Dennis, I highly recommend an article on Liquid cooling that was
published in this year's  January issue of Sport Aviation. Dr. Neal
does an excellent job of explaining it and I have exchanged e mails
with him on a few points.  He makes reference to a spreadsheet he
created that will provide the coordinates for the StreamLine Ducts
published by K&W.  This duct if done perfect produces an 84% pressure
recovery - about the best around.  The trouble is like all ducts, it
wants a foot or two of inlet.  However, I concluded (and Dr Neal
agreed) that if you truncate the duct from the inlet end you preserve
much of the goodness.  I estimated that my 6" duct length reduce
pressure recovery by approx 20% BUT that still gave me much better
cooling than a conventional duct would.

Using the Streamline (truncated in my case) duct I was able to reduce
my radiator intake area from a total of 48 sq inches to 33 sq inches
and I do not believe that is the limit.  The next time I fly I'll find
out as I have now reduced my intake area (for both GM cores) to a
total of 28 sq inches which is slightly over 1/2 of what I have flown
most of my hours with.  We will see within a month if all goes well.

I'll keep doing the math until someone who knows (easy to find) more
AND wants (harder to find) to hang it out.  Thanks for the comments
and for pointing out the error.  I think what happened was I was
thinking of the cp of water 1.0 and of air 0.25 and subconsciously
noted the  0.75 difference and therefore that became my 1.75 product
instead of the 4 it should have been..  I should really proof it a bit
better, but I don't think I would have caught it without you pointing
it out.

If you REALLY want to dig into cooling try Aerodynamics of Propulsion
by Kuchman and Weber (K&W) Aerodynamics of propulsion       . By: Dietrich Küchemann;  Johanna Weber       . Publisher: New York, McGraw-Hill, 1953.  This is the one I would recommend.  Unfortunately its very difficult
to find a copy - I had to get a photo copy of one.

ED

Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
  ----- Original Message -----   From: Dennis Haverlah   To: Rotary motors in aircraft   Sent: Sunday, August 15, 2004 10:35 PM
  Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re:
  DeltaT Coolant was : [FlyRotary] Re: coolant temps


  Ed,

  Please- Please - Don't quit doing the math!!!  I am an engineer but
  its been many years - so I really appreciate your working the
  problems!!.  I print and save all of them.  I will be doing my
  engine this winter and am itching to get going on it.  Right now I'm
  half way through the wing construction.  This last math "problem"
  had lots of good info.  I hope to expand on it to figure out cooling
  air inlet requirements for high power climb configurations.  Also, I
  want to look at water velocities in the hoses.  I am not sure 3/4
  inch hose is big enough for cooling a Renesis engine at full
  throttle.  Any comments??

  Also - could you give me a good reference for inlet duct design?

  Thanks in advance -

  Dennis H.

  Ed Anderson wrote:


    Dennis, you are absolutely correct about my error.  In fact, you
    pointing it out had me looking for any other errors and I am
    embarrassed to say I found another one - but not one of a math
    nature - it turns out both errors sort of offset each other so the
    answer 47 mph for air velocity I got with the errors is not much
    different than the 42 mph after doing it correctly {:>)

    Thanks for pointing out my error and getting me to examine the
    work again.

    I think Rusty is right - spending too much time on  math.  You
    will beat Rusty, I won't be able to get started until next week
    end and by that time you 'll be finished {:<{,

    Oh, the other error?

    When I did my calculations for the air with my head up and locked,
     I used the same temperature increase for the air that Tracy saw       decrease in his coolant.  Well, of course  DUH!. the temperature
     of the air will increase considerably more than that for the same
     BTU absorbed.  In fact for the GM cores the typical Delta T
     measured and reported  for the increase in air temps range from
     20-30F.  In fact if you look at the static situation it only
     takes 0.25 BTU to raise the temperature of a lbm of air 1 degree
     F.  So you could expect the temperature of a lbm of air to be 4
     times higher than that of water in the static situation (for the
     same BTU).      Unfortunately its not quite that simple with flowing air.      Although if we continued to slow the air flow through the core
     the temperature of the air would continue to increase - but like
     the old boys and the radiator you would reach a point where the
     air temps would be high due to the slowing flow - but the mass
     flow rate would decrease to the point that less and less Heat was
     being removed.  So again a balance is needed.

