I, too, wondered what my compression ratio would be when at full boogie running my turbo boost. I looked around for a formula that was already in existance, but, while looking I derived it anyway, independently. There really isn't anything complicated about it. However, there are a few things that one needs to understand when thinking about compression ratio and cylinder pressure. First thing is that the compression ratio is a mechanical number based on engine measurements. Nothing magical here, but everything needs to be considered. I will develop a model that you can plug your numbers into in a bit, so bear with me. Another thing is atmospheric pressure. The most important thing is to get the largest MASS of air into the cylinder for the most power. At a low atmospheric pressure in a normally aspirated engine you will get less mass of air at higher elevations than at sea level. Another related item is that boost is generally at a higher temperature and is less dense than the ambient air conditions so you get less mass than expected. Another consideration is the engine's efficiency in filling the cylinders with air. Mechanical Compression ratio. This is pretty straight forward in that you need to know the swept volume and the static volume of the cylinder and head. The formula is: CR = (V + Vcl) / Vcl where: V = swept volume Vcl = clearance volume You will need to know the following about your engine to accurately determine mechanical compression ratio(an example is included): Mechanical Compression Ratio Whew! That is a lot of stuff to do, but necessary, especially if you are running a high compression engine and need to know because of detonation, etc. However, in most instances, boosted engines are of a much lower compression ratio: usually around 8.5 to 1. But those top fuel dragsters have CR anound 5! The rest of this is easy! Effective CRboosted = ((Boost Pressure / Atmospheric Pressure) + 1) x Mechanical CR So, if we have a mechanical compression ratio of 8.5, a boost pressure of 14.7 psig and the atmospheric pressure is 14.7 psi, then Effective CRboosted = (( 14.7 / 14.7) + 1) x 8.5 = 17.0 to 1 Now some racers use boost of around 25 psig, so then.... Effective CRboosted = (( 25.0 / 14.7) + 1) x 8.5 = 22.9 to 1 Ok, so that is pretty high. Now what would the corresponding cylinder pressure be? Again, there are additional things to consider. Volumetric efficiency in filling the cylinders, and temperature of the air going into the cylinders. Most street engines have a volumetric efficiency of around 85%, some higher, some lower. I suspect that the newer high tech engines are in the 90's percent range. What this means is that it is hard for the air to get inside your cylinders due to flow restrictions of some sort or another. Casting flash, rough port walls, mismatch on gaskets, etc. A well tuned race motor at it's design speed can achieve volumetric efficiencies or 110% or more. That is due to ram tuning of the pressure pulses of the air column. The next item to really consider is the temperature of the air going into the cylinders. It is really important to get the air temperature down to below ambient because you wil then getting the most MASS of air in the cylinders that can be gotten! So Intercooling is needed if the boost pressure goes much above ambient. A boost pressue ratio of 2 (ie, 14.7 psig boost pressure) will raise the air temperature to around 250 degrees which means that the density is less. Means that your are getting the pressure, but not the mass, so not a doubling of horsepower. Intercool and get the temp down and the mass of air goes up and horsepower goes up right along with it. Adding all this stuff in... Cylinder Pressure = Effecive CR x Efficiency x Atmospheric Pressure = 22.9 x 0.90 x 14.7 = 303.7 psig
