> 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

Ok, now I lost 2 hours work because my computer crashed, but here goes again.  

Im about to design 2 cooling systems: one thick and one thin rad.   The engine I will be using will require 600 cu in of radiator to cool.  

One radiator is 10”x10” and 6 inches deep.  The other is 24.5” x 24.5” and 1 inch deep. Both have the same volume.  Through proper sizing of the inlets, I am now going to allow 1800 cu in of air to flow through each rad every second.  This is 3 times the rad volume, same for both rads.  Since the volume of air within each rad is the same, each will be replacing the entire volume within each rad 3 times each second.  This means that each incoming cu. in. of  air will spend 0.33 seconds in the rad, regardless of which rad.  

Every packet of air is now spending the same amount of time in contact with radiator surface (0.33 sec), and each cubic inch of rad is seeing the same volume of air (3 cubic inches/sec) – regardless of radiator.  Average “pre-heating” of the air is the same.  Thus, the exit air is the same temperature for both rads and both rads are cooling with the same effectiveness.  In both cases we are using the same amount of high energy incoming air.

Now, the ONLY difference is the speed of the air.  Since each packet of air must get out of the rad in .33 seconds, the thick rad is flowing at 18”/sec while the thin rad is flowing at 3”/sec.  That is a 6 fold difference in speed through the rad.  Since turbulent drag is a function of speed cubed that is a 216 fold increase in drag through the radiator!   Drag is energy.  If it takes 1/4 horse power to push the air through the thin rad, it will take 54hp to push it through the extra thick rad in this example. That extra drag energy is coming from somewhere, and that somewhere is the forward motion of the aircraft.  How is this extra drag manifest on the aircraft?  The thin rad is showing a smaller pressure drop across its surfaces, leaving more pressure available to accelerate the outgoing air.

You may argue as above that this exit speed is insignificant.  But your example is probably for an air cooled engine.  They use very small cooling surfaces and thus very large velocities.  It is no wonder that they chew up all the energy in the incoming air.  I submit that water cooling is far superior because the velocities can be made much smaller (making it all the more important for us to have efficient exit ducts).  And again I submit that a thin rad maximizes this advantage.  

Be careful with that outgoing air, it is now going much faster than in other planes.  It must now be ducted as though it were high energy incoming air and treated with much care.

David Leonard