Sender: <marv@lancaironline.net> (Marvin Kaye)
To: flyrotary
Date: Thu, 27 Mar 2003 10:58:11 -0500
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From: "William" <wschertz@ispwest.com>
References: <list-2088107@logan.com>
Subject: Re: [FlyRotary] Re: Thick vs Thin
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----- Original Message -----
From: <daveleonard@cox.net>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Sent: Thursday, March 27, 2003 1:52 AM
Subject: [FlyRotary] Re: Thick vs Thin


> 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.
----------------------------------------------
I think I have to make a couple of comments here regarding the assumptions
above.

The rotary engine groups (both of them) have focused on the cubic inches of
radiator volume as the "defining parameter" for what works and what doesn't
in cooling the engine.  This is because it is the easiest to measure, not
because it is the best representation of the cooling phenomina.

Heat transfer from the water to the air has several resistances. They are:

Getting the heat from the bulk of the cooling fluid to the metal surface of
the radiator.  This is enhanced by having the fluid be turbulent, and having
a fluid with good heat capacity, high thermal conductivity, and low
viscosity. The amount of heat transferred Q = h*A*delT1 where h is the
convective heat transfer coefficient, A is the area wetted by the fluid, and
delT1 is the temperature difference across the water-to-metal film.

The next resistance is the metal of the radiator.  The key properties are a
high thermal conductivity, and the heat transfer is
Q = (k/x)*A*delT2 where k is the thermal conductivity and x is the thickness
of the metal. You want high k (copper or aluminum typically) and low x (thin
metal). delT2 is the temperature drop across the metal surface.

The final resistance is that from the metal tube surface to the air. This
resistance is typically the highest of all three, (although it may not be in
the case of the oil cooler, the liquid-side resistance is much greater than
for a water radiator).  The air side heat transfer is Q = ha*Aa*delT3, where
"ha" is the air side film coefficient, Aa is the area (may be different that
A in the previous two cases because of the addition of fins), and delT3 is
the difference between the metal surface and the air.

The "units" of 'h', 'ha', and '(k/x)' are Btu/hr*ft^2

Now, for a given area of tubing/fins, the value of Q (Btu/hr) must be the
same for all three areas, that is the heat given up by the liquid = the heat
transferred through the wall = the heat taken up by the air.  So,

h*A*delT1 = (k/x)*A*delT2 = ha*Aa*delT3 = Q.

Also, delT1 + delT2 + delT3 = the overall drop from the liquid to the air.
This comes into play several ways -- as the air moves through the radiator,
it is heating up and the liquid is cooling down (our desired outcome).  So
the temperature difference between the liquid and the air is decreasing, and
the Q for a given area decreases as the temperature of the air rises towards
the temperature of the water. So for any radiator (thick or thin) the
element of the radiator at the exit end is less effective than the element
at the entrance end. The difference is a matter of degree.

The heat being dumped to the air heats the air, and the ability of the air
to absorb the heat is given by Q = m*Cp*delT4, where m is the mass flow rate
(#/hr) of the air, Cp is the heat capacity (Btu/#degF), and delT4 is the
temperature rise of the air.  Now for a thin radiator, delT4 is kept small,
because the area of fin exposed to the air is less (shorter path), therefore
for the same heat release, the mass flow rate must be larger.  For a thick
radiator, the delT4 is larger (air is coming closer to temperature of the
water), so the mass of air required is less.

What this means is that there is some "optimum" thickness of radiator.  As
you go from very thick, where the mass of air would be the minimum, the
frontal area would be minimum, but the effectiveness of the tail end of the
radiator would be small, and the pressure drop to force the air through the
radiator would be large.  As you get thinner and thinner, the area
increases, the pressure drop decreases, the effectiveness of the metal
surface increases, but the mass flow rate of air required increases (and
must be distributed evenly over the surface).  In the extreme of a *very*
thin radiator, the area would be very large with attendant
distribution/space requirements, and the temperature rise of the air would
be very small.

I am not sure how to arrive at the optimum thickness, but in practical
terms, the "thick" radiators are about 3.5" and the "thin" radiators that
fit in a cowling are about 2" thick.

Bill Schertz