adiabatic), you assume no temperature or air leaves the system.

For an isothermal process, you assume the temperature doesn't change - this is the formula you used before. It works, but only for constant temperature.

You can also calculate an Isovolumetric process (constant volume) or an Isobaric process (constant pressure). They all work well, but only if your assumptions are 'close enough'.