Return-Path: Received: from [24.25.9.100] (HELO ms-smtp-01-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.2) with ESMTP id 364426 for flyrotary@lancaironline.net; Sat, 14 Aug 2004 01:51:08 -0400 Received-SPF: none receiver=logan.com; client-ip=24.25.9.100; envelope-from=eanderson@carolina.rr.com Received: from EDWARD (cpe-069-132-183-211.carolina.rr.com [69.132.183.211]) by ms-smtp-01-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id i7E5oYPg029744 for ; Sat, 14 Aug 2004 01:50:36 -0400 (EDT) Message-ID: <002701c481c2$a1f7b2e0$2402a8c0@EDWARD> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps Date: Sat, 14 Aug 2004 01:50:41 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0024_01C481A1.1AAF4B70" 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_0024_01C481A1.1AAF4B70 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable 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_0024_01C481A1.1AAF4B70 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
 
Dennis, you are absolutely correct = about my=20 error.  In fact, you pointing it out had me looking for any other = errors=20 and I am embarrassed to say I found another one - but not one of a math = nature -=20 it turns out both errors sort of offset each other so the answer 47 mph = for air=20 velocity I got with the errors is not much different than the 42 mph = after doing=20 it correctly {:>)
 
Thanks for pointing out my error and = getting me to=20 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 = until=20 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=20 up and locked,  I used the same temperature increase for the air = that Tracy=20 saw   decrease in his coolant.  Well, of course  = DUH!. the=20 temperature of the air will increase considerably more than that for the = same=20 BTU absorbed.  In fact for the GM cores the typical Delta T = measured and=20 reported  for the increase in air temps range from 20-30F.  In = fact if=20 you look at the static situation it only takes 0.25 BTU to raise the = temperature=20 of a lbm of air 1 degree F.  So you could expect the temperature of = a lbm=20 of air to be 4 times higher than that of water in the static situation = (for the=20 same BTU). 
 
 Unfortunately its not quite that = simple with=20 flowing air.  Although if we continued to slow the air flow through = the=20 core the temperature of the air would continue to increase - but like = the old=20 boys and the radiator you would reach a point where the air temps would = be high=20 due to the slowing flow - but the mass flow rate would decrease to the = point=20 that less and less Heat was being removed.  So again a balance is=20 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 = more=20 realistic increase in air temps of say 25F.  we have air mass flow =  W=20 =3D 2400/(25*.25) (note I am using  the Cp 0.25 for air early in = the problem)=20 =3D 384 lbm/minute of air to remove the 2400 BTU.
 
This requires 384/0.076 =3D 5052 CFM of = air at sea=20 level or 5052/60 =3D 84 cubic feet/sec.  Taking our 1.32 sq ft of = GM core=20 surface we find the air velocity required is 84/ 1.32 =3D 63 ft/sec or = 42 mph for=20 our evaporator cores.   The approach with the error gave 47 = MPH which=20 seemed reasonable to me so I didn't even consider any = errors.
 
  Gotta be more careful.  = Gotta stop this=20 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
----- Original Message -----
From:=20 Dennis Haverlah
Sent: Friday, August 13, 2004 = 11:37=20 PM
Subject: [FlyRotary] Re: Answer = to when=20 is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant=20 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 it=20 another way air has 1/4 the heat capacity of the same mass of = water. =20 Hence you need 4 times the mass of water to get the same heat content = capacity=20 in air.
In you calculation for air, you multiplied the water mass = by 1.75=20 to get the equivilant air mass required..  Shouldn't it be = multiplied by=20 4.  

Dennis H.

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

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

Mark=20 S.


     At 09:58 PM 8/12/2004 = -0400, you=20 wrote:
Right=20 you are, Dave
 
Below  is=20 one semi-official definition of BTU in English units.  1 = BTU is=20 amount of heat to raise 1 lb of water 1 degree = Fahrenheit.  =20
 
So with Tracy's = 30 gpm flow=20 of water =3D 240 lbs/min.  Since its temperature is raised = 10 degree=20 F we have
 
BTU = =3D 240 * 10 * 1=20 =3D 2400 BTU/min
 
I know I'm=20 ancient and  I should move into the new metric world, but = at least=20 I didn't do it in Stones and Furlongs = {:>)
 
Ed
 
The Columbia = Encyclopedia,=20 Sixth Edition.  2001.
 
British=20 thermal unit
 
 
abbr. Btu,=20 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 = occurs=20 at a temperature of 39.1 degrees Fahrenheit (=B0F) ] by 1=B0F. = The Btu may=20 also be defined for the temperature difference between 59=B0F = and 60=B0F.=20 One Btu is approximately equivalent to the following: 251.9 = calories;=20 778.26 foot-pounds; 1055 joules; 107.5 kilogram-meters; = 0.0002928=20 kilowatt-hours. A pound (0.454 kilogram) of good coal when = burned should=20 yield 14,000 to 15,000 Btu; a pound 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 in=20 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 know=20 that the specific heat of water is 1.0 cal/(deg = Celsius*gram). =20 Does that also work out to 1.0 BTU/(deg. Farhengight * Lb.) = ?=20
Dave Leonard
Tracy my calculations shows = your coolant=20 temp drop is where it should be:=20
My calculations show that at 7 = gph fuel=20 burn you need to get rid of 2369 BTU/Min through your=20 coolant/radiators.  I rounded it off to 2400 = BTU/min.=20
Q =3D W*DeltaT*Cp  Basic = Heat/Mass Flow=20 equation  With water as the mass with a weight of 8 lbs/ = gallon=20 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 30*8=20 =3D 240 lb/min mass flow. specific heat of water  Cp =3D = 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 the=20 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 = something=20 closer to pure water.  But in any case, certainly in the = ball=20 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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