    So round 2,  taking the 240 lbm of coolant that conveyed 2400 BTU
    with a temp drop of 10F.  We find that for a more realistic
    increase in air temps of say 25F.  we have air mass flow  W =
    2400/(25*.25) (note I am using  the Cp 0.25 for air early in the
    problem) = 384 lbm/minute of air to remove the 2400 BTU.
          This requires 384/0.076 = 5052 CFM of air at sea level or 5052/60
    = 84 cubic feet/sec.  Taking our 1.32 sq ft of GM core surface we
    find the air velocity required is 84/ 1.32 = 63 ft/sec or 42 mph
    for our evaporator cores.   The approach with the error gave 47
    MPH which seemed reasonable to me so I didn't even consider any
    errors.

      Gotta be more careful.  Gotta stop this math, Gotta get started
      putting my aircraft back together, gotta go to bed
    .

    Again, thanks, Dennis

    Ed.




    So     Ed Anderson
    RV-6A N494BW Rotary Powered
    Matthews, NC
      ----- Original Message -----       From: Dennis Haverlah       To: Rotary motors in aircraft       Sent: Friday, August 13, 2004 11:37 PM
      Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re:
      DeltaT Coolant was : [FlyRotary] Re: coolant temps


      Ed:       Not to pick too much but I believe there is a problem with the
      math for the air cooling calculation.  Water has a specific heat
      of 1 BTU/lb-deg.F. and air is 0.25 BTU/lb-deg.F. or to say it
      another way air has 1/4 the heat capacity of the same mass of
      water.  Hence you need 4 times the mass of water to get the same
      heat content capacity in air. In you calculation for air, you
      multiplied the water mass by 1.75 to get the equivilant air mass
      required..  Shouldn't it be multiplied by 4.         Dennis H.

      Ed Anderson wrote:

        I'm sorry, Mark, I did not show that step.  You are correct
        the weight (mass) of water(or any other cooling medium) is an
        important factor as is its specific heat.

         In the example you used  - where we have a static 2 gallons
         capacity of water, It would actually only take 8*2 = 16 lbms
         *10 = 160 BTU to raise the temp of the water 1 degree F.  The
         difference is in one case we are talking about raising the
         temperature of a fixed static amount of water which can not
         readily get rid of the heat, in the other (our radiator
         engine case) we are talking about how much heat the coolant
         can transfer from engine to radiator. Here the flow rate is
         the key factor.          But lets take your typical 2 gallon cooling system capacity
        and see what we can determine.

        If we take our 2 gallons and start moving it from engine to
        radiator and back we find that each times the 2 gallons
        circulates it transfers 160 BTU (in our specific example!!).
        So at our flow rate of 30 gpm we find it will move that 160
        BTU 15 times/minute (at 30 gallons/minute the 2 gallons would
        be transferred 15 times).  So taking our 160 BTU that it took
        to raise the temp of our 2 gallons of static water 10F that we
        now have being moved from engine to radiator 15 times a minute
        = 160*15 = 2400 BTU/Min. Amazing isn't it?   So no magic, just
        math {:>).  So that is how our 2 gallons of water can transfer
        2400 BTU/min from engine to radiator.  It also shows why the
        old wives tale about "slow water" cooling better is just that
        (another story about how that got started)


        In the  equation Q = W*deltaT*cp that specifies how much heat
        is transferred ,we are not talking about capacity such as 2
        gallons capacity of a cooling system but instead are talking
        about mass flow.  As long as we reach that flow rate  1 gallon
        at 30 gpm or 1/ gallon at 60 gpm or 1/4 gallon at 120 gpm all
        will remove the same amount of heat.  However if you keep
        increasing the flow rate and reducing the volume you can run
        into other problems - like simply not enough water to keep
        your coolant galleys filled {:>), so there are limits.

