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In a message dated 10/21/2002 7:07:07 AM Central Daylight Time, n268bl@charter.net writes:
Subj: [FlyRotary] Re: Electric Water Pumps
Date: 10/21/2002 7:07:07 AM Central Daylight Time
From: n268bl@charter.net (Bill Eslick)
Sender: flyrotary@lancaironline.net
Reply-to: <A HREF="mailto:flyrotary@lancaironline.net">
flyrotary@lancaironline.net</A>
To: flyrotary@lancaironline.net
..............................So to remove the waste heat from the engine,
operating at 180 horsepower,
there needs to be a 47 Degree temperature rise across the engine and a 47
Degree temperature drop across the radiator.
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This may not be the case in actual practice. I estimate my engine at about
140 HP at this point, and running this thru the equations gives me about a
36 degree drop required. (Did I do that correctly?)
I can climb out on 100+ degree days and the temp drop across my (GM) cores
is only 10-15 degrees. Cooling is stable at that point at around 190
degrees. In my first 100 hours at many temps and altitudes, the drop is
always around 10 to 15 degrees.
I am using the CHT probes of my EIS clamped to the exit tubes of the cores
to get these temps. They are wrapped with firesleeve to ward off other
sources of heat.
Bill Eslick
>> Homepage: http://www.flyrotary.com/
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Your measurements of the delta T indicate that you have much greater flow rate than 80 liters per minute. A higher flow rate means that the delta T will be less. Your arithmetic is correct, at 80 liter/min flow, you should expect 36 degree delta T. Since you are getting 15 degree delta T, your flow rate is about 36*80/15 = 192 liter/minute, or 50 gallons/minute (which is about what Powersport found they needed).
Everyone please bear in mind that I have nothing against Electric Water Pumps, but I firmly believe in things like thermodynamics, heat capacity, and pump pressure curves. Too many engineering courses I guess -- I have only used the data published by Davies-Craig and Mazda in my calculations.
Bill Schertz
KIS Cruiser #4045
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