Cfm At Wot
Cfm At Wot
Postmaster
Joined: Dec 2003
Posts: 4,196
Likes: 11
From: Arizona
The formula to calculate it is as follows.
Air Flow (CFM) = [Displacement (cubic inches)/1728] x [RPM/Cycle Factor] x Volumetric Efficiency
Cycle Factor:
2 (for 4 cycle gas or diesel engine)
1 (for 2 cycle gas or diesel engine)
Volumetric Efficiency (approximate average):
Naturally Aspirated Gasoline & Diesel Engines = 0.8
Supercharged Diesel Engines = 1.30
Turbocharged Diesel Engines = 1.75
Note: The 1.75 volumetric efficiency is applicable only at top governed engine speed under full load conditions.
So based on a 5.4L(329.53CI) two valve, normally aspirated with max rating at 4500 RPM
The formula results in 686 CFM
Hope this helps
Air Flow (CFM) = [Displacement (cubic inches)/1728] x [RPM/Cycle Factor] x Volumetric Efficiency
Cycle Factor:
2 (for 4 cycle gas or diesel engine)
1 (for 2 cycle gas or diesel engine)
Volumetric Efficiency (approximate average):
Naturally Aspirated Gasoline & Diesel Engines = 0.8
Supercharged Diesel Engines = 1.30
Turbocharged Diesel Engines = 1.75
Note: The 1.75 volumetric efficiency is applicable only at top governed engine speed under full load conditions.
So based on a 5.4L(329.53CI) two valve, normally aspirated with max rating at 4500 RPM
The formula results in 686 CFM
Hope this helps
Last edited by blackhat620; Jul 12, 2007 at 10:29 PM.
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Joined: Jun 2007
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Awesome, yeah that helps i bought the truck and the guy before me put an attwood bilge blower inline with the intake thinkin it would help at wot it's rated 230 cfm and now i think i need to take it out and just put in some 4"ID intake hose instead. If i leave it in there not using it would it restrict my airflow too much?
Sit. Stay.
Joined: Oct 2001
Posts: 18,308
Likes: 20
From: Washington State
Originally Posted by blackhat620
The formula results in 686 CFM
However, I got a different number and want to check my math with you. I came up with 343CFM.
I took 329 cubic inches and divided it by 1728. (Why 1728, BTW??)
Air Flow (CFM) = [Displacement (cubic inches)/1728]
I then multiplied .1903 by 2250 since 2250 is one half of 4500. BTW, redline is around 5200-5400rpm
x [RPM/Cycle Factor]
428.175 times .8 = 342.54
x Volumetric Efficiency
Postmaster
Joined: Dec 2003
Posts: 4,196
Likes: 11
From: Arizona
Originally Posted by Monsta
Great formula. Thanks for posting!
However, I got a different number and want to check my math with you. I came up with 343CFM.
I took 329 cubic inches and divided it by 1728. (Why 1728, BTW??) I came up with .1903.
I then multiplied .1903 by 2250 since 2250 is one half of 4500. BTW, redline is around 5200-5400rpm I ended up with 428.175.
428.175 times .8 = 342.54
It appears as if you didn't take into the fact that the 5.4L is a 4-cycle.
However, I got a different number and want to check my math with you. I came up with 343CFM.
I took 329 cubic inches and divided it by 1728. (Why 1728, BTW??) I came up with .1903.
I then multiplied .1903 by 2250 since 2250 is one half of 4500. BTW, redline is around 5200-5400rpm
I ended up with 428.175. 428.175 times .8 = 342.54
It appears as if you didn't take into the fact that the 5.4L is a 4-cycle.
You are correct, i screwed up the math out of the calulator
I should write down my numbers instead of just punching them into the calculator.Thank you for correcting my error. The CFM of the 5.4L should be 343.
