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View Full Version : Question on the turbo formula sticky....


Addict
01-16-2002, 12:39 PM
I'm not sure I follow the formula that's listed.
I used THIS CALCULATOR (http://familyinternet.about.com/library/conv/blvolume.htm) to get my CID.

1.6 Liters = 97.63200 Cubic Inches

(97.63200 x RPM x 0.5 x VE) / 1728 = CFM

What do I enter for my RPM?
How do I figure my Volumetric Efficency? I know the average is 0.8-0.9.

(97.63200 x 3000 x 0.5 x 0.8) / 1728 = 67.8

Alright. I'm guessing this is correct. So moving on to the next part.
Knowing the desired boost level, calculate airflow rate under
boost by multiplying the pressure ratio by the airflow rate
(NA-CFM). To figure out the pressure ratio take (14.7 + boost)
divided by 14.7

This lost me. Let's say my boost desired was 10psi.

(14.7 + 10)/14.7 = 1.6802721088435374149659863945578

That doesn't seem right. Though I guess it could be. What is the (NA-CFM)? Is this a formula too? Also where can I find a compressor map?

Heres a low and a high range for horsepower expectations

Low: 0.052 x CID x (psi boost + 14.7) = bhp

High 0.077 x CID x (psi boost + 14.7) = bhp


Low : 0.052 x 67.8 x (10 + 14.7) = 87.08232
High : 0.077 x 67.8 x (10 + 14.7) = 128.94882

Sorry if I sound stupid, I'm just trying to make sure I understand this correctly.

Boosted3g
01-16-2002, 04:24 PM
OK the reason it sounds wrong is because it is;) Let me see if i can help you.

To get the NA-CFM which is just a measure of the volume of air that your engine can consume naturally aspirated. Since you have a B16 im going to assume you rev to 7500 (i dont know what the redline is ill just pick a # as an example). So this is what your formula will look like.

97.632 X 7500 X .5 X .85 / 1728

Notice i used 7500 as the rpm being that is the engines maximum rpm. Also .5 because it is a 4 stroke engine and i guessed .85 as the VE which is reasonable.

When you do the equation you come up with 180.09

That 180.09 is how many cfm your motor consumed NA.

now for the next part all it is basically doing is calcualting a pressure ratio. We will make this easy and say in boosting 14.7 psi (AKA 1 bar)

so (14.7 + 14.7) / 14.7 = 2

Basically all that means is in pushing 2 times the atmospheric pressure into the engine so theoretically the engine should double in CFM. Lets find out,this is where there is an extra step. If its not listed ill fix the sticky

2 X 180.09 = 360.18

That 360.18 is the amount of CFM the engine consumes while exposed to 14.7 psi of boost. This # will be useful when looking at compressor maps.



For a list of compresor maps there are a few listed at
www.turboneticsinc.com (http://www.turboneticsinc.com)


If you want to know any other formulas for engines just let me know. I have one for everything from calculating piston speeds to figuing out compresssion ratios.

Addict
01-16-2002, 06:30 PM
Thanks man! I knew you'd be able to shed some light on this.