Return-Path: Received: from [24.25.9.102] (HELO ms-smtp-03-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.2) with ESMTP id 366310 for flyrotary@lancaironline.net; Sun, 15 Aug 2004 23:27:51 -0400 Received-SPF: none receiver=logan.com; client-ip=24.25.9.102; envelope-from=eanderson@carolina.rr.com Received: from EDWARD (cpe-069-132-183-211.carolina.rr.com [69.132.183.211]) by ms-smtp-03-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id i7G3RHiB018662 for ; Sun, 15 Aug 2004 23:27:18 -0400 (EDT) Message-ID: <003e01c48340$f3c64760$2402a8c0@EDWARD> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Cooling Article and Spreadsheet Date: Sun, 15 Aug 2004 23:27:26 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_003B_01C4831F.6C76B400" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1409 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2800.1409 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_003B_01C4831F.6C76B400 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Here is the URL for Dr. Neal Willfords cooling spreadsheet - really a = good thing to play with. http://www.eaa.org/benefits/sportaviation/liquidcooling5.xls Really need to read his well written article as well Ed 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 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).=20 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 =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. 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=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 Ed:=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 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 =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 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 =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) 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. 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 =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. Then using the definition of the BTU we have 240 lbs of water = that must be raised 10F. Using our heat transfer equation=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 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 {:>). 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. =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? 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 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. 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=20 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 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. =20 Hope this helped clarify the matter. Ed 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 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 =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 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.) ?=20 Dave Leonard=20 Tracy my calculations shows your coolant temp drop is = where it should be:=20 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.=20 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=20 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=20 Solving for DeltaT =3D Q/(W*Cp) =3D 2400/(240*1) =3D = 2400/240 =3D 10 or 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=20 Ed=20 Ed Anderson=20 RV-6A N494BW Rotary Powered=20 Matthews, NC=20 ------=_NextPart_000_003B_01C4831F.6C76B400 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Here is the URL for Dr. Neal Willfords = cooling=20 spreadsheet - really a good thing to play with.
 
htt= p://www.eaa.org/benefits/sportaviation/liquidcooling5.xls
 
Really need to read his well written = article as=20 well
 
 
Ed
 
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
From:=20 Dennis Haverlah
Sent: Sunday, August 15, 2004 = 10:35=20 PM
Subject: [FlyRotary] Re: Answer = to when=20 is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant=20 temps

Ed,

Please- Please - Don't quit doing the = math!!! =20 I am an engineer but its been many years - so I really appreciate your = working=20 the problems!!.  I print and save all of them.  I will be = doing my=20 engine this winter and am itching to get going on it.  Right now = I'm half=20 way through the wing construction.  This last math "problem" had = lots of=20 good info.  I hope to expand on it to figure out cooling air = inlet=20 requirements for high power climb configurations.  Also, I want = to look=20 at water velocities in the hoses.  I am not sure 3/4 inch hose is = big=20 enough for cooling a Renesis engine at full throttle.  Any=20 comments??

Also - could you give me a good reference for inlet = duct=20 design?

Thanks in advance -

Dennis H.

Ed Anderson = wrote:
 
Dennis, you are absolutely correct = about my=20 error.  In fact, you pointing it out had me looking for any = other=20 errors and I am embarrassed to say I found another one - but not one = of a=20 math nature - it turns out both errors sort of offset each other so = the=20 answer 47 mph for air velocity I got with the errors is not much = different=20 than the 42 mph after doing it correctly {:>)
 
Thanks for pointing out my error = and getting me=20 to examine the work again.
 
I think Rusty is right - spending = too much time=20 on  math.  You will beat Rusty, I won't be able to get = started=20 until next week end and by that time you 'll be finished=20 {:<{,
 
Oh, the other error?
 
When I did my calculations for the = air with my=20 head up and locked,  I used the same temperature increase for = the air=20 that Tracy saw   decrease in his coolant.  Well, of=20 course  DUH!. the temperature of the air will increase = considerably=20 more than that for the same BTU absorbed.  In fact for the GM = cores the=20 typical Delta T measured and reported  for the increase in air = temps=20 range from 20-30F.  In fact if you look at the static situation = it only=20 takes 0.25 BTU to raise the temperature of a lbm of air 1 degree = F.  So=20 you could expect the temperature of a lbm of air to be 4 times = higher than=20 that of water in the static situation (for the same = BTU). 
 
