CFM Calucations

[TheSilentChamber]

Anyone know how to calculate how many cubic feet of air will be traveling though a 3" diamater hole at X MPH? This without factoring pressure drop from aerodynamics and such. Also without any resriction on the other side of the hole. This isnt for intake so I dont want any pointless responces about "ram air".

[beef_bourito]

find the area of the circle (pi*radius(in feat)*2) then multiply by whatever X miles is in feat, then divide by 60mins.

Edit: changed inches to feet

[TheSilentChamber]

yeah I'v done figured it out... had a brain freeze for a minute. Used a 3"x3" square for simplicity sake.

came up with with 22cfm at 1mph.

[Schister66]

Anyways here are the numbers I came up with. The velocities are in miles per hour and mach, and the flow rates are in cfm. Measurements for the piping are in inches.

0.4 mach = 304 MPH

2" piping
1.57 x 2 = 3.14 sq in
300 cfm = 156 mph = 0.20 mach
400 cfm = 208 mph = 0.27 mach
500 cfm = 261 mph = 0.34 mach
585 cfm max = 304 mph = 0.40 mach


2.25" piping
3.9740625 sq in = 1.98703125 x 2
300 cfm = 123 mph = 0.16 mach
400 cfm = 164 mph = 0.21 mach
500 cfm = 205 mph = 0.26 mach
600 cfm = 247 mph = 0.32 mach
700 cfm = 288 mph = 0.37 mach
740 cfm max = 304 mph = 0.40 mach


2.5" piping
4.90625 sq in = 2.453125 x 2
300 cfm = 100 mph = 0.13 mach
400 cfm = 133 mph = 0.17 mach
500 cfm = 166 mph = 0.21 mach
600 cfm = 200 mph = 0.26 mach
700 cfm = 233 mph = 0.30 mach
800 cfm = 266 mph = 0.34 mach
900 cfm = 300 mph = 0.39 mach
913 cfm max = 304 mph = 0.40 mach


2.75" piping
5.9365625 sq in = 2.96828125 x 2
300 cfm = 82 mph = 0.10 mach
400 cfm = 110 mph = 0.14 mach
500 cfm = 137 mph = 0.17 mach
600 cfm = 165 mph = 0.21 mach
700 cfm = 192 mph = 0.25 mach
800 cfm = 220 mph = 0.28 mach
900 cfm = 248 mph = 0.32 mach
1000 cfm = 275 mph = 0.36 mach
1100 cfm max = 303 mph = 0.40 mach


3.0" piping
7.065 sq in = 3.5325 x 2
300 cfm = 69 mph = 0.09 mach
400 cfm = 92 mph = 0.12 mach
500 cfm = 115 mph = 0.15 mach
600 cfm = 138 mph = 0.18 mach
700 cfm = 162 mph = 0.21 mach
800 cfm = 185 mph = 0.24 mach
900 cfm = 208 mph = 0.27 mach
1000 cfm = 231 mph = 0.30 mach
1100 cfm = 254 cfm = 0.33 mach
1200 cfm = 277 mph = 0.36 mach
1300 cfm max= 301 mph = 0.39 mach

[Schister66]

Figuring out your Needs!

first you need to know how much air it will need to flow to reach your target horsepower. to figure that, you would use the following formula:
(CID x RPM) / 3456 = CFM

here's an example of a B16:
(97Ci x 8000) / 3456 = 225CFM

of course, if your engine is bored or stroked, you will have to compensate the CID.

the engine will flow 225CFM at 100% volumetric effeciency (VE). great, in a perfect world. actually VE is about 80-90%. so you'll need to adjust to the VE. 85% is a good number to work with. so addjust your CFM to 191CFM

next up is the pressure ratio. the pressure ratio is basically the pressure of the air going into the turbo in comparison to the pressure coming out. unless you are running sequential turbos, the inlet pressure will be the atmospheric pressure, which is an average of 14.7. so if you want 12psi, here's the formula:
(12 + 14.7) / 14.7 = 1.82:1

now you need the temperature rise. as the compressor compresses the air, it will raise the temperature. there is a formula to figure that rise! there is an ideal temperature rise to where the rise is equivelant to the amount of work it takes to compress the air. here's the formula!
T2 = T1 (P2 ÷ P1)0.283

confused yet? of course not! but lets break it down with some back spins and stuff.

T2 = Outlet Temperature in °R
T1 = Inlet Temperature in °R
°R = °F + 460
P1 = Inlet Pressure Absolute
P2 = Outlet Pressure Absolute

easier now huh?

assuming it's 80º outside and we're shooting for 12psi, your inlet temperature (T1) = 80º + 460 = 540ºR

the P1 inlet pressure will be atmospheric in our case and the P2 outlet pressure will be 12psi. atmospheric pressure is about 14.7 psi (as mentioned earlier), so the inlet pressure will be 14.7 psi, to figure the outlet pressure add the boost pressure to the inlet pressure.
P2 = 14.7 + 12 = 26.7 psi

we now have everything we need to figure out the ideal outlet temperature. now take this info into our original formula ( T2 = T1 (P2 ÷ P1)0.283 ) to figure out T2:
T2 = 540ºR(26.7 ÷ 14.7)0.283 = 676ºR

676ºR = 216ºF = ideal oulet temperature. that's a 136º temperature rise.

once again, in a perfect world, these formulas work grear. unfortunately, there's our old friend adiabetic effeciency (AE). a 136ºF temperature rise is at 100% AE. AE of the compressor is usually 65-75%. so you would use 70% for average. so to figure out the actual temperature rise from the ideal temperature rise, you can use this:
Ideal Outlet Temperature Rise ÷ AE = Actual Outlet Temperature Rise

so, 136º ÷ .7 = 194º

then you add the actual temperature rise to the intake temperature (80º) = 274º

now you can figure out your density ratio! as the air is heated, it expands and increases the volume and flow. to compare the inlet and outlet flow, you must know the density ratio. the formula for that is:
(Inlet °R ÷ Outlet °R) × (Outlet Pressure ÷ Inlet Pressure) = Density Ratio

ok, so our example formula would be:
(540ºR ÷ 676ºR) × (26.7 ÷ 14.7) = 1.46 Density Ratio

with all this crap, you can figure out what the actual inlet flow is in CFM. to do this, use this:

Outlet CFM × Density Ratio = Actual Inlet CFM

so!