        Our  2 gallon capacity is, of course, simply recirculated at
        the rate of 30 gpm through our engine (picking up heat- approx
        2400 BTU/min in this specific example) and then through our
        radiator (giving up heat of 2400 BTU/Min  to the air flow
        through the radiators) assuming everything works as planned.         IF  the coolant does not give up as much heat in the radiators
        (to the air stream) as it picks up in the engine then you will
        eventually (actually quite quickly) over heat your engine.

        The 240 lb figure I used in the previous example comes from
        using 8 lb/gal (a common approximation, but not precise as you
        point out) to calculate the mass flow.

        The mass flow = mass of the medium (8 lbs/gallon for water) *
        Flow rate(30 gpm) =240 lbs/min mass flow. Looking at the units
        we have(8 lbs/gallon)*(30 Gallon/minute) canceling out the
        like units (gallons) leaves us with 240 lb/minute which is our
        mass flow in this case.

        Then using the definition of the BTU we have 240 lbs of water
        that must be raised 10F.  Using our heat transfer equation         Q = W*deltaT*cp, we have Q = 240*10*1 = 2400 BTU/minute is
        required to increase the temperature of this mass flow by 10F

        Using the more accurate weight of water we would have  8.34*30
        =  250.2 lbm/minute  so the actual BTU required is closer to
        2502 BTU/min instead of my original 2400 BTU/Min, so there is
        apporx a 4% error in using 8 lbs/gallon.  If we could ever get
        accurate enough where this 4% was an appreciable part of the
        total errors in doing our back of the envelope thermodynamics
        then it would pay to use 8.34 vice 8, but I don't think we are
        there, yet {:>).

        Now the same basic equation applies to the amount of heat that
        the air transfers away from out radiators.  But here the mass
        of air is much lower than the mass of water so therefore it
        takes a much higher flow rate to equal the same mass flow.         What makes it even worse is that the specific heat of air is
        only 0.25 compared to water's 1.0.  So a lb of air will only
        carry approx 25% the heat of a lb of water, so again for this
        reason you need more air flow.          if 30 gpm of water will transfer 2400 bTu of engine heat
        (using Tracy's fuel burn of 7 gallon/hour), how much air does
        it require to remove that heat from the radiators?

        Well  again we turn to our equation and with a little algebra
        we have W = Q/(DeltaT*Cp) = 2400/(10*1) = 240 lbm/min. Not a
        surprise as that is what we started with.         But now taking the 240 lbm/min mass flow and translating that
        into Cubic feet/minute of air flow.  We know that a cubic foot
        of air at sea level weighs approx 0.076 lbs.  So 240
        lbm/(0.076 lbm/Cubic foot) = 3157 cubic feet/min to equal  the
        same mass as the coolant. But since the specific heat of air
        is lower (0.25) that water, we actually need 75% more air mass
        or 1.75 * 3157 = 5524.75 CFM air flow at sea level. Now I know
        this sounds like a tremendous amount of air but stay with me
        through the next step.

        Taking two GM evaporator cores with a total frontal area of
        2*95 = 190 sq inches and turning that in to square feet = 1.32
        sq ft we take our 5524.75 cubic feet minute and divide by 1.32
        sq ft = 4185 ft/min for the required air velocity to move that
        much air volume through our two evaporator cores.  To get the
        air velocity in ft/sec divide 4185/60 = 69.75 ft/sec airflow
        velocity through our radiators  or 47.56 Mph.  Now that sounds
        more reasonable doesn't it??          Now all of this is simply a first order estimate.  There are
        lots of factors such as the density of the air which unlike
        water changes with altitude, the temperature of the air, etc.
        that can change the numbers a bit.  But, then there is really
        not much point in trying to be more accurate given the
        limitations of our experimentation accuracy {:>).


        Also do not confuse the BTUs required to raise the temperature
        of 1 lb of water 1 degree F with that required to turn water
        in to vapor - that requires orders of magnitude more BTU.          Hope this helped clarify the matter.

