In a four-stroke naturally aspirated engine, the theoretical maximum volume of air that each cylinder can take in during the intake cycle is equal to the swept volume of that cylinder (0.7854 x bore x bore x stroke).
Since each cylinder has one intake stroke every two revolutions of the crankshaft, then the theoretical maximum volume of air it can ingest during each rotation of the crankshaft is equal to one-half its displacement. The actual amount of air the engine takes in compared to the theoretical maximum is called volumetric efficiency (VE). An engine operating at 100% VE is using its total displacement every two crankshaft revolutions.
Its all about the air pimp as I said many times. Its how much air / fuel mixture you can stuff into the cylinder... That mass is directly proportional to (a) the air density and (b) the volumetric efficiency. There is a VERY close relationship between an engine's VE curve and its torque curve.
In Old tech naturally-aspirated, two-valve-per-cylinder, pushrod engines a VE over 95% is excellent, and 100% is achievable, but extremely difficult to attain. Only full blown race prepped engine reach 110%, and that is by means of extremely expensive specialized intake manifolds, combustion chambers, exhaust header technology and super trick valve system components. The practical limit for normally-aspirated engines, typically DOHC layout with four or more valves per cylinder, is about 115%, which can only be achieved under the most highly-developed conditions, with precise intake and exhaust passage tuning.
Generally, the RPM at peak VE is same as the RPM at the torque peak. Automotive engines rarely exceed 90% VE. There is a lot of good reasons for that performance, including the design requirements for automotive engines (good low-end torque, good throttle response, high mileage, low emissions, low noise, inexpensive production costs, restrictive form factors, etc.), as well as the allowable tolerances for components in high-volume production.
For a known engine displacement and RPM, you can calculate the engine airflow at 100% VE, in sea-level-standard-day cubic feet per minute (scfm) as follows:
100% VE AIRFLOW (scfm) = DISPLACEMENT (ci) x RPM / 3456
Using that equation to evaluate a 540 cubic-inch engine operating 2700 RPM reveals that, at 100% VE, the engine will flow 422 SCFM.
If we know how to calculate the fuel flow required for a given amount of power produced, we can calculate the mass airflow required for that amount of fuel, then by using that calculated airflow along with the engine displacement, the targeted operating RPM, and the achievable VE values, we can pretty much determine the resulting performance.
Once we know the required fuel flow, we can determine the required airflow. Assumes 12.5 air fuel mixture.
Using that best-power air-to-fuel ratio, you calculate required airflow:
MASS AIRFLOW (pph) = 12.5 (Pounds-per-Pound) x FUEL FLOW (pph)
But airflow is usually discussed in terms of volume flow (Standard Cubic Feet per Minute, SCFM). One cubic foot of air at standard atmospheric conditions (29.92 inches of HG absolute pressure, 59°F temperature) weighs 0.0765 pounds. So the volume airflow required is:
AIRFLOW (scfm) = 12.5 (ppp) x FUEL FLOW (pph) / (60 min-per-hour x 0.0765 pounds per cubic foot)
That equation reduces to:
REQUIRED AIRFLOW (scfm) = 2.723 x FUEL FLOW (pph)
OK, hang on. The really useful stuff is next.
If I know my FUEL FLOW, I get:
FUEL FLOW (pph) = HP x BSFC
Replacing "FUEL FLOW" in with "HP x BSFC" :
REQUIRED AIRFLOW (scfm) = 2.723 x HP x BSFC
Now, we can estimate the airflow required for a given amount of horsepower, we can calculate the 100% VE airflow your engine can generate at a known RPM.
Combining those equations will give us one equation which tells us how close we are to our performance goal by knowing Requited HP, RPM, CID and BSFC
REQUIRED VE = ( 9411 x HP x BSFC ) / (DISPLACEMENT x RPM)
Here is an example of how useful that relationship can be. Suppose you decide that a certain 2.2 liter engine would make a great Mini Stock powerplant. You decide that 300 HP is a nice number, and 5200 RPM produces an acceptable mean piston speed
The required VE for that engine will be:
Required VE = (9411 x 300 x .45 ) / (134 x 5200 ) = 1.82 (182 %) no way..no how...
Clearly that's not going to happen with a normally aspirated engine. Here's another example. Suppose you want 300 HP from a 540 cubic inch engine at 2700 RPM, and assume a BSFC of 0.45. Plugging the known values into the equation =
Required VE = (9411 x 300 x .45 ) / (540 x 2700) = 0.87 (87 %)
hope this clears things up a little