 Unfortunately its not quite = that simple=20 with flowing air.  Although if we continued to slow the air = flow=20 through the core the temperature of the air would continue to = increase - but=20 like the old boys and the radiator you would reach a point where the = air=20 temps would be high due to the slowing flow - but the mass flow rate = would=20 decrease to the point that less and less Heat was being = removed.  So=20 again a balance is needed.
 
So round 2,  taking the 240 = lbm of coolant=20 that conveyed 2400 BTU with a temp drop of 10F.  We find that = for a=20 more realistic increase in air temps of say 25F.  we have air = mass flow=20  W =3D 2400/(25*.25) (note I am using  the Cp 0.25 for air = early in=20 the problem) =3D 384 lbm/minute of air to remove the 2400 = BTU.
 
This requires 384/0.076 =3D 5052 = CFM of air at=20 sea level or 5052/60 =3D 84 cubic feet/sec.  Taking our 1.32 sq = ft of GM=20 core surface we find the air velocity required is 84/ 1.32 =3D 63 = ft/sec or 42=20 mph for our evaporator cores.   The approach with the = error gave=20 47 MPH which seemed reasonable to me so I didn't even consider any=20 errors.
 
  Gotta be more careful.  = Gotta stop=20 this math, Gotta get started putting my aircraft back together, = gotta go to=20 bed
.
 
Again, thanks, Dennis
 
Ed.
 
 
 
 
So
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, = NC
-----=20 Original Message ----- From:=20 Dennis Haverlah = To:=20 Rotary motors in = aircraft=20 Sent:=20 Friday, August 13, 2004 11:37 PM Subject:=20 [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT = Coolant was=20 : [FlyRotary] Re: coolant temps

Ed:

Not to pick too much but I believe = there is a=20 problem with the math for the air cooling calculation.  Water = has a=20 specific heat of 1 BTU/lb-deg.F. and air is 0.25 BTU/lb-deg.F. or = to say=20 it another way air has 1/4 the heat capacity of the same mass of=20 water.  Hence you need 4 times the mass of water to get the = same heat=20 content capacity in air.
In you calculation for air, you = multiplied the=20 water mass by 1.75 to get the equivilant air mass required..  = Shouldn't it be multiplied by 4.  

Dennis = H.

Ed=20 Anderson wrote:
I'm sorry, Mark, I did not show = that=20 step.  You are correct the weight (mass) of water(or any = other=20 cooling medium) is an important factor as is its specific=20 heat.
 
 In the example you = used  - where=20 we have a static 2 gallons capacity of water, It would actually = only=20 take 8*2 =3D 16 lbms *10 =3D 160 BTU to raise the temp of the = water 1 degree=20 F.  The difference is in one case we are talking about = raising the=20 temperature of a fixed static amount of water which can not = readily get=20 rid of the heat, in the other (our radiator engine case) we are = talking=20 about how much heat the coolant can transfer from engine to = radiator.=20 Here the flow rate is the key factor. 
 
But lets take your typical 2 = gallon cooling=20 system capacity and see what we can determine.
 
If we take our 2 gallons and = start moving=20 it from engine to radiator and back we find that each times the = 2=20 gallons circulates it transfers 160 BTU (in our specific = example!!). So=20 at our flow rate of 30 gpm we find it will move that 160 BTU 15=20 times/minute (at 30 gallons/minute the 2 gallons would be = transferred 15=20 times).  So taking our 160 BTU that it took to raise the = temp of=20 our 2 gallons of static water 10F that we now have being = moved from=20 engine to radiator 15 times a minute =3D 160*15 =3D 2400=20 BTU/Min. Amazing isn't it?   So no magic, just = math=20 {:>).  So that is how our 2 gallons of water can = transfer 2400=20 BTU/min from engine to radiator.  It also shows why the old = wives=20 tale about "slow water" cooling better is just that (another = story about=20 how that got started)
 
 
In the  equation Q =3D = W*deltaT*cp that=20 specifies how much heat is transferred ,we are = not=20 talking about capacity such as 2 gallons = capacity=20 of a cooling system but instead are talking about mass=20 flow.  As long as we reach that flow rate  1 = gallon=20 at 30 gpm or 1/ gallon at 60 gpm or 1/4 gallon at 120 gpm all = will=20 remove the same amount of heat.  However if you=20 keep increasing the flow rate and reducing the volume you = can run=20 into other problems - like simply not enough water to keep = your=20 coolant galleys filled {:>), so there are = limits.
 
