How much CFM of air does my engine use? I think 2.0 liter engine, 4 cylinders means 1 liter of air per revolution. So 4000 RPM= 4000 liters/minute. Convert liters to cubic feet 4000*0.0353= 141 CFM.
This look right to you?
That means it “displaces” 282 CFM. The amount it actually uses will depsnd upon the postion of your throttle plate and the capacity of your engine to “breath”. Valves, manifolds, cam profiles, port anomolies, exhaust restrictions, and other variables affect how much air your engine actually uses.
Sounds about right. If you look in a catalogue for performance equipment, a good carburetor might be rated at 350-400 CFM. That would be a V8 revving up to 5000+ rpm.
There is a small item called “volumetric efficiency” which car designers have to deal with. The air going through without a boost would be less than atmospheric pressure. The rate by which it falls short determines the figure. If you only get 90% of the air (at atmospheric) your volumetric efficiency would be 90%.
Turbo- or super-charging puts more air through and thus increases ouput. Your Volumetric efficiency with a turbo would be over 100% of atmospheric.
Great! It uses 141 CFM and displaces 282 CFM. How does boost work? Is it a straight line graph or 20% boost needs 35% more air?
If displacement = cylinder volume, then each cylinder is sucking in and combusting 0.5L of air every other revolution, so at 4000 rpm you are combusting (4000/2)40.5=4000 liters of air. I don’t understand the difference between “uses” and “displaces” and I don’t understand mountainbike’s answer.
Now I need to introduce a new term, “SFCM” or “standard cubic feet per minute” (or the equivalent SLPM, standard liters per minute). One standard cubic foot of air is one cubic foot of air at standard temperature and pressure (STP), which is 1 atmosphere of pressure and 68 degrees F. (There are several slightly different definition of STP, but let’s keep this simple.) One standard cubic foot of air contains 36 grams of gas molecules, and one cylinder (0.5 L) contains 0.63 grams of air.
Since your engine is sucking air in through the intake, the pressure is lower than one atmosphere, and since the engine is hot, the temperature is higher. So the air you draw in is less dense than standard conditions, and contains less than 0.63 grams per cylinder. WIth Turbocharging, air is forced into the cylinder at atmospheric or higher pressure, so you can combust more than 0.63 grams of air.
In other words, at 4000 rpm you will always use 141 CFM, but you might use less or more than 141 SCFM.
141 CFM looks right to me.
Volumetric Efficiency can exceed 100% in a Naturally Aspirated (no boost) engine at certain speeds through the effects of intake and exhaust manifold tuning.
The difference is simply that even though the volume of the cylinder multiplied by the number of piston cycles may equal a specific amount, that does not mean that the amount of air drawn in will equal that. Doc explained it well as volumetric efficiency. My attempt to explain it in lay terms was apparently wanting.
SFCM complicates the issue by assuming 29.92 in. hg. (the accepted standard for ambient pressure at sea level) and then comparing the air used by the engine at its moment of operation to a standard. I think the question was looking at it as being relative to ambient at the moment of engine operation. The original question simply multiplied C.I.D. by 1/2 RPM (only on the intake stroke does the piston draw air) and assumed that as the amount of air used. My point was that inherant restictions to the engine’s breathing in make that assumption incorrect.
I think Doc explained it best. My hat’s off to him.
Both your points are equally important if you’re interested in what really matters. Static displacement is nothing really if you’re not going to use it properly. There is no replacement for displacement but you have to use it wisely too. What matters most after the bucket size is the number of oxygen molecules you can get into the cylinder because that determines how much fuel you can combust and therefore how much power you can make. Atmospheric pressure and air temperature determines the number of oxygen molecules available per given volume of air.
Increase power by injecting extra oxygen (nitrous), maximizing the VE (some like to call it mass efficiency) of the engine, or both. Maximize VE by compressing the atmospheric intake charge (turbo or supercharger) or by careful engine design that takes advantage of pulse charging (air mass and inertia), attention to valve rate and overlap etc. As mentioned it is not that difficult to achieve >100% VE on a normally aspirated engine with careful attention to various design aspects.
Interesting point about the effect of N2O on the density of the air…and thus the volume being drawn into the “bucket” (I liked that term). As I know you’re aware, N2O also packs more oxygen in. Air is roughly 22% oxygen, and N2O is 33% oxygen (50% more oxygen per volume)…and the N2O is fed in under pressure, air isn’t (assuming no boost device like a turbo or supercherger).
The more oxygen, the hotter the fire…just like a bellows in the fireplace.
I’m curious…is the purpose for the question to determine the correct specs for a new carb or fuel system?
Yes, a narrow range. Chrysler had these long intake manifolds years ago to improve the breathing of their V8s.
You can also ram the air in like on the original 300SL Mercedes. But that needs speed and won’t help in city traffic when you want fast power.