        Ed


        Ed Anderson
        RV-6A N494BW Rotary Powered
        Matthews, NC
          ----- Original Message -----           From: Mark Steitle           To: Rotary motors in aircraft           Sent: Friday, August 13, 2004 8:32 AM
          Subject: [FlyRotary] Re: DeltaT Coolant was : [FlyRotary]
          Re: coolant temps


          Ed,
          Please humor me (a non-engineer) while I ask a dumb
          question.  If it takes 1BTU to raise 1lb of water 1 degree,
          and you factor in 30 gpm flow to come up with a 2400 BTU
          requirement for a 10 degree rise for 1 lb of water, where
          does the number of pounds of water figure into the equation,
          or do we just ignore that issue?  Water is 8.34 lbs/gal, and
          say you have 2 gallons of coolant, that would be 16.68 lbs.           Seems that we would need to multiply the 2400 figure by
          16.68 to arrive at a total system requirement of 40,032
          BTU/min.  What am I missing here?

          Mark S.


               At 09:58 PM 8/12/2004 -0400, you wrote:

            Right you are, Dave
                         Below  is one semi-official definition of BTU in English
            units.  1 BTU is amount of heat to raise 1 lb of water 1
            degree Fahrenheit.                            So with Tracy's 30 gpm flow of water = 240 lbs/min.  Since
            its temperature is raised 10 degree F we have
                         BTU = 240 * 10 * 1 = 2400 BTU/min
                         I know I'm ancient and  I should move into the new metric
            world, but at least I didn't do it in Stones and Furlongs
            {:>)
                         Ed
                         The Columbia Encyclopedia, Sixth Edition.  2001.
                         British thermal unit
                                      abbr. Btu, unit for measuring heat quantity in the
            customary system of English units of measurement, equal to
            the amount of heat required to raise the temperature of
            one pound of water at its maximum density [which occurs at
            a temperature of 39.1 degrees Fahrenheit (°F) ] by 1°F.
            The Btu may also be defined for the temperature difference
            between 59°F and 60°F. One Btu is approximately equivalent
            to the following: 251.9 calories; 778.26 foot-pounds; 1055
            joules; 107.5 kilogram-meters; 0.0002928 kilowatt-hours. A
            pound (0.454 kilogram) of good coal when burned should
            yield 14,000 to 15,000 Btu; a pound of gasoline or other .
                                                                                                       Ed Anderson
            RV-6A N494BW Rotary Powered
            Matthews, NC               ----- Original Message -----               From: DaveLeonard               To: Rotary motors in aircraft               Sent: Thursday, August 12, 2004 8:12 PM               Subject: [FlyRotary] Re: DeltaT Coolant was :
              [FlyRotary] Re: coolant temps


              Ed, are those units right.  I know that the specific
              heat of water is 1.0 cal/(deg Celsius*gram).  Does that
              also work out to 1.0 BTU/(deg. Farhengight * Lb.) ? Dave
              Leonard Tracy my calculations shows your coolant temp
              drop is where it should be: My calculations show that at
              7 gph fuel burn you need to get rid of 2369 BTU/Min
              through your coolant/radiators.  I rounded it off to
              2400 BTU/min. Q = W*DeltaT*Cp  Basic Heat/Mass Flow
              equation  With water as the mass with a weight of 8 lbs/
              gallon and a specific heat of 1.0 Q = BTU/min of heat
              removed by coolant mass flow                Assuming 30 GPM coolant flow = 30*8 = 240 lb/min mass
               flow. specific heat of water  Cp = 1.0 Solving for
               DeltaT = Q/(W*Cp) = 2400/(240*1)  =  2400/240 = 10 or                your delta T for the parameters specified should be
               around 10F               Assuming a 50/50 coolant mix with a Cp  of 0.7 you would
              have approx 2400/(240 *0.7) = 2400/168 = 14.2F so I
              would say you do not fly with                a 50/50 coolant mix but something closer to pure water.
                But in any case, certainly in the ball park.               You reported 10-12F under those conditions, so I would
              say condition is 4. Normal operation Ed               Ed Anderson               RV-6A N494BW Rotary Powered               Matthews, NC