How to determine amount of head flow to feed specific engine size?
How to determine amount of head flow to feed specific engine size? I am interested in calculating the minimum amount of air flow that you need to support a specific engine size. For example, lets discuss a LSX 440 cu. in. motor. How much air flow do I need to feed this monster? Do I need a head that flows 300 CFM at .60" valve lift or 340 CFM, or more.
Having said that, I have the following:
440 cubes is 55 cubes per cylinder, with a 4 cycle engine running at 6000 RPM, each cylinder needs 165,000 cu.in of air per minute, thats, 55 x 6000/2.
So, we need a cylinder head that can support that flow volume. I am assuming that we should actually talk about area under curve for the cylinder head. But to simplify stuff, maybe we can talk about the head flow at .30" valve lift.
I have a PP LS6 head at 209 CFM at .3 valve lift. This head will supply 361,152 cu.in. of air per minute per intake port. It seems that 361,152 is way overkill, but nobody would say that a head that flows 322 CFM at .60" valve lift is overkill on a 440.... So what am I doing wrong?
Having said that, I have the following:
440 cubes is 55 cubes per cylinder, with a 4 cycle engine running at 6000 RPM, each cylinder needs 165,000 cu.in of air per minute, thats, 55 x 6000/2.
So, we need a cylinder head that can support that flow volume. I am assuming that we should actually talk about area under curve for the cylinder head. But to simplify stuff, maybe we can talk about the head flow at .30" valve lift.
I have a PP LS6 head at 209 CFM at .3 valve lift. This head will supply 361,152 cu.in. of air per minute per intake port. It seems that 361,152 is way overkill, but nobody would say that a head that flows 322 CFM at .60" valve lift is overkill on a 440.... So what am I doing wrong?
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The intake duration is not 360 degrees, the head doesn't flow near that number when assembled on the engine, etc., etc. Interesting none the less, and there may be some empirical formulas out there for this sort of thing.
I would also point out that any "restriction" calculated this way would only be in effect at the very top of the RPM range. So say you flow 330 but the engine could theoeretically use 350. You would see a theoretical power loss of 350/330 or about 6% - but only when the RPMs passed the 330 "point". Losses lower down from "over-heading" with the 350 head might make up more than that.
I would also point out that any "restriction" calculated this way would only be in effect at the very top of the RPM range. So say you flow 330 but the engine could theoeretically use 350. You would see a theoretical power loss of 350/330 or about 6% - but only when the RPMs passed the 330 "point". Losses lower down from "over-heading" with the 350 head might make up more than that.
Last edited by Gannet; Oct 3, 2007 at 08:37 PM.
After posting on the tire stuff, I realized that the valve is NOT open 360 degrees, in fact, one full cycle is 720 degrees, so for a 230/230 cam, Does this mean I should factor in 230/720 times the 361,152 cu.in. (and get 115368 cu.in.)... So the desired (required) flow was 165,000 cu.in. per minute of air. Therefore, these heads would not supply enough air.
I think that I am getting closer to the right way to view this. Of course, there is efficientcies to consider also.
I think that I am getting closer to the right way to view this. Of course, there is efficientcies to consider also.
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From: Upstate NY
After posting on the tire stuff, I realized that the valve is NOT open 360 degrees, in fact, one full cycle is 720 degrees, so for a 230/230 cam, Does this mean I should factor in 230/720 times the 361,152 cu.in. (and get 115368 cu.in.)... So the desired (required) flow was 165,000 cu.in. per minute of air. Therefore, these heads would not supply enough air.
I think that I am getting closer to the right way to view this. Of course, there is efficientcies to consider also.
I think that I am getting closer to the right way to view this. Of course, there is efficientcies to consider also.
It's not the size of the dog in the fight...power output determines the need for air. A 600 fwhp engine is going to require about the same air if it's 300 cubes, 400 cubes or 600 cubes.
You would be better served using a good engine simulator. They do a lot of very complicated calculations to determine the amount of air needed to produce the torque at every rpm (therefore power).
FWIW, few folks work in cubic inches of airflow per minute, but rather cubic feet per minute (CFM). Divide your numbers by 1728.
Airflow thru the intake valve in an operating engine is not just about valve lift. It's all about pressure differential and the curtain area presented by the valve and valve seat. Air only moves from a higher pressure area to a lower pressure area. The pressures inside an operating engine continuously vary during the 720° of one cycle and they also vary with changing rpm.
OK, I was being kind above. No offense intended, but I think you are on the wrong track on a handcart with a 100 car freight train bearing down on you at 100 mph.
Old SStroker: what you say is correct (especially the first statement), and your last paragraph is pretty funny.
