which CFM formula to use for turbo sizing?
#1
which CFM formula to use for turbo sizing?
What formula should I use to determine the approximate CFM a 13B will flow at 100% efficiency?
The straight up one for a four-stroke of:
CI * RPM / 3456 = CFM
I don't think will work. If I plug into it, I get:
80 * 7000 / 3456 = ~162 CFM at 7000 RPM
Take into account the restrictions in the intake and you get much less than that even. SO.. what formula should I use? A rotary is much different than a piston motor, so what should I do?
The straight up one for a four-stroke of:
CI * RPM / 3456 = CFM
I don't think will work. If I plug into it, I get:
80 * 7000 / 3456 = ~162 CFM at 7000 RPM
Take into account the restrictions in the intake and you get much less than that even. SO.. what formula should I use? A rotary is much different than a piston motor, so what should I do?
#2
A 13B is 160cid, not 80. That formula assumes the displacement is, uh, displaced, ( ) over a period of 2 revolutions. Over a period of two revolutions, a 13B displaces 160cid, since its rated displacement is per SINGLE revolution.
So, plugging 160 into your formula (or rather, multiplying your result by two) gives us 324cfm. Much more closerful, although I suspect that formula assumes 100% VE. Still, it's as good a number as any to work with, certainly better than 162cfm.
So, plugging 160 into your formula (or rather, multiplying your result by two) gives us 324cfm. Much more closerful, although I suspect that formula assumes 100% VE. Still, it's as good a number as any to work with, certainly better than 162cfm.
#3
80cid per one revolution of the eccentric shaft (output RPM, as measured on the tach). That means 160cid for that measurement, ok that makes more sense.
So measure it like a 160cid engine. That makes more sense, ok. So now I can do the calculations at about 3000 RPM, all the way to redline, and see what it makes.
So for a S5 TurboII motor, it should flow 160 * 7000 / 3456 = 324 at 100% VE. So assume about 280 to 300 CFM?
So measure it like a 160cid engine. That makes more sense, ok. So now I can do the calculations at about 3000 RPM, all the way to redline, and see what it makes.
So for a S5 TurboII motor, it should flow 160 * 7000 / 3456 = 324 at 100% VE. So assume about 280 to 300 CFM?
#4
Using these calculations..
n(lbs/min)= P(psia) x V(cu.ft./min) x 29 / (10.73 x T(deg R))
I came up with this
f(x) = ((14.7 * ((160 * x) / 3456) * 29) / (10.73 * 520)) *.85
assuming 85% VE at 7000 RPM, a rotary flows 21 lb/hr of air.
using same at 4000 RPM a rotary flows 11.7 lb/hr of air.
Does this seem incredibly low? I guess for a N/A these are correct, but from here on out, how do I match THESE numbers to a turbo? From what it looks like, even a GT35 is WAY too big for a rotary to stay anywhere near the efficient range of that turbo.
What am I doing wrong here? Should I be plugging in 14.7 + 10psi (randomly picked desired boost level) for P(psia)? Or what's going wrong here?
n(lbs/min)= P(psia) x V(cu.ft./min) x 29 / (10.73 x T(deg R))
I came up with this
f(x) = ((14.7 * ((160 * x) / 3456) * 29) / (10.73 * 520)) *.85
assuming 85% VE at 7000 RPM, a rotary flows 21 lb/hr of air.
using same at 4000 RPM a rotary flows 11.7 lb/hr of air.
Does this seem incredibly low? I guess for a N/A these are correct, but from here on out, how do I match THESE numbers to a turbo? From what it looks like, even a GT35 is WAY too big for a rotary to stay anywhere near the efficient range of that turbo.
What am I doing wrong here? Should I be plugging in 14.7 + 10psi (randomly picked desired boost level) for P(psia)? Or what's going wrong here?
#5
Originally posted by Barwick
f(x) = ((14.7 * ((160 * x) / 3456) * 29) / (10.73 * 520)) *.85
f(x) = ((14.7 * ((160 * x) / 3456) * 29) / (10.73 * 520)) *.85
f(x) = (((14.7+ b) * ((160 * x) / 3456) * 29) / (10.73 * 520)) *.85
Where b = boost pressure in PSI
#6
Originally posted by Evil Aviator
The original formula asks for Absolute pressure, and you only put the Ambient pressure in the above equation. You need to add your boost pressure. Using your numbers, you should have 24.7 (ie 14.7+10) instead of 14.7. I don't have time right now to troubleshoot the whole thing, but try this and see if it works better:
f(x) = (((14.7+ b) * ((160 * x) / 3456) * 29) / (10.73 * 520)) *.85
Where b = boost pressure in PSI
The original formula asks for Absolute pressure, and you only put the Ambient pressure in the above equation. You need to add your boost pressure. Using your numbers, you should have 24.7 (ie 14.7+10) instead of 14.7. I don't have time right now to troubleshoot the whole thing, but try this and see if it works better:
f(x) = (((14.7+ b) * ((160 * x) / 3456) * 29) / (10.73 * 520)) *.85
Where b = boost pressure in PSI
I think I kinda see how corrected pressure (on the left side) fits into this.. makes a little more sense.
btw this IS Barwick, I'm on Raptor13x's computer.
#7
So by that formula.. what turbos are a good size? I was looking through the GT turbos and at even 15 psi (I was looking at 10 ideally) most of those are either too small or too big, nothing fits very well.
