Sender: <marv@lancaironline.net> (Marvin Kaye)
To: flyrotary
Date: Thu, 27 Mar 2003 02:52:05 -0500
Message-ID: <redirect-2088105@logan.com>
From: <daveleonard@cox.net>
Subject: Re: [FlyRotary] Re: Thick vs Thin
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> 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 coo=
l the
> engine is lost in either thick or thin radiator.  There is no significa=
nt
> energy to recover.  This is easily verified by measuring the velocity o=
f the
> air coming out of the rad.  Typical values are on the order of 20 mph o=
r
> 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 obvio=
us
> 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) =3D 40,000.
> Now do it for 20 mph
> (2/2) * (20 * 20) =3D 400
> =

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

> Tracy

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


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


One radiator is 10=94x10=94 and 6 inches deep.  The other is 24.5=94 x 24=
=2E5=94 and 1 inch deep. Both have the same volume.  Through proper sizin=
g of the inlets, I am now going to allow 1800 cu in of air to flow throug=
h each rad every second.  This is 3 times the rad volume, same for both r=
ads.  Since the volume of air within each rad is the same, each will be r=
eplacing the entire volume within each rad 3 times each second.  This mea=
ns 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 wi=
th radiator surface (0.33 sec), and each cubic inch of rad is seeing the =
same volume of air (3 cubic inches/sec) =96 regardless of radiator.  Aver=
age =93pre-heating=94 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 ef=
fectiveness.  In both cases we are using the same amount of high energy i=
ncoming air.

Now, the ONLY difference is the speed of the air.  Since each packet of a=
ir must get out of the rad in .33 seconds, the thick rad is flowing at 18=
=94/sec while the thin rad is flowing at 3=94/sec.  That is a 6 fold diff=
erence in speed through the rad.  Since turbulent drag is a function of s=
peed cubed that is a 216 fold increase in drag through the radiator!   Dr=
ag is energy.  If it takes 1/4 horse power to push the air through the th=
in 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 somewh=
ere is the forward motion of the aircraft.  How is this extra drag manife=
st on the aircraft?  The thin rad is showing a smaller pressure drop acro=
ss its surfaces, leaving more pressure available to accelerate the outgoi=
ng air.

You may argue as above that this exit speed is insignificant.  But your e=
xample 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 f=
ar superior because the velocities can be made much smaller (making it al=
l the more important for us to have efficient exit ducts).  And again I s=
ubmit that a thin rad maximizes this advantage.  =


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

David Leonard