The internal combustion and its flow charteristics, etc, etc, is very complicated. I realize that. However, I wanted to break this down to a handy calculation, similar in vain to the one I remember from one of my engine books. To determine the power output of an specific size engine you do this:
The ideal CFM of a 440 cu.in engine @ 6,000 RPM is: (440/1728) x 6,000/2 which equals 763.9 CFM. The author claims that it takes about 1.3 CFM to make 1.0 Hp. Therefore, a 440 should make about 588 Hp @ the flywheel.
So, my goal was to come up with a calculation that we can use to determine the amount of flow that the cylinder head must deliver to attain that 588 Hp.
Now before we get into an argument about the 1.3 CFM to make 1.0 Hp, this guy's figuring is for NA street/strip engine, not blown engines or even Pro Stock or racing engines. And I think that he is referring to Gen I and Gen II engines.
I think that I am on the right track, provided that I take into account the duration of the cam. But, I can use the ideal CFM to get at the expected Hp, but still I want to be able to get at the required flow of the cylinder head. Can you help me out?
The internal combustion and its flow charteristics, etc, etc, is very complicated. I realize that. However, I wanted to break this down to a handy calculation, similar in vain to the one I remember from one of my engine books. To determine the power output of an specific size engine you do this:
The ideal CFM of a 440 cu.in engine @ 6,000 RPM is: (440/1728) x 6,000/2 which equals 763.9 CFM. The author claims that it takes about 1.3 CFM to make 1.0 Hp. Therefore, a 440 should make about 588 Hp @ the flywheel.
So, my goal was to come up with a calculation that we can use to determine the amount of flow that the cylinder head must deliver to attain that 588 Hp.
Now before we get into an argument about the 1.3 CFM to make 1.0 Hp, this guy's figuring is for NA street/strip engine, not blown engines or even Pro Stock or racing engines. And I think that he is referring to Gen I and Gen II engines.
I think that I am on the right track, provided that I take into account the duration of the cam. But, I can use the ideal CFM to get at the expected Hp, but still I want to be able to get at the required flow of the cylinder head. Can you help me out?
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From: Upstate NY
The internal combustion and its flow charteristics, etc, etc, is very complicated. I realize that. However, I wanted to break this down to a handy calculation, To determine the power output of an specific size engine you do this:
The ideal CFM of a 440 cu.in engine @ 6,000 RPM is: (440/1728) x 6,000/2 which equals 763.9 CFM. The author claims that it takes about 1.3 CFM to make 1.0 Hp. Therefore, a 440 should make about 588 Hp @ the flywheel.
So, my goal was to come up with a calculation that we can use to determine the amount of flow that the cylinder head must deliver to attain that 588 Hp.
Now before we get into an argument about the 1.3 CFM to make 1.0 Hp, this guy's figuring is for NA street/strip engine, not blown engines or even Pro Stock or racing engines. And I think that he is referring to Gen I and Gen II engines.
I think that I am on the right track, provided that I take into account the duration of the cam. But, I can use the ideal CFM to get at the expected Hp, but still I want to be able to get at the required flow of the cylinder head. Can you help me out?
The ideal CFM of a 440 cu.in engine @ 6,000 RPM is: (440/1728) x 6,000/2 which equals 763.9 CFM. The author claims that it takes about 1.3 CFM to make 1.0 Hp. Therefore, a 440 should make about 588 Hp @ the flywheel.
So, my goal was to come up with a calculation that we can use to determine the amount of flow that the cylinder head must deliver to attain that 588 Hp.
Now before we get into an argument about the 1.3 CFM to make 1.0 Hp, this guy's figuring is for NA street/strip engine, not blown engines or even Pro Stock or racing engines. And I think that he is referring to Gen I and Gen II engines.
I think that I am on the right track, provided that I take into account the duration of the cam. But, I can use the ideal CFM to get at the expected Hp, but still I want to be able to get at the required flow of the cylinder head. Can you help me out?
Put down the calculator. It's time to get empirical. Gather information from engines similar to the one you are "desigining": Displacement, fwhp (or convert from rwhp), maximum head flow from a realistic (not "happy") flowbench @ 28 in H2O.
Now calculate the HP/CFM for these engines and graph it. I'd suggest power on the X-axis and CFM on the Y-axis. Don't start with zero on either. Use say 200 to 400 CFM. Plot as many engines for which you can find reliable info. Now try to fit a straight line to the data. This might be the tricky part and should show you something. You then have a way to relate CFM to HP.
Relating HP to displacement is a little more complex, but that's for another time. Don't throw away the data you collected.
You'd make a pretty good straight man.
Great idea, of course the problem with doing what you suggest is: I have to graph "believable data",,, not someone's second hand knowledge or as you say happy numbers, or "I have a buddy who got xxx RWHp with such and such". First data point is my own '99 LS1 car. I had it on a chassis dyno and have flow data from Patriot Performance. I'm assuming that my data is solid. What do you have to offer as realiable data? I will also search signatures for data.
BTW: I love empirical !!!!!
BTW: I love empirical !!!!!
