Guest bastaad525 Posted April 7, 2004 Share Posted April 7, 2004 How would you calculate your motor's effective displacement when it's running boosted? is there a set equation? I'm my motor is a 2.8L, and I'm running 10psi of boost, how much air is being displaced? Quote Link to comment Share on other sites More sharing options...
Brad-ManQ45 Posted April 7, 2004 Share Posted April 7, 2004 Look here: http://www.hammondsplains.com/zclub/techtips/turbo/turbomaps/index.htm About 1.68 times the displacement....1.68 times the airflow from 0 boost. Figure about 4.7 litres...thats the volume of air the displacement of the engine and 10 lbs. of boost will produce. Other factors like intercooling have an effect also.... Quote Link to comment Share on other sites More sharing options...
Scottie-GNZ Posted April 7, 2004 Share Posted April 7, 2004 That is a pretty good tech article. Another simple way to look at it is as follows: For every 14.7psi of boost, the engine is capable of increasing the volume it can force into the cylinders equal to its original displacement. So, a L28ET running about 15psi is capable of stuffing 5.6L into its cylinders. Apply this simple formula to one running 10psi and you get the same result as Brad specified. Note I said "is capable of" because that assumes almost 100% efficiency and as Brad pointed out, there are several factors that have an effect on the results. E.g., 10psi from a stock T3 is not the same as 10psi from a T04. Likewise, 10psi from any turbo through a Starion I/C is not the same as through a very efficient I/C, yada yada. It is all about moving large volumes of air eficiently. Whatever you end up with in your cylinders will make for a big bang and explains why turbo engines make so much torque. Quote Link to comment Share on other sites More sharing options...
Guest bastaad525 Posted April 7, 2004 Share Posted April 7, 2004 Cool thanks for the info guys. The answer was actually right under my nose if I'd stopped to think about it long enough. 14.7psi = 1 bar = twice normal atmospheric pressure = twice as much air forced in = twice the effective displacement. Duh The numbers are definately pretty interesting, as far as 2.8L @ 10psi = about 4.7L. Very close to a Mustang 4.9L, and interestingly enough, my hp and torque numbers are almost dead on with what stock '5.0' mustangs usually put down, shy of some torque but with some added hp (stock 5.0's making 225hp and 300ft lbs at the flywheel, whereas my numbers are about 235 and 275 if you account for 15% drivetrain loss)... which also makes sense, shy of some torque because I'm effectively .2L less, but with some added hp because the L28et makes effective torque higher into the rev range. Of course lots of things could come into play to make that 4.9L make a hell of a lot more power but it is a trip how close the numbers are. Turbo's are friggin cool... displacement on demand Quote Link to comment Share on other sites More sharing options...
Pyro Posted April 7, 2004 Share Posted April 7, 2004 (14.7 + boost pressure)/14.7 = "turbo muliplier" Quote Link to comment Share on other sites More sharing options...
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