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Discussion Starter #1
Hi guys

I'm trying to wrap my brain around calculating the CFM needs for an engine at full RPM but for some reason the answer is eluding me and the fact that I usually live in a metric world doesn't help of course

As you can probably guess this is for my "other" project but once I understand how it works I can extrapolate what I need from an Opel based engine

So let's say a slightly warmed over 2.0 peaking at 6k rpm, that way the answer would be helpful to other people on here, using one of the calculators I posted in the link here a while a go I've found this engine would need 138 cfm at 6k rpm and a VE of 100 ....

Does this sound right ?

I'm trying to understand the connection between theoretical flow of ports and the actual needs of the engine, as well as chosing carburettor sizes

Tried Googling it but you get too many hits these days and there are so many opinions you don't know who to listen to
 

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The math is easy, it's the assumptions that are hard!

Hey ggl, As it says above! The math looks like displacement per revolution times revolutions per minute -- displacement per minute. For a 2.0L four stroke motor you have 122 cu in/2 (it takes 2 revs to complete one cycle in a 4 stroke) so you have 61 cu in per rev. At 6000 rpm the motor requires 366000 cu in per min. We don't usually talk about cim so divide that number by 1728 for cu ft per cu in and you get 212 cfm. Easy!

The assumptions you can make on what else drives the real number are numerous and complicated. You will have to ask Wrench what his were and why (and what software he is referring to). Various sim programs will use more than 100% Ve as dynamic ram will increase the efficiency as will a properly designed exhaust system. The porting is still another study. Lots more things on top of those! -- Good luck, Doug
 

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Discussion Starter #4
That's very close to the [email protected] I get using the simplified formula I found on the Edelbrock site

CFM = CID x RPM x VE ÷ 3456

Thx Dan, at least I'm getting closer to a point where I can figure things out

And thx to slracer who posted while I was busy looking up some more stuff, looks like I'm on the right track here
 

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Discussion Starter #6
Now if only people could be consistent when it comes to CFM rating carbs they test I could actually make an informed choice.
 

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Discussion Starter #8 (Edited)
Goy, I use the swept volume of one cylinder to calculate the fuel required for
a given A/F ratio.
At this point I'm not too worried about fuel Dan. As I mentioned earlier this is for the TR6 and the general consensus among lay people is that the carbs on it are "more than big enough"

Well, best as I can figure from the formula that seems to agree with you both is that even in stock form it gulps air at a rate of around 220cfm in stock form at 5k rpm, assuming 100% Ve, which is probably a stretch but still

I've seen claims for the stock Zenith Stromberg 175 carbs on there now as high as 200 cfm pr carb but also similar sized SU carbs with the standard air box flow as little as 110 cfm pr carb and at that point I'm already fairly close and I was hoping for a nice 20% increase in power before I swap out my carbs for something more suitable a bit further down the line

I understand.
But how does the manufacturer knows if your using there carb in a SSD or DSD/ITB config.
The vacuum ratings change..don't they?
Exactly, so if they could agree on a set vacuum rating, or at least provide the rating they tested under then I'd have a fighting chance

I need to sleep on this, after a beer
 

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Well, best as I can figure from the formula that seems to agree with you both is that even in stock form it gulps air at a rate of around 220cfm in stock form at 5k rpm, assuming 100% Ve, which is probably a stretch but still..
I need to sleep on this, after a beer
I've noticed that also.
I think that the engineers measures the ICE at 85% efficient around that rpm.
To them it's 1.0...100%
 

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Hello. it may be okay to specify at which vacuum the measurement is at. Americans use 10 "wc, and in Norway we tend to use 28" wc.
Just a given cfm number does tell us anything.

and by the way, you'll find just where the torque curve will turn. BHP is a result of torque x rpm, and the Hp will continue to rise even after this value.
 

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Here's one of my observations
BSFC <.600
VE% > 90% ..Yep can go over 100% at times
219 cfm @ 5700rpm-247 cfm @ 7050 rpm
TQ peaks at 5700 142lbs
HP peaks at 7050 147
 

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Here's one of my observations
BSFC <.600
VE% > 90% ..Yep can go over 100% at times
219 cfm @ 5700rpm-247 cfm @ 7050 rpm
TQ peaks at 5700 142lbs
HP peaks at 7050 147
Wow. You clearly run a very nerrow, peaky, upper rpm powerband type cam!
 

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Number check

Here's one of my observations
BSFC <.600
VE% > 90% ..Yep can go over 100% at times
219 cfm @ 5700rpm-247 cfm @ 7050 rpm
TQ peaks at 5700 142lbs
HP peaks at 7050 147
Wrench, If I convert the 147hp at 7050rpm to torque using HP=Torque*rpm/5252 (which I think is the right equation), I get 109.5 lbf-ft. All is good!

However when I calculate HP from your peak torque numbers, HP = 142*5700/5252, the answer is I get is 154+. As this is HIGHER than your PEAK HP number, I am at a loss to understand. Is one of the above numbers a typo? I was going to make a point about a wide, flat HP curve isn't that peaky, or even that narrow, but these numbers don't help me! :sigh: -- Doug
 

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Wrench, If I convert the 147hp at 7050rpm to torque using HP=Torque*rpm/5252 (which I think is the right equation), I get 109.5 lbf-ft. All is good!

...
Doug you are correct
The HP/TQ curve needs to cross "X" at 5252.
given that Horse power is a human term
Torque can be plotted easily
 

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However when I calculate HP from your peak torque numbers, HP = 142*5700/5252, the answer is I get is 154+. As this is HIGHER than your PEAK HP number, I am at a loss to understand. Is one of the above numbers a typo? I was going to make a point about a wide, flat HP curve isn't that peaky, or even that narrow, but these numbers don't help me! :sigh: -- Doug
Your math is incorrect. HP=142x5252/5700 =130.8 hp
 
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