Chris, I also have no idea about how to figure how much air an engine uses.
Then I typed "air consumption engine".
In the
first attempt I got a table for what I'd call "pollution calculation" purposes. I'd guess we don't know if air is at atmospheric pressure when it enters the cylinder, so probably this kind of calculation is wrong. Anyway, this is a first estimate for your engine:
First formula, if you only know the displacement
Air used = Displacement/2 * rpm
Air used = 3.5 liters/2 * 7000 rpm = 12250 liters/minute
As a liter/minute is like 0.0353 cfm, then that's like
433 cfm.
However, in the
second attempt I found that you could use this approach:
You know that air weighs 1.2 grams/liter and a gallon of gasoline weighs 6 pounds (that's like 2700 grams). So, if you have a fuel:air ratio of 1:14, for every gallon of gas you use 14 * 2700 = 37800 grams of air.
If you divide by 1.2 grams that weighs one liter of air, that's 31500 liters of air you have to use per gallon, which means you get more or less 1100 cubic feet of air per gallon of gasoline.
Second formula if you know the fuel you used
Air used (cfm) = fuel used (in gallons/minute) * 1100
I don't know. Something's wrong here.
You have to use 0.38 gallons/minute to get the 433 cfm of the first approach. If you check somehow how much fuel your engine uses (you can download the information from the ECU of your car at your local workshop, btw) you can find how much air you need.
So, for the
third attempt, let's go with SZ approach (welcome, SZ!), which seems to be right (unless your car uses 23 gallons/hour at that rpm regime!). I take the approach "for-dummies", step by step.
Power of a V6, 3.5 liters at 7000 rpm. Shut. Let's see.
My first "googling" tells me that Ford offers a V6, 3.5 liters rated at 265 hp. Honda gives you the same V6/3.5 at 275 hp. Give or take, that's like 200 Kwatts. You have peak power at 5000 rpm if you get a Ford, or at 6200 rpm in the case of Honda.
I see a problem here: according to SZ assumptions, we should use
less fuel at 7000 rpm, when we have
less power, but let's move on over the "little problem" of calculating thermal efficiency for every speed of the engine... ehem.
I assume a thermal efficiency of 30%, that gives me 666 Kwatts (nice number!
).
If you run your engine for one hour you have 200 Kwatts-hour, that's like 720 Mjoules (at 3.6 Mjoules per Kwatt-hour).
In one minute you get 12 Mjoules of work (that is, 720/60).
Knowing that gasoline gives you 47 Mjoules per kilogram, you have to use .255 Kg of gasoline (that I get from 12/47). That's a little over half a pound of gas per minute, which means less than 0.09 gallons/minute (dividing by 6 pounds per gallon), a figure very different from the "volumetric" answer I got in the first attempt.
To sum up:
Fuel used (gallons/minute) = Power (HP) * 0.000351
As I know already (from the second attempt) that you have to multiply the fuel used by 1100, I get
Third formula if you know the power of your engine
Air used (cfm) = Power (HP) * 0.39
In your case:
Air used = 275 HP * 0.39 =
105 cfm (give or take)
That's like 1/4th of the first, "volumetric" answer I got.
Can anyone explain where I made a mistake? Or, in the improbable case I made no mistakes (in the 15 minutes I spent on this), why the air seems to be at 1/4th of the "normal" density inside the cylinder?