Realistically, what's the VE of a stock ported 13bt?
Realistically, what's the VE of a stock ported 13bt?
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#9
Use this calculator to help ease the math. Just make sure that you use 2.6 liters instead of 1.3.
http://www.squatto.com/turbomath.php
http://www.squatto.com/turbomath.php
#10
Racing Rotary Since 1983
iTrader: (6)
Joined: Oct 2001
Posts: 6,136
Likes: 563
From: Florence, Alabama
here's another way to back into the answer to your question.
it is generally accepted that it takes 1.92 cfm to make 1 rwhp w the turboed rotary. a nicely done 20b w the proper sized turbo can make 650 rwhp without breaking a sweat. 650 rwhp times 1.92 equals 1248 cfm. generally compressor maps are stated in pounds per minute.... 10 lbs/minute equals 144.71 cfm. you should be looking for a turbo that will flow about 86 lbs/minute efficiently. my friends pettit built 20b w a t66 on it is underturboed as working out the flow map using the above calculations we end up w 515 rwhp. that makes for an exciting ride but isn't fully developed for the motor.
howard coleman
it is generally accepted that it takes 1.92 cfm to make 1 rwhp w the turboed rotary. a nicely done 20b w the proper sized turbo can make 650 rwhp without breaking a sweat. 650 rwhp times 1.92 equals 1248 cfm. generally compressor maps are stated in pounds per minute.... 10 lbs/minute equals 144.71 cfm. you should be looking for a turbo that will flow about 86 lbs/minute efficiently. my friends pettit built 20b w a t66 on it is underturboed as working out the flow map using the above calculations we end up w 515 rwhp. that makes for an exciting ride but isn't fully developed for the motor.
howard coleman
#12
Thanks, that page is kinda useful for quick stuff I guess.
I like having a graph to look at though so I can figure out what I'm going to be working with at all the different RPM. I use a free program called MathGV to graph using the equations above...
I like having a graph to look at though so I can figure out what I'm going to be working with at all the different RPM. I use a free program called MathGV to graph using the equations above...
#13
Yea, I've always heard and used 1.9 (which I guess 1.92 is pretty much the same). SO if you were looking at a compressor map and it shows that it would be making 70lbs/min in the area you'd be running than you'd do 70/.069 to convert to cfm and get 1015 cfm then you'd divide that by 1.90 (or 1.92 I guess) and get 534rwhp.
I've compared that with actuall results of a few single turbo cars that I know what boost, hp, and turbo they were running and it seemed to be fairly accurate. I think all the examples I was looking at acutally came out to within about 25rwhp of what the calculations showed.....which I was happy with it being that accurate.
STEPHEN
I've compared that with actuall results of a few single turbo cars that I know what boost, hp, and turbo they were running and it seemed to be fairly accurate. I think all the examples I was looking at acutally came out to within about 25rwhp of what the calculations showed.....which I was happy with it being that accurate.
STEPHEN
#15
Racing Rotary Since 1983
iTrader: (6)
Joined: Oct 2001
Posts: 6,136
Likes: 563
From: Florence, Alabama
yes, as my post stated, the 1.92 cfm relationship/rwhp is for the turbo'd ported 13b rotary. as to VE, search around and you will come up w the number... personally i can't think of a reason to spend any time with VE since you have what you need w the 1.92 number to calculate airflow, hp etc...
i spent a moment looking at your graph and it is a linear construction which does not accurately portray the non linear aspects of hp/torque production in the real world. your graph shows a linear relationship between boost & hp and rpm & hp.( 295 hp at 7500 at no boost and 590 at 14.7 lbs--- 100 hp at 2500 rpm and 200 at 5000rpm) the logic behind the graph assumes standard temperature and pressure (which doesn't happen except in fantasyland) and the same flow harmonics at different boost levels (ditto) no electronic degradation etc etc. ditto for doubling the rpm doubling the hp. you'd find more insight looking at a compressor map. what you want is a compressor that is most efficient between max torque rpm and max hp. you can plot a real world demand curve relating to rpm and boost from the map. i suggest you use approx 630 rwhp as a target. and around 4500 rpm max torque.
howard coleman
i spent a moment looking at your graph and it is a linear construction which does not accurately portray the non linear aspects of hp/torque production in the real world. your graph shows a linear relationship between boost & hp and rpm & hp.( 295 hp at 7500 at no boost and 590 at 14.7 lbs--- 100 hp at 2500 rpm and 200 at 5000rpm) the logic behind the graph assumes standard temperature and pressure (which doesn't happen except in fantasyland) and the same flow harmonics at different boost levels (ditto) no electronic degradation etc etc. ditto for doubling the rpm doubling the hp. you'd find more insight looking at a compressor map. what you want is a compressor that is most efficient between max torque rpm and max hp. you can plot a real world demand curve relating to rpm and boost from the map. i suggest you use approx 630 rwhp as a target. and around 4500 rpm max torque.
howard coleman
#16
I used the maps to guess what it might be flowing at a specific RPM, like when I'm doing 3000 RPM, do I want to be able to step on it and get boost, or is it ok if the compressor can't make boost that low?
And similarly, to match up the peak output at redline to still be somewhere in the ballpark of "efficient" on the compressor map.
And similarly, to match up the peak output at redline to still be somewhere in the ballpark of "efficient" on the compressor map.
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