225CFM × 1.46 = 328.5CFM

that's a 31% increase in CFM, which is a potential for 31% increase in power. ei, 160hp = 209.6hp. of course, that number is directly effected by intercooler, downpipe, exhaust, fuel flow, etc.

so now you know you need about 328.5CFM to reach your target of 12psi, you now can find compressor maps for different turbos to select the compressor that would best suit your needs. some maps are in CFM, and some maps are in lbs/min. to convert CFM to lbs/min, you would multiply CFM x .069.

when looking at a compressor map, you match the corrected air flow (22.7lbs/min in our case) to the pressure ratio (1.82:1 in our case). what you are looking to do is plot your graph where it would be most efficient for the turbo. anywhere below 60%, your turbo will spin entirely too fast of a shaft speed rpm and burn itself up.

A special thanks to Club Si for that bit of info. See the original article at http://forums.clubsi.com/showflat.php?Number=4847585




This is all from my write up in the Honda section if you care to read the whole thing....


SOme of that info won't apply, but try to take what does and use it...

[Schister66]

http://www.automotiveforums.com/vbul...d.php?t=490574

original thread.....and its easier to read than the stuff i pasted in here

[TheSilentChamber]

Yeah, thanks, but like I said... not for intake.

[curtis73]

Schister! Stop being so darn smart. You make me look bad in my own forum

Thanks for the good info! Do you happen to remember the MPH or ft/sec where velocity is generally considered peaked? I remember reading that a velocity of X ft/sec or MPH is about the peak for head flow or exhaust flow. Above that it starts being a restriction... or am I talking out my @ss?

[TheSilentChamber]

Originally Posted by curtis73

Schister! Stop being so darn smart. You make me look bad in my own forum

Thanks for the good info! Do you happen to remember the MPH or ft/sec where velocity is generally considered peaked? I remember reading that a velocity of X ft/sec or MPH is about the peak for head flow or exhaust flow. Above that it starts being a restriction... or am I talking out my @ss?

In an enclosed area there is a point at which the speed of the air (or whatever) will become so turbulent that it starts to restrict itself. But I'm not sure if thats what your asking or not.

[curtis73]

I guess... For instance, a head's intake port. As RPMs increase, the velocity of the air increases. There reaches a speed where (as a vague general rule) the speed of the intake charge won't go any faster and power starts to drop off. Its at this point where the power curve has peaked and starts dropping off.

Porting that same head means that you can get more CFMs before that speed is attained. Basically, you've slowed down the flow which shifts that peak speed to a higher RPM. I just remember that there is a generally accepted speed at which it peaks.

[Alastor187]

Originally Posted by TheSilentChamber

In an enclosed area there is a point at which the speed of the air (or whatever) will become so turbulent that it starts to restrict itself. But I'm not sure if thats what your asking or not.

And wouldn't that velocity be dependent on the size of the cross-section? Shouldn't a cross-section with a smaller area be more sensitive to velocity as compared to a section with a larger area?

[Alastor187]

Originally Posted by curtis73

I guess... For instance, a head's intake port. As RPMs increase, the velocity of the air increases. There reaches a speed where (as a vague general rule) the speed of the intake charge won't go any faster and power starts to drop off. Its at this point where the power curve has peaked and starts dropping off.

Is it really fair to compare the peak velocity through the intake with that through a similarly shaped pipe? While the pipe will experience a continuous flow of fluid, the intake on other hand is experiencing more of a pulsed flow. So the inertial effects of the working fluid will being to have a significant impact on the characteristic of the flow through the intake.

Unless of course that is not what you were implying.

[TheSilentChamber]

Originally Posted by curtis73

I guess... For instance, a head's intake port. As RPMs increase, the velocity of the air increases. There reaches a speed where (as a vague general rule) the speed of the intake charge won't go any faster and power starts to drop off. Its at this point where the power curve has peaked and starts dropping off.

Porting that same head means that you can get more CFMs before that speed is attained. Basically, you've slowed down the flow which shifts that peak speed to a higher RPM. I just remember that there is a generally accepted speed at which it peaks.

I forget the exact measurment and how to get it, but I also remember reading an artical on it a while back. Talked about a engine builder who used to port out heads to the same size as the valve, the motors had pretty much zero bottom end power but made serious high end.


By the way incase anyone is wandering I'm calculating how much air will theoretically flow though a brake cooling duct.

[Schister66]

0.4 mach = 304 MPH

Your actual results may vary (ie when you add an IC) so just use my numbers as sort of a bench mark to compare different piping sizes. One thing you guys will notice is that none of the velocities goes above 304 MPH or 0.4 mach. According to Corky Bell, Maximum Boost pg 61, 304 MPH or 0.4 mach is the point at which airflow meets increased resistance (drag) and flow losses are experienced.

[curtis73]

Ah, thanks! I have that book, so that must be where I read it. I'll look it up.

Thanks for letting me pirate a thread