> > Tracy replies:
> >
> >  OK here is the guts of what I blathered on for two hours about:
> >
> > You are correct about the importance of the velocity of air in the
> drag
> > equation.  But the air velocity through the rad is of no (or very
> minor)
> > importance.
> >
> > The velocity that matters is the velocity of the air at it's *source*.
> > That, of course, is the speed of the air rushing past the airplane
> that we
> > snagged to run through our cooling system.  Since this speed is the
> same
> > regardless of what radiator is sitting inside the cowl, the *only*
> factor
> > that makes any difference is how much of that air we used.  More air =
> > more
> > energy = more drag.
> >
> > Tracy
> >

David said:

> Yes, except.  The forward energy used to slow down that incoming air is
> offset by any thrust you may be able to obtain from the exiting air (or,
> if you prefer, there is also significant drag at the exit point, this
> drag is inversely proportional to the speed of the exiting air).  Like
> in a jet engine, what matters is the difference between the intake and
> exhaust velocity.  And this is where I disagree with your first point,
> the rad is the main determinant of this difference in velocity between
> intake and exhaust (given optimum ducting- which I'll admit is darn near
> impossible).  The more energy you loose passing air through the rad, the
> less you will be able to speed it for smooth exit.  And it is velocity
> through the rad (because of its cubed effect) that is the primary
> determinant of energy loss there.

Tracy again:
 I promise not to risk 2 hours of typing this time :-)

Correct, much of the debate hinges on this point.  It is my contention that
90+ percent of the energy represented by the high speed air used to cool the
engine is lost in either thick or thin radiator.  There is no significant
energy to recover.  This is easily verified by measuring the velocity of the
air coming out of the rad.  Typical values are on the order of 20 mph or
less.  To verify, Calculate the velocity of 3000 cfm (typical value for a
180 HP engine) going through an 18"x 20" rad.
 Since energy is equal to (1/2 Mass) x (Velocity *squared*) is is obvious
that recovered energy is a lost cause.

But let's calculate it for fun.  Set mass arbitrarily at 2.
First calculate energy at 200 mph
 (2/2) * (200 * 200) = 40,000.
Now do it for 20 mph
(2/2) * (20 * 20) = 400

This shows that 99% of the energy in that 200 mph air is lost going through
the rad (only 1% is left).

Tracy