turbos and math
#1
turbos and math
Hi all, I am trying to figure out how to calculate how boost translates into cubic inches or liters. For example, Kenne Bell at one time made a supercharger for the Ford v10. It was, if I recall, 2.2 liters and I would guess that would be at full boost which I think was 8lbs. So at 8lbs of boost I would think that would make the engine 9.0 liters (6.8 plus 2.2). Am I correct in this assumption? So now what about a 6.0 Diesel at 28lbs of boost, how much air would the engine be getting. I hope you understand what I am trying to say, it is not always easy to write what I am thinking.
#2
Not quite. The 2.2l quoted is (roughly) the volume of air moved by the supercharger's screws in one revolution. The KB's superchargers spin at several times the motor's speed to achieve a volume is excess of the motor’s displacement.
Here's how you want to think about it in gross terms: At sea level atmospheric pressure is about 14.7 psi. Boost is relative to atmospheric pressure and is typically measured in psi, but for this exercise we'll start by using a unit of measure called Atmospheres or ATM. 14.7 psi is one ATM, 29.4 is 2 ATM, and 7.35 is 0.5 ATM. A theoretically perfect normally aspirated engine will move its displacement (6.8l in your example) with each revolution. (In actuality it will move slightly more or less depending on the elasticity of air molecules, restrictions, etc. and is why tuned intakes and headers improve power). Anyhow, if your blower is capable of 1 ATM you are moving approximately twice as much air (like running a 13.6l normally aspirated motor) of 2 ATM is three times as much air (like running a 20.4l normally aspirated motor).
Now if we go back to 8psi, which is 8/14.7 or 0.544 (ATM), which is like running a 6.8l * 1.544 = 10.5l normally aspirated motor. Make sense? A 6.0l diesel at 28 psi is like a 28/14.7 = 1.90 ATM which is running like a 6.0l * 2.90 = 17.48l normally aspirated motor.
Here's how you want to think about it in gross terms: At sea level atmospheric pressure is about 14.7 psi. Boost is relative to atmospheric pressure and is typically measured in psi, but for this exercise we'll start by using a unit of measure called Atmospheres or ATM. 14.7 psi is one ATM, 29.4 is 2 ATM, and 7.35 is 0.5 ATM. A theoretically perfect normally aspirated engine will move its displacement (6.8l in your example) with each revolution. (In actuality it will move slightly more or less depending on the elasticity of air molecules, restrictions, etc. and is why tuned intakes and headers improve power). Anyhow, if your blower is capable of 1 ATM you are moving approximately twice as much air (like running a 13.6l normally aspirated motor) of 2 ATM is three times as much air (like running a 20.4l normally aspirated motor).
Now if we go back to 8psi, which is 8/14.7 or 0.544 (ATM), which is like running a 6.8l * 1.544 = 10.5l normally aspirated motor. Make sense? A 6.0l diesel at 28 psi is like a 28/14.7 = 1.90 ATM which is running like a 6.0l * 2.90 = 17.48l normally aspirated motor.
Last edited by FTE Herman; 02-06-2008 at 08:42 PM.
#3