X-Mozilla-Keys:                                                                                 

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:

  A: The cooling HP  needs of the engine is fixed at roughly 100% of
  the output HP and not a function of airspeed. Particularly
  with a constant speed prop. One HP is 2545 BTU's per hour
  or 745 watts. Normally aspirated output HP of course decreases
  with altitude.

Rodger Hilyard wrote:

     Taking the Mazda research  word, 20% or 5735 BTU's go
 into the water radiator, 2294 BTU's go into the oil radiator
 PER Minute.

      So we put 5735 BTU's in to the water.......
      Now we need to get it out of the engine.

      I have seen data that indicates that the cooling flow is on the
 order of 13 GPM.  Lots of things affect this but I have no data to
 give any other expected values.  More flow could be better to reduce the
temperature drop across the radiator.

      Specific heat is the ratio of the Btu's required to raise a lb of
 a material 1 deg , to that required to raise a lb of water 1 deg , at
 some specified temperature.

   The Btu is defined as the heat necessary to raise 1 lb of water
 1 deg Fahrenheit

    It is expected that a 50/50 mixture by volume of distilled water
 and Ethylene Glycol will be the transport medium.

   Water ( specific heat of 1.006 Btu per lb @ 200 F )
   (  8.33 lbs per gallon )

    Ethylene Glycol ( specific heat of 0.67 Btu per lb @ 212 F )
    (  9.16 lbs per gallon )

    So 1 gallon of E/G will absorb 0.67 Btu per lb x 9.16  lb per
 gallon per1 F temp rise. or   ( 6.17 Btu per gallon, per minute, per 1
deg F rise)


  And 1 gallon of Water will absorb 1 Btu x 8.33 lb per gallon per
 1 F temp rise. ( 8.33 Btu per  gallon, per minute, per 1 deg F )

  So we add the heat removed by each fluid and divide by our
 two gallon total.

    6.17 Btu per gallon E/G, Plus 8.33 Btu per gallon Water , Equals
 14.5 Btu per minute, per 1 F, per 2 gallons of mixture.
   Or 7.25 Btu  Per Minute ,per 1 deg F, per gallon.

   Now we take the heat removal ability of one gallon of our
 fluid mixture and multiply by the number of gallons per minute
 we are flowing ,to find  the heat removal of our total flow
 for each deg  F of fluid temp rise.

    13 GPM X 7.25 Btu per GPM per 1 deg F  Equals  94.25 Btu for
 every 1 deg F temp rise.

   Therefore: 5735 Btu heat input divided by 94.25 Btu per 1 deg F
Equals   60.85 deg F Temperature rise at engine block exit to
 radiator, at a flow of 13 gallons per Minute.

  If your thermostat opens at 185 deg F and is fully open  at 200 deg
 F, your water inlet temp. to the engine must be 61 deg F lower than
 the exit temp , or 139 deg F out of the radiator.

   Well we got the heat out of the engine and into the radiator and
 we know the heat content ( 5735 Btu's per minute ) the temperature
 drop the radiator must provide ( 61 deg F ), and
 the radiator outlet temp. ( 139 deg F ).
 Now we must couple this into the AIR.

  To be continued............

Sounds good so far but I wonder if anybody is getting only 139 degrees F
out
of the rad with 200 F in. Anybody have a water-out-of-the-rad temp gauge
as well as a water in temp gauge?
 
Paul Lamar
 
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