Ram air

On Jun 2, 10:01 am, "Ken S. Tucker" wrote:

Tina, I think this analysis you posted is good,
" It's only a 360 cubic inch engine turning at 2300 RPM or so. Isn't
that a demand of, let's see, at 23 inches mp at sea level that's
23/30 * 2300/2 * 360 / 12^3 or 180 cubic feet a minute? "

Can't do it that way. You're assuming a volumetric efficiency
of 100% which we never attain without considerable boosting. The
volumetric efficiency at full throttle and redline RPM at sea level
isn't likely to be much more than 50 or 60%.
Got to do it using fuel flow. Best power mixture comes at
around 12:1 (pounds of air to pounds of fuel) and stoichiometric
mixture (no wasted air or fuel) is 15:1. Weight of air at sea level is
about .078 pounds per cubic foot, and weight of gasoline is 6 lb per
US gallon.
An O-320 @ 2700 RPM @ S.L. = Displacement of 15,000 cubic feet
per hour.
Full throttle fuel flow of 10.3 GPH @ 12:1 best power = 9434
cu. ft./hr (with fixed-pitch prop).
9434 ÷ 15,000 = .629 (62.9%) volumetric efficiency @ sea
level.

Not very good, is it? Air has viscosity and the drag of the
entire induction system, even with the throttle wide open, is
significant. Add to that the inertia of the air, and the intake
valve's opening and closing causing the stop-go action of the air in
the system, and things get slowed down considerably.
It's worse in auto engines that turn at high RPM. That's why
many have four valves per cylinder, or turbos, or both.

Dan