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 363556 for flyrotary@lancaironline.net; Fri, 13 Aug 2004 10:15:57 -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 i7DEEsPg009316 for ; Fri, 13 Aug 2004 10:14:55 -0400 (EDT) Message-ID: <000c01c4813f$eb3b20a0$2402a8c0@EDWARD> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Answer to when is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps Date: Fri, 13 Aug 2004 10:14:59 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0009_01C4811E.63DAC460" 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_0009_01C4811E.63DAC460 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable 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. = 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 =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 =20 Dave Leonard=20 Tracy my calculations shows your coolant temp drop is where it = should be:=20 =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 =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 =20 Q =3D BTU/min of heat removed by coolant mass flow=20 =20 Assuming 30 GPM coolant flow =3D 30*8 =3D 240 lb/min mass flow. = specific heat of water Cp =3D 1.0=20 =20 =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 =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 =20 a 50/50 coolant mix but something closer to pure water. But in = any case, certainly in the ball park.=20 =20 You reported 10-12F under those conditions, so I would say = condition is 4. Normal operation=20 =20 Ed=20 =20 Ed Anderson=20 RV-6A N494BW Rotary Powered=20 Matthews, NC=20 ------=_NextPart_000_0009_01C4811E.63DAC460 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
I'm sorry, Mark, I did not show that = step. =20 You are correct the weight (mass) of water(or any other cooling = medium) is=20 an important factor as is its specific heat.
 
 In the example you used  - = where we have=20 a static 2 gallons capacity of water, It would actually only take 8*2 = =3D 16 lbms=20 *10 =3D 160 BTU to raise the temp of the water 1 degree F.  The = difference is=20 in one case we are talking about raising the temperature of a fixed = static=20 amount of water which can not readily get rid of the heat, in the other = (our=20 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=20 capacity and see what we can determine.
 
If we take our 2 gallons and start = moving it from=20 engine to radiator and back we find that each times the 2 gallons = circulates it=20 transfers 160 BTU (in our specific example!!). So at our flow rate of 30 = gpm we=20 find it will move that 160 BTU 15 times/minute (at 30 gallons/minute the = 2=20 gallons would be transferred 15 times).  So taking our 160 BTU that = it took=20 to raise the temp of our 2 gallons of static water 10F that we now = have=20 being moved from 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 = BTU/min=20 from engine to radiator.  It also shows why the old wives tale = about "slow=20 water" cooling better is just that (another story about how that got=20 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 long=20 as we reach that flow rate  1 gallon at 30 gpm or 1/ gallon at 60 = gpm or=20 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 = volume=20 you can run 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 = course,=20 simply recirculated at the rate of 30 gpm through our engine (picking up = heat-=20 approx 2400 BTU/min in this specific example) and then through our = radiator=20 (giving up heat of 2400 BTU/Min  to the air flow through the=20 radiators) assuming everything works as planned.  IF  the = coolant does=20 not give up as much heat in the radiators (to the air stream) as it = picks up in=20 the engine then you will eventually (actually quite quickly) over heat = your=20 engine.
 
The 240 lb figure I used in the = previous example=20 comes from using 8 lb/gal (a common approximation, but not precise as = you point=20 out) to calculate the mass flow.
 
The mass flow =3D mass of the medium (8 = lbs/gallon=20 for water) * Flow rate(30 gpm) =3D240 lbs/min mass flow. Looking at the = units we=20 have
(8 lbs/gallon)*(30 Gallon/minute) = canceling out the=20 like units (gallons) leaves us with 240 lb/minute which is our mass flow = in this=20 case.
 
Then using the definition of the BTU we = have 240=20 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=20 BTU/minute is required to increase the temperature of this mass flow by=20 10F
 
Using the more accurate weight of water = we would=20 have  8.34*30 =3D  250.2 lbm/minute  so the actual BTU = required is=20 closer to 2502 BTU/min instead of my original 2400 BTU/Min, so there is = apporx a=20 4% error in using 8 lbs/gallon.  If we could ever get accurate = enough where=20 this 4% was an appreciable part of the total errors in doing our back of = the=20 envelope thermodynamics then it would pay to use 8.34 vice 8, but I = don't think=20 we are there, yet {:>).
 
Now the same basic equation applies to = the amount=20 of heat that the air transfers away from out radiators.  But here = the mass=20 of air is much lower than the mass of water so therefore it takes a much = higher=20 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. =20 So a lb of air will only carry approx 25% the heat of a lb of water, so = again=20 for this reason you need more air flow. 
 
if 30 gpm of water will transfer 2400 = bTu of engine=20 heat (using Tracy's fuel burn of 7 gallon/hour), how much air does it = require to=20 remove that heat from the radiators?
 
