Return-Path: Received: from [65.54.241.217] (HELO hotmail.com) by logan.com (CommuniGate Pro SMTP 4.2) with ESMTP id 364815 for flyrotary@lancaironline.net; Sat, 14 Aug 2004 11:52:23 -0400 Received-SPF: none receiver=logan.com; client-ip=65.54.241.217; envelope-from=lors01@msn.com Received: from hotmail.com ([65.54.168.114]) by hotmail.com with Microsoft SMTPSVC(5.0.2195.6713); Sat, 14 Aug 2004 08:51:53 -0700 Received: from mail pickup service by hotmail.com with Microsoft SMTPSVC; Sat, 14 Aug 2004 08:51:53 -0700 Received: from 4.174.4.62 by bay3-dav10.bay3.hotmail.com with DAV; Sat, 14 Aug 2004 15:51:53 +0000 X-Originating-IP: [4.174.4.62] X-Originating-Email: [lors01@msn.com] X-Sender: lors01@msn.com From: "Tracy Crook" 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 11:39:43 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_003A_01C481F3.648EE960" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: MSN 9 X-MimeOLE: Produced By MSN MimeOLE V9.10.0006.2205 Seal-Send-Time: Sat, 14 Aug 2004 11:39:43 -0400 Message-ID: X-OriginalArrivalTime: 14 Aug 2004 15:51:53.0680 (UTC) FILETIME=[9EB57500:01C48216] Return-Path: lors01@msn.com This is a multi-part message in MIME format. ------=_NextPart_000_003A_01C481F3.648EE960 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Just another data point. I am routinely seeing 50 - 55 deg Delta = between OAT air and air coming out back side of rads. Higher delta T = (of air) is one of my main arguments for using thick heat exchangers in = aircraft. Tracy ----- Original Message -----=20 From: Ed Anderson=20 To: Rotary motors in aircraft=20 Sent: Saturday, August 14, 2004 1:50 AM Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT = Coolant was : [FlyRotary] Re: coolant temps 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_003A_01C481F3.648EE960 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Just another data point.  I am routinely seeing 50 - 55 = deg Delta=20 between OAT air and air coming out back side of rads.  Higher = delta T=20 (of air) is one of my main arguments for using thick heat exchangers in=20 aircraft.
 
Tracy
----- Original Message -----
From: Ed Anderson
To: Rotary motors in = aircraft
Sent: Saturday, August 14, 2004 = 1:50=20 AM
Subject: [FlyRotary] Re: Answer = to when=20 is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant=20 temps

 
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=20 air velocity I got with the errors is not much different than the 42 = mph after=20 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 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=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 more=20 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 = through=20 the core the temperature of the air would continue to increase - but = like the=20 old boys and the radiator you would reach a point where the air temps = would be=20 high due to the slowing flow - but the mass flow rate would decrease = to the=20 point 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=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 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=20 for our evaporator cores.   The approach with the error gave = 47 MPH=20 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
----- Original Message -----
From:=20 Dennis Haverlah
To: Rotary motors in = aircraft=20
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 = 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=20 capacity in air.
In you calculation for air, you multiplied the = water=20 mass by 1.75 to get the equivilant air mass required..  = Shouldn't it be=20 multiplied by 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=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 take=20 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. Here=20 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 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=20 flow rate of 30 gpm we find it will move that 160 BTU 15 = times/minute (at=20 30 gallons/minute the 2 gallons would be transferred 15 = times).  So=20 taking 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=20 of water can transfer 2400 BTU/min from engine to radiator.  = It also=20 shows why the old wives tale about "slow water" cooling better is = just=20 that (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=20 talking about capacity such as 2 gallons = capacity of=20 a cooling system but instead are talking about mass=20 flow.  As long as we reach that flow rate  1 = gallon at=20 30 gpm or 1/ gallon at 60 gpm or 1/4 gallon at 120 gpm all will = remove the=20 same amount of heat.  However if you = keep increasing the=20 flow rate and reducing the volume you can run into = other problems -=20 like simply not enough water to keep your coolant galleys filled = {:>),=20 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 air=20 flow through the radiators) assuming everything works as = planned. =20 IF  the coolant does not give up as much heat in the = radiators (to=20 the air stream) as it picks up in the engine then you will = eventually=20 (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 = 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=20 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 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=20 get accurate enough where this 4% was an appreciable part of the = total=20 errors in doing our back of the envelope thermodynamics then it = would pay=20 to use 8.34 vice 8, but I don't think we are there, yet=20 {:>).
 
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=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 of=20 engine heat (using Tracy's fuel burn of 7 gallon/hour), how much = air does=20 it 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=20 lbm/min. 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=20 240 lbm/(0.076 lbm/Cubic foot) =3D 3157 cubic feet/min to = equal  the=20 same mass as the coolant. But since the specific heat of air is = lower=20 (0.25) that water, we actually need 75% more air mass or 1.75 * = 3157 =3D=20 5524.75 CFM air flow at sea level. Now I know this sounds like a=20 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 =3D=20 1.32 sq ft we take our
5524.75 cubic feet minute and = divide by 1.32=20 sq ft =3D 4185 ft/min for the required air velocity to move that = much air=20 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, etc.=20 that can change the numbers a bit.  But, then there is really = not=20 much point in trying to be more accurate given the limitations of = our=20 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 = 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)=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 BTU=20 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 ignore=20 that issue?  Water is 8.34 lbs/gal, and say you have 2 = gallons of=20 coolant, that would be 16.68 lbs.  Seems that we would need = to=20 multiply the 2400 figure by 16.68 to arrive at a total system=20 requirement of 40,032 BTU/min.  What am I missing = here?

Mark=20 S.


     At 09:58 PM 8/12/2004 = -0400, you=20 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.  Since=20 its temperature is raised 10 degree F we=20 have
 
BTU =3D = 240 * 10 * 1 =3D=20 2400 BTU/min
 
I = know I'm=20 ancient and  I should move into the new metric world, but = at=20 least I didn't do it in Stones and Furlongs=20 {:>)
 
Ed
 
The Columbia Encyclopedia, = Sixth=20 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=20 should yield 14,000 to 15,000 Btu; a pound of gasoline or = other=20 =
 
 
 
 
 
 
 Ed=20 Anderson
RV-6A N494BW Rotary Powered
Matthews, NC =
----- 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:=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=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 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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