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Ed,
It's a lot more efficient to publish the mistake. A lot more people
will figure out you made a mistake than will figure out the problem in
the first place. It saves you time, and it gets the rest of us
thinking. :)
Bob White
On Sun, 15 Aug 2004 23:22:42 -0400
"Ed Anderson" <eanderson@carolina.rr.com> wrote:
> 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
>
>
--
http://www.bob-white.com
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