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From: Upstate NY
Great idea, of course the problem with doing what you suggest is: I have to graph "believable data",,, not someone's second hand knowledge or as you say happy numbers, or "I have a buddy who got xxx RWHp with such and such". First data point is my own '99 LS1 car. I had it on a chassis dyno and have flow data from Patriot Performance. I'm assuming that my data is solid. What do you have to offer as realiable data? I will also search signatures for data.
BTW: I love empirical !!!!!
BTW: I love empirical !!!!!
If you have a data point way off the normal scatter you can suspect it was BS (high) or engine problems (low). You might look on other boards for dyno info.
Have fun!
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This sound like a great excercise. I would make one suggestion. Keep it to m6 cars. The auto cars throw too much chance for change into the mix, especially since 99% of cars with the typical H/C combo have a higher stall converter.
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Joined: Jun 2002
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From: Upstate NY
I have been collecting some data for plotting purposes, but that is boring, too many unknowns --- real data VS "happy" data.
I do not want this thread to die too early. How about some more feedback.
For a 440 cu.in motor that is going to use a cam that makes peak power at 6500 RPM, I figure that I need an intake runner of about 264 CFM flow at "average valve lift" or "average flow during valve opening" or Average Runner Flow (ARF).
Check this out:
440 / 1728 = 0.2546 cu.ft,,,,,, times 6500/2 = 827 CFM (called "ideal CFM" by a technical author). Each cylinder needs about 103 CFM.
Now I have for a 230/230 cam:
230/720 * ARF = 103 CFM ..... ARF=322 CFM
For a 240/240 cam:
240/720 * ARF = 103 CFM ..... ARF=309 CFM
But how do we get at the ARF of 322 CFM,,, what does it represent, or do I have nothing with this long winded ordeal? For 322 CFM that is what some heads can flow at full valve lift. So I think that I am missing a factor.
I do not want this thread to die too early. How about some more feedback.
For a 440 cu.in motor that is going to use a cam that makes peak power at 6500 RPM, I figure that I need an intake runner of about 264 CFM flow at "average valve lift" or "average flow during valve opening" or Average Runner Flow (ARF).
Check this out:
440 / 1728 = 0.2546 cu.ft,,,,,, times 6500/2 = 827 CFM (called "ideal CFM" by a technical author). Each cylinder needs about 103 CFM.
Now I have for a 230/230 cam:
230/720 * ARF = 103 CFM ..... ARF=322 CFM
For a 240/240 cam:
240/720 * ARF = 103 CFM ..... ARF=309 CFM
But how do we get at the ARF of 322 CFM,,, what does it represent, or do I have nothing with this long winded ordeal? For 322 CFM that is what some heads can flow at full valve lift. So I think that I am missing a factor.
TECH Fanatic
Joined: Jun 2002
Posts: 1,979
Likes: 3
From: Upstate NY
I have been collecting some data for plotting purposes, but that is boring, too many unknowns --- real data VS "happy" data.
I do not want this thread to die too early. How about some more feedback.
For a 440 cu.in motor that is going to use a cam that makes peak power at 6500 RPM, I figure that I need an intake runner of about 264 CFM flow at "average valve lift" or "average flow during valve opening" or Average Runner Flow (ARF).
Check this out:
440 / 1728 = 0.2546 cu.ft,,,,,, times 6500/2 = 827 CFM (called "ideal CFM" by a technical author). Each cylinder needs about 103 CFM.
Now I have for a 230/230 cam:
230/720 * ARF = 103 CFM ..... ARF=322 CFM
For a 240/240 cam:
240/720 * ARF = 103 CFM ..... ARF=309 CFM
But how do we get at the ARF of 322 CFM,,, what does it represent, or do I have nothing with this long winded ordeal? For 322 CFM that is what some heads can flow at full valve lift. So I think that I am missing a factor.
I do not want this thread to die too early. How about some more feedback.
For a 440 cu.in motor that is going to use a cam that makes peak power at 6500 RPM, I figure that I need an intake runner of about 264 CFM flow at "average valve lift" or "average flow during valve opening" or Average Runner Flow (ARF).
Check this out:
440 / 1728 = 0.2546 cu.ft,,,,,, times 6500/2 = 827 CFM (called "ideal CFM" by a technical author). Each cylinder needs about 103 CFM.
Now I have for a 230/230 cam:
230/720 * ARF = 103 CFM ..... ARF=322 CFM
For a 240/240 cam:
240/720 * ARF = 103 CFM ..... ARF=309 CFM
But how do we get at the ARF of 322 CFM,,, what does it represent, or do I have nothing with this long winded ordeal? For 322 CFM that is what some heads can flow at full valve lift. So I think that I am missing a factor.
There are many factors which determine power output. You are attempting to find a simple rule of thumb which really doesn't exist. You will continue to frustrate yourself going about it this way. That may be better than boring.
If you are thinking about being an engine builder, don't give up your day job.