Our  2 gallon = capacity is, of=20 course, simply recirculated at the rate of 30 gpm through our = engine=20 (picking up heat- approx 2400 BTU/min in this specific example) = and then=20 through our radiator (giving up heat of 2400 BTU/Min =  to the=20 air flow through the radiators) assuming everything works as=20 planned.  IF  the coolant does not give up as much = heat in the=20 radiators (to the air stream) as it picks up in the engine then = you will=20 eventually (actually quite quickly) over heat your = engine.
 
The 240 lb figure I used in the = previous=20 example comes from using 8 lb/gal (a common approximation, but = not=20 precise as you point out) to calculate the mass = flow.
 
The mass flow =3D mass of the = medium (8=20 lbs/gallon for water) * Flow rate(30 gpm) =3D240 lbs/min mass = flow.=20 Looking at the units we have
(8 lbs/gallon)*(30 = Gallon/minute) canceling=20 out the like units (gallons) leaves us with 240 lb/minute which = is our=20 mass flow in this case.
 
Then using the definition of = the BTU we=20 have 240 lbs of water that must be raised 10F.  Using our = heat=20 transfer equation
 
Q =3D W*deltaT*cp, we have Q = =3D 240*10*1 =3D=20 2400 BTU/minute is required to increase the temperature of this = mass=20 flow by 10F
 
Using the more accurate weight = of water we=20 would have  8.34*30 =3D  250.2 lbm/minute  so the = actual=20 BTU required is closer to 2502 BTU/min instead of my original = 2400=20 BTU/Min, so there is apporx a 4% error in using 8 = lbs/gallon.  If=20 we could ever get accurate enough where this 4% was an = appreciable part=20 of the total errors in doing our back of the envelope = thermodynamics=20 then it would pay to use 8.34 vice 8, but I don't think we are = there,=20 yet {:>).
 
Now the same basic equation = applies to the=20 amount of heat that the air transfers away from out = radiators.  But=20 here the mass of air is much lower than the mass of water so = therefore=20 it takes a much higher flow rate to equal the same mass = flow.  What=20 makes it even worse is that the specific heat of air is = only 0.25=20 compared to water's 1.0.  So a lb of air will only carry = approx 25%=20 the heat of a lb of water, so again for this reason you need = more air=20 flow. 
 
if 30 gpm of water will = transfer 2400 bTu=20 of engine heat (using Tracy's fuel burn of 7 gallon/hour), how = much air=20 does it require to remove that heat from the = radiators?
 
Well  again we turn to our = equation=20 and with a little algebra we have W =3D Q/(DeltaT*Cp) =3D = 2400/(10*1) =3D 240=20 lbm/min. Not a surprise as that is what we started with. =
 
But now taking the 240 lbm/min = mass flow=20 and translating that into Cubic feet/minute of air flow.  = We know=20 that a cubic foot of air at sea level weighs approx 0.076=20 lbs.  So 240 lbm/(0.076 lbm/Cubic foot) =3D 3157 cubic = feet/min to=20 equal  the same mass as the coolant. But since the specific = heat of=20 air is lower (0.25) that water, we actually need 75% more air = mass or=20 1.75 * 3157 =3D 5524.75 CFM air flow at sea level. Now I know = this sounds=20 like a tremendous amount of air but stay with me through the = next=20 step.
 
Taking two GM evaporator cores = with a total=20 frontal area of 2*95 =3D 190 sq inches and turning that in to = square feet=20 =3D 1.32 sq ft we take our
5524.75 cubic feet minute and = divide by=20 1.32 sq ft =3D 4185 ft/min for the required air velocity to move = that much=20 air volume through our two evaporator cores.  To get = the=20 air velocity in ft/sec divide 4185/60 =3D 69.75 ft/sec = airflow=20 velocity through our radiators  or 47.56 Mph.  Now = that sounds=20 more reasonable doesn't it?? 
 
Now all of this is simply a = first order=20 estimate.  There are lots of factors such as the density of = the air=20 which unlike water changes with altitude, the temperature of the = air,=20 etc. that can change the numbers a bit.  But, then there is = really=20 not much point in trying to be more accurate given the = limitations of=20 our experimentation accuracy {:>).
 
 
Also do not confuse the BTUs = required to=20 raise the temperature of 1 lb of water 1 degree F with that = required to=20 turn water in to vapor - that requires orders of magnitude more=20 BTU. 
 
Hope this helped clarify the=20 matter.
 