Well  again we turn to our = equation and with a=20 little algebra we have W =3D Q/(DeltaT*Cp) =3D 2400/(10*1) =3D 240 = lbm/min. Not a=20 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 mass as=20 the coolant. But since the specific heat of air is lower (0.25) that = water, we=20 actually need 75% more air mass or 1.75 * 3157 =3D 5524.75 CFM air flow = at sea=20 level. Now I know this sounds like a tremendous amount of air but stay = with me=20 through the next step.
 
Taking two GM evaporator cores with a = total frontal=20 area of 2*95 =3D 190 sq inches and turning that in to square feet =3D = 1.32 sq ft we=20 take our
5524.75 cubic feet minute and divide by = 1.32 sq ft=20 =3D 4185 ft/min for the required air velocity to move that much air = volume through=20 our two evaporator cores.  To get the air velocity in = ft/sec=20 divide 4185/60 =3D 69.75 ft/sec airflow velocity through our = radiators =20 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=20 unlike water changes with altitude, the temperature of the air, etc. = that can=20 change the numbers a bit.  But, then there is really not much point = in=20 trying to be more accurate given the limitations of our experimentation = accuracy=20 {:>).
 
 
Also do not confuse the BTUs required = to raise the=20 temperature of 1 lb of water 1 degree F with that required to turn water = in to=20 vapor - that requires orders of magnitude more BTU. 
 
Hope this helped clarify the = matter.
 
Ed
 
 
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
From:=20 Mark=20 Steitle
Sent: Friday, August 13, 2004 = 8:32=20 AM
Subject: [FlyRotary] Re: DeltaT = Coolant=20 was : [FlyRotary] Re: coolant temps

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

Mark S.


     At 09:58 PM = 8/12/2004=20 -0400, you wrote:
Right you=20 are, Dave
 
Below  is one=20 semi-official definition of BTU in English units.  1 BTU is = amount of=20 heat to raise 1 lb of water 1 degree Fahrenheit.  =20
 
So with Tracy's 30 = gpm flow of=20 water =3D 240 lbs/min.  Since its temperature is raised 10 = degree F we=20 have
 
BTU =3D 240 * = 10 * 1 =3D 2400=20 BTU/min
 
I know I'm = ancient=20 and  I should move into the new metric world, but at least I = didn't do=20 it in Stones and Furlongs {:>)
 
Ed
 
The Columbia Encyclopedia, Sixth=20 Edition.  2001.
 
British thermal=20 unit
 
 
abbr. Btu, = unit for=20 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 at a=20 temperature of 39.1 degrees Fahrenheit (=B0F) ] by 1=B0F. The Btu = may also be=20 defined for the temperature difference between 59=B0F and 60=B0F. = One Btu is=20 approximately equivalent to the following: 251.9 calories; 778.26=20 foot-pounds; 1055 joules; 107.5 kilogram-meters; 0.0002928 = kilowatt-hours. A=20 pound (0.454 kilogram) of good coal when burned should yield 14,000 = to=20 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 that the specific heat = of=20 water is 1.0 cal/(deg Celsius*gram).  Does that also work out = to 1.0=20 BTU/(deg. Farhengight * Lb.) ?=20
=20
Dave Leonard
Tracy my calculations shows your coolant temp drop is where it = should=20 be:=20
=20
My calculations show that at 7 gph fuel burn you need to get = rid of=20 2369 BTU/Min through your coolant/radiators.  I rounded it = off to=20 2400 BTU/min.=20
=20
Q =3D W*DeltaT*Cp  Basic Heat/Mass Flow equation  = With water=20 as the mass with a weight of 8 lbs/ gallon and a specific heat of = 1.0=20
=20
Q =3D BTU/min of heat removed by coolant mass flow=20
=20
 Assuming 30 GPM coolant flow =3D 30*8 =3D 240 lb/min = mass flow.=20 specific heat of water  Cp =3D 1.0=20
=20
=20
 Solving for DeltaT =3D Q/(W*Cp) =3D 2400/(240*1)  = =3D =20 2400/240 =3D 10 or  your delta T for the parameters specified = should be=20 around 10F=20
=20
Assuming a 50/50 coolant mix with a Cp  of 0.7 you would = have=20 approx 2400/(240 *0.7) =3D 2400/168 =3D 14.2F so I would say you = do not fly=20 with=20
=20
 a 50/50 coolant mix but something closer to pure = water. =20 But in any case, certainly in the ball park.=20
=20
You reported 10-12F under those conditions, so I would say = condition=20 is 4. Normal operation=20
=20
Ed=20
=20
Ed Anderson=20
RV-6A N494BW Rotary Powered=20
Matthews, NC=20
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