Return-Path: Received: from ns5.rokland.us ([67.15.10.31] verified) by logan.com (CommuniGate Pro SMTP 4.2) with ESMTP-TLS id 366361 for flyrotary@lancaironline.net; Sun, 15 Aug 2004 23:57:37 -0400 Received-SPF: none receiver=logan.com; client-ip=67.15.10.31; envelope-from=bob@bob-white.com Received: from bgp01386623bgs.brodwy01.nm.comcast.net ([68.35.161.221] helo=localhost.localdomain) by ns5.rokland.us with smtp (Exim 4.34) id 1BwYcK-0008Ni-Dz for flyrotary@lancaironline.net; Sun, 15 Aug 2004 22:57:05 -0500 Date: Sun, 15 Aug 2004 21:57:02 -0600 From: Bob White To: "Rotary motors in aircraft" Subject: Re: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps Message-Id: <20040815215702.0ffd5ec2.bob@bob-white.com> In-Reply-To: References: X-Mailer: Sylpheed version 0.9.12 (GTK+ 1.2.10; i686-pc-linux-gnu) Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable X-AntiAbuse: This header was added to track abuse, please include it with any abuse report X-AntiAbuse: Primary Hostname - ns5.rokland.us X-AntiAbuse: Original Domain - lancaironline.net X-AntiAbuse: Originator/Caller UID/GID - [47 12] / [47 12] X-AntiAbuse: Sender Address Domain - bob-white.com X-Source: X-Source-Args: X-Source-Dir: 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" 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. >=20 > 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. >=20 > 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. >=20 > If you REALLY want to dig into cooling try Aerodynamics of Propulsion > by Kuchman and Weber (K&W) Aerodynamics of propulsion=20 > . By: Dietrich K=FCchemann; Johanna Weber=20 > . Publisher: New York, McGraw-Hill, 1953. =20 > This is the one I would recommend. Unfortunately its very difficult > to find a copy - I had to get a photo copy of one. >=20 > ED >=20 > Ed Anderson > RV-6A N494BW Rotary Powered > Matthews, NC > ----- Original Message -----=20 > From: Dennis Haverlah=20 > To: Rotary motors in aircraft=20 > 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 >=20 >=20 > Ed, >=20 > 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?? >=20 > Also - could you give me a good reference for inlet duct design? >=20 > Thanks in advance - >=20 > Dennis H. >=20 > Ed Anderson wrote: >=20 >=20 > 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 {:>) >=20 > Thanks for pointing out my error and getting me to examine the > work again. >=20 > 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 {:<{, >=20 > Oh, the other error? >=20 > 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 =20 > 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).=20 >=20 > Unfortunately its not quite that simple with flowing air.=20 > 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. >=20 > 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 =3D > 2400/(25*.25) (note I am using the Cp 0.25 for air early in the > problem) =3D 384 lbm/minute of air to remove the 2400 BTU. > =20 > This requires 384/0.076 =3D 5052 CFM of air at sea level or 5052/60 > =3D 84 cubic feet/sec. Taking our 1.32 sq ft of GM core surface we > find the air velocity required is 84/ 1.32 =3D 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. >=20 > Gotta be more careful. Gotta stop this math, Gotta get started > putting my aircraft back together, gotta go to bed > . >=20 > Again, thanks, Dennis >=20 > Ed. >=20 >=20 >=20 >=20 > So=20 > Ed Anderson > RV-6A N494BW Rotary Powered > Matthews, NC > ----- Original Message -----=20 > From: Dennis Haverlah=20 > To: Rotary motors in aircraft=20 > 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 >=20 >=20 > Ed:=20 >=20 > 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. =20 >=20 > Dennis H. >=20 > Ed Anderson wrote: >=20 > 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. >=20 > In the example you used - where we have a static 2 gallons > capacity of water, It would actually only take 8*2 =3D 16 lbms > *10 =3D 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. =20 >=20 > But lets take your typical 2 gallon cooling system capacity > and see what we can determine. >=20 > 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 > =3D 160*15 =3D 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) >=20 >=20 > In the equation Q =3D 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. >=20 > 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.