Ed
 
 
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, = NC
-----=20 Original Message ----- From:=20 Mark Steitle = To:=20 Rotary motors in=20 aircraft Sent:=20 Friday, August 13, 2004 8:32 AM Subject:=20 [FlyRotary] Re: DeltaT Coolant was : [FlyRotary] Re: coolant=20 temps

Ed,
Please humor me (a = non-engineer)=20 while I ask a dumb question.  If it takes 1BTU to raise = 1lb of=20 water 1 degree, and you factor in 30 gpm flow to come up with = a 2400=20 BTU requirement for a 10 degree rise for 1 lb of water, where = does the=20 number of pounds of water figure into the equation, or do we = just=20 ignore that issue?  Water is 8.34 lbs/gal, and say you = have 2=20 gallons of coolant, that would be 16.68 lbs.  Seems that = we would=20 need to multiply the 2400 figure by 16.68 to arrive at a total = system=20 requirement of 40,032 BTU/min.  What am I missing=20 here?

Mark S.


     At = 09:58 PM=20 8/12/2004 -0400, you wrote:
Right you are, Dave
 
Below  is one semi-official definition of BTU = in English=20 units.  1 BTU is amount of heat to raise 1 lb of water = 1 degree=20 Fahrenheit.  
 
So with Tracy's 30 gpm flow of water =3D 240 = lbs/min. =20 Since its temperature is raised 10 degree F we=20 have
 
BTU =3D = 240 * 10 * 1=20 =3D 2400 BTU/min
 
I know=20 I'm ancient and  I should move into the new metric = world, but=20 at least I didn't do it in Stones and Furlongs=20 {:>)
 
Ed
 
The Columbia Encyclopedia, = Sixth=20 Edition.  2001.
 
British=20 thermal unit
 
 
abbr.=20 Btu, unit for measuring heat quantity in the customary = system of English = units of=20 measurement, equal to the amount of heat required to = raise the=20 temperature of one pound of water at its maximum density = [which=20 occurs at a temperature of 39.1 degrees Fahrenheit (=B0F) ] = by 1=B0F.=20 The Btu may also be defined for the temperature difference = between=20 59=B0F and 60=B0F. One Btu is approximately equivalent to = the following:=20 251.9 calories; 778.26 foot-pounds; 1055 joules; 107.5=20 kilogram-meters; 0.0002928 kilowatt-hours. A pound (0.454 = kilogram)=20 of good coal when burned should yield 14,000 to 15,000 Btu; = a pound=20 of gasoline or other=20 = .
 
 
 
 
 
 
 Ed=20 Anderson
RV-6A N494BW Rotary Powered
Matthews, = NC=20
----- Original Message -----=20
From: DaveLeonard=20
To: Rotary motors=20 in aircraft=20
Sent: Thursday, August 12, 2004 8:12 PM=20
Subject: [FlyRotary] Re: DeltaT Coolant was : = [FlyRotary] Re:=20 coolant temps

Ed, are those units = right.  I=20 know that the specific heat of water is 1.0 cal/(deg=20 Celsius*gram).  Does that also work out to 1.0 = BTU/(deg.=20 Farhengight * Lb.) ?=20
Dave Leonard
Tracy my calculations = shows your=20 coolant temp drop is where it should be:=20
My calculations show that = at 7 gph=20 fuel burn you need to get rid of 2369 BTU/Min through your = coolant/radiators.  I rounded it off to 2400 = BTU/min.=20
Q =3D W*DeltaT*Cp  = Basic Heat/Mass=20 Flow equation  With water as the mass with a weight = of 8 lbs/=20 gallon and a specific heat of 1.0
Q =3D BTU/min of heat = removed by coolant=20 mass flow=20
 Assuming 30 GPM = coolant flow =3D=20 30*8 =3D 240 lb/min mass flow. specific heat of = water  Cp =3D=20 1.0=20
 Solving for DeltaT = =3D Q/(W*Cp) =3D=20 2400/(240*1)  =3D  2400/240 =3D 10 or  your = delta T for=20 the parameters specified should be around 10F=20
Assuming a 50/50 coolant = mix with a=20 Cp  of 0.7 you would have approx 2400/(240 *0.7) =3D = 2400/168 =3D=20 14.2F so I would say you do not fly with=20
 a 50/50 coolant mix = but=20 something closer to pure water.  But in any case, = certainly=20 in the ball park.=20
You reported 10-12F under = those=20 conditions, so I would say condition is 4. Normal = operation=20
Ed=20
Ed Anderson
RV-6A N494BW Rotary = Powered
Matthews, NC=20 =

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