X-Mozilla-Keys:
Rodger Hilyard wrote:
Paul wrote:
I have been giving a lot of thought lately to the cooling
problem. So far we have been going on pure experiment only.
I thought I would run this by you for comments.
Preliminary givens:
C: Given a perfect duct or 100% dynamic pressure recovery,
the mass flow of air passing through a radiator is
proportional
to dynamic pressure which is proportional to the square of
speed and inversely proportional to the thickness
of the radiator.
D: The temperature increase of the cooling air exiting
the radiator is directly proportional to the square
root of the thickness of the radiator.
Jeff questioned this and it was a guess on my part.
It needs a little more thought or experiment to really
come up with the right function. Also there appears to
be a practical limit of about 3.5 inches. Probably
less than that if you climb at less than 100 MPH.
E: The first half of the radiator does 3/4 of the cooling.
-------------------------------------------------------------
So now we know :
Heat to get rid of.............................5735 Btu's
Radiator inlet temps........................200 deg F
Radiator outlet temp........................139 deg F
Minimum air flow @ 50 deg F
exit air rise...................................5770 CFM
Minimum Inlet duct size
at 80 MPH......................................118 sq. in.
Cooling Flow..................................... 13 GPM
Inlet Air temp......................................100 deg F
Next we will look at the radiator parameters that will
meet these conditions. Probably more fun than a fellow
can stand....
To be
continued........................................................
If I could make the perfect A/C radiator , I would make
it so that the energy lost flowing air through it would be
recovered by the energy, added to the air flow, from the
energy removed from the cooling fluid. Thus making the
radiator "transparent " to air flow.
Let us look at the heat transfer process of air flowing over
a surface, i.e. the core of your radiator.
Air is a good Insulator !!
That's right , in small cells where circulation is restrained,
Air has a Thermal Conductivity of 0.16 Btu per HOUR through
a volume of the dimensions of 1 inch thick by an area of
1 square foot.
As compared to
Water which is 4.3 Btu X inch / Hr X ft sq. X deg F,
Or Copper which is 2600 Btu X inch / Hr X ft sq. X deg F
So how on earth are we going to remove 5735 Btu's per
Minute........
One way to help is to cut that 1 Inch thickness down ,as
an example, if we reduce it to 0.1 inch thick, we would
increase the heat transfer by the ratio of 1.00 Inch divided
by 0.10 inch ,or 1.0 / 0.1 =10 ( 10 times the heat removal ! )
Lets do that again, but this time we'll squeeze it down
to 0.001 inch thick. 1.0 / 0.001 =1000 ! ( 1000 times )
At this point lets reduce that T/C Btu figure to units that we
have been using in this discussion .
0.16 Btu per inch per hour per square foot per deg F
Is 0.00001852 Btu per inch per minute per square inch
per deg F.
So that 0.001 inch thick layer will transfer 1000 times
as much, or 0.01852 Btu per minute per sq. in per deg F.
How can I get such a small thickness of air you ask ?
Do you remember back in Aeronautics 101 where you
were told that air flowing over the surface of your airfoil
came to a stop AT the surface of the airfoil ? Well that's
where this thin layer is.
It's the thickness of this insulative layer that is going to
dominate our heat transfer process.
Well what happens to the heat after it migrates through
our thin layer ?
Imagine that your that little air molecule easing your way
down to the surface of our radiator, your gona get hot and
jump right up off there because you picked up some energy,
made you spin up and you even moved out from that quiet
thin insulative layer right into the edge of that wind storm
flowing by. You guessed it , you and all your heat were
swept away to mix in and transfer some of your new found
heat to others of your chilly friends.
Well if the thickness of this layer is so important, How do
you know if it's not 0.002" or even 0.0005 " thick , a factor
of two in our determination of surface area we need.....
Well " I " Don't. But others do and they have made all
sorts of measurements and published them under the
term film coefficients.
To be continued....................................
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