=20 > 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. >=20 > 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. >=20 > The mass flow =3D mass of the medium (8 lbs/gallon for water) * > Flow rate(30 gpm) =3D240 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. >=20 > Then using the definition of the BTU we have 240 lbs of water > that must be raised 10F. Using our heat transfer equation=20 >=20 > Q =3D W*deltaT*cp, we have Q =3D 240*10*1 =3D 2400 BTU/minute is > required to increase the temperature of this mass flow by 10F >=20 > Using the more accurate weight of water we would have 8.34*30 > =3D 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 {:>). >=20 > 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.=20 > 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. =20 >=20 > 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? >=20 > Well again we turn to our equation and with a little algebra > we have W =3D Q/(DeltaT*Cp) =3D 2400/(10*1) =3D 240 lbm/min. Not a > surprise as that is what we started with.=20 >=20 > 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) =3D 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 =3D 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. >=20 > Taking two GM evaporator cores with a total frontal area of > 2*95 =3D 190 sq inches and turning that in to square feet =3D 1.32 > sq ft we take our 5524.75 cubic feet minute and divide by 1.32 > sq ft =3D 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 =3D 69.75 ft/sec airflow > velocity through our radiators or 47.56 Mph. Now that sounds > more reasonable doesn't it?? =20 >=20 > 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 {:>). >=20 >=20 > 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. =20 >=20 > Hope this helped clarify the matter. >=20 > Ed >=20 >=20 > Ed Anderson > RV-6A N494BW Rotary Powered > Matthews, NC > ----- Original Message -----=20 > From: Mark Steitle=20 > To: Rotary motors in aircraft=20 > Sent: Friday, August 13, 2004 8:32 AM > Subject: [FlyRotary] Re: DeltaT Coolant was : [FlyRotary] > Re: coolant temps >=20 >=20 > 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.=20 > 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? >=20 > Mark S. >=20 >=20 > At 09:58 PM 8/12/2004 -0400, you wrote: >=20 > Right you are, Dave > =20 > 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. =20 > =20 > So with Tracy's 30 gpm flow of water =3D 240 lbs/min. Since > its temperature is raised 10 degree F we have > =20 > BTU =3D 240 * 10 * 1 =3D 2400 BTU/min > =20 > 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 > {:>) > =20 > Ed > =20 > The Columbia Encyclopedia, Sixth Edition. 2001. > =20 > British thermal unit > =20 > =20 > 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 (=B0F) ] by 1=B0F. > The Btu may also be defined for the temperature difference > between 59=B0F and 60=B0F. 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 . > =20 > =20 > =20 > =20 > =20 > =20 > =20 > Ed Anderson > RV-6A N494BW Rotary Powered > Matthews, NC=20 > ----- Original Message -----=20 > From: DaveLeonard=20 > To: Rotary motors in aircraft=20 > Sent: Thursday, August 12, 2004 8:12 PM=20 > Subject: [FlyRotary] Re: DeltaT Coolant was : > [FlyRotary] Re: coolant temps >=20 >=20 > 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 =3D 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 =3D BTU/min of heat > removed by coolant mass flow=20 > Assuming 30 GPM coolant flow =3D 30*8 =3D 240 lb/min mass > flow. specific heat of water Cp =3D 1.0 Solving for > DeltaT =3D Q/(W*Cp) =3D 2400/(240*1) =3D 2400/240 =3D 10= or=20 > your delta T for the parameters specified should be > around 10F=20 > Assuming a 50/50 coolant mix with a Cp of 0.7 you would > have approx 2400/(240 *0.7) =3D 2400/168 =3D 14.2F so I > would say you do not fly with=20 > a 50/50 coolant mix but something closer to pure water. > But in any case, certainly in the ball park.=20 > You reported 10-12F under those conditions, so I would > say condition is 4. Normal operation Ed=20 > Ed Anderson=20 > RV-6A N494BW Rotary Powered=20 > Matthews, NC=20 >=20 >=20 --=20 http://www.bob-white.com N93BD - Rotary Powered BD-4 (soon)