Return-Path: Received: from [24.93.47.42] (HELO ms-smtp-03-eri0.texas.rr.com) by logan.com (CommuniGate Pro SMTP 4.2) with ESMTP id 366278 for flyrotary@lancaironline.net; Sun, 15 Aug 2004 22:36:37 -0400 Received-SPF: none receiver=logan.com; client-ip=24.93.47.42; envelope-from=clouduster@austin.rr.com Received: from austin.rr.com (cs6625183-144.austin.rr.com [66.25.183.144]) by ms-smtp-03-eri0.texas.rr.com (8.12.10/8.12.7) with ESMTP id i7G2a4o7001224 for ; Sun, 15 Aug 2004 21:36:05 -0500 (CDT) Message-ID: <41201D66.6010800@austin.rr.com> Date: Sun, 15 Aug 2004 21:35:18 -0500 From: Dennis Haverlah User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.4) Gecko/20030624 Netscape/7.1 (ax) X-Accept-Language: en-us, en MIME-Version: 1.0 To: Rotary motors in aircraft Subject: Re: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps References: In-Reply-To: Content-Type: multipart/alternative; boundary="------------060309050704000702090707" X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. --------------060309050704000702090707 Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 8bit 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). > > 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 = 2400/(25*.25) (note I am > using the Cp 0.25 for air early in the problem) = 384 lbm/minute of > air to remove the 2400 BTU. > > This requires 384/0.076 = 5052 CFM of air at sea level or 5052/60 = 84 > cubic feet/sec. Taking our 1.32 sq ft of GM core surface we find the > air velocity required is 84/ 1.32 = 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 > Ed Anderson > RV-6A N494BW Rotary Powered > Matthews, NC > > ----- Original Message ----- > From: Dennis Haverlah > To: Rotary motors in aircraft > 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: > > 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. > > 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 = 16 lbms *10 >> = 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 = >> 160*15 = 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 = 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 = mass of the medium (8 lbs/gallon for water) * >> Flow rate(30 gpm) =240 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 >> >> Q = W*deltaT*cp, we have Q = 240*10*1 = 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 = >> 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. >> >> 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 = Q/(DeltaT*Cp) = 2400/(10*1) = 240 lbm/min. Not a >> surprise as that is what we started with. >> >> 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) = 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 = >> 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 >> = 190 sq inches and turning that in to square feet = 1.32 sq ft >> we take our >> 5524.75 cubic feet minute and divide by 1.32 sq ft = 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 = 69.75 ft/sec airflow velocity through our >> radiators or 47.56 Mph. Now that sounds more reasonable doesn't >> it?? >> >> 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. >> >> Hope this helped clarify the matter. >> >> Ed >> >> >> Ed Anderson >> RV-6A N494BW Rotary Powered >> Matthews, NC >> >> ----- Original Message ----- >> From: Mark Steitle >> To: Rotary motors in aircraft >> >> 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 >>> >>> 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. >>> >>> So with Tracy's 30 gpm flow of water = 240 lbs/min. Since >>> its temperature is raised 10 degree F we have >>> >>> BTU = 240 * 10 * 1 = 2400 BTU/min >>> >>> 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 {:>) >>> >>> Ed >>> >>> The Columbia Encyclopedia, Sixth Edition. 2001. >>> >>> British thermal unit >>> >>> >>> 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 (°F) ] by 1°F. The >>> Btu may also be defined for the temperature difference >>> between 59°F and 60°F. 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 . >>> >>> >>> >>> >>> >>> >>> >>> Ed Anderson >>> RV-6A N494BW Rotary Powered >>> Matthews, NC >>> >>> ----- Original Message ----- >>> From: DaveLeonard >>> To: Rotary motors in aircraft >>> >>> Sent: Thursday, August 12, 2004 8:12 PM >>> 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.) ? >>> Dave Leonard >>> Tracy my calculations shows your coolant temp drop is >>> where it should be: >>> My calculations show that at 7 gph fuel burn you need to >>> get rid of 2369 BTU/Min through your coolant/radiators. >>> I rounded it off to 2400 BTU/min. >>> Q = W*DeltaT*Cp Basic Heat/Mass Flow equation With >>> water as the mass with a weight of 8 lbs/ gallon and a >>> specific heat of 1.0 >>> Q = BTU/min of heat removed by coolant mass flow >>> Assuming 30 GPM coolant flow = 30*8 = 240 lb/min mass >>> flow. specific heat of water Cp = 1.0 >>> Solving for DeltaT = Q/(W*Cp) = 2400/(240*1) = >>> 2400/240 = 10 or your delta T for the parameters >>> specified should be around 10F >>> Assuming a 50/50 coolant mix with a Cp of 0.7 you would >>> have approx 2400/(240 *0.7) = 2400/168 = 14.2F so I >>> would say you do not fly with >>> a 50/50 coolant mix but something closer to pure >>> water. But in any case, certainly in the ball park. >>> You reported 10-12F under those conditions, so I would >>> say condition is 4. Normal operation >>> Ed >>> Ed Anderson >>> RV-6A N494BW Rotary Powered >>> Matthews, NC >>> --------------060309050704000702090707 Content-Type: text/html; charset=us-ascii Content-Transfer-Encoding: 7bit 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). 
 
 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 = 2400/(25*.25) (note I am using  the Cp 0.25 for air early in the problem) = 384 lbm/minute of air to remove the 2400 BTU.
 
This requires 384/0.076 = 5052 CFM of air at sea level or 5052/60 = 84 cubic feet/sec.  Taking our 1.32 sq ft of GM core surface we find the air velocity required is 84/ 1.32 = 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
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
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:

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.  

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 = 16 lbms *10 = 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 = 160*15 = 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 = 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 = mass of the medium (8 lbs/gallon for water) * Flow rate(30 gpm) =240 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
 
Q = W*deltaT*cp, we have Q = 240*10*1 = 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 =  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. 
 
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 = Q/(DeltaT*Cp) = 2400/(10*1) = 240 lbm/min. Not a surprise as that is what we started with.
 
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) = 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 = 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 = 190 sq inches and turning that in to square feet = 1.32 sq ft we take our
5524.75 cubic feet minute and divide by 1.32 sq ft = 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 = 69.75 ft/sec airflow velocity through our radiators  or 47.56 Mph.  Now that sounds more reasonable doesn't it?? 
 
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. 
 
Hope this helped clarify the matter.
 
Ed
 
 
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
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
 
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.  
 
So with Tracy's 30 gpm flow of water = 240 lbs/min.  Since its temperature is raised 10 degree F we have
 
BTU = 240 * 10 * 1 = 2400 BTU/min
 
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 {:>)
 
Ed
 
The Columbia Encyclopedia, Sixth Edition.  2001.
 
British thermal unit
 
 
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 (°F) ] by 1°F. The Btu may also be defined for the temperature difference between 59°F and 60°F. 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 .
 
 
 
 
 
 
 
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
From: DaveLeonard
To: Rotary motors in aircraft
Sent: Thursday, August 12, 2004 8:12 PM
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.) ?
Dave Leonard
Tracy my calculations shows your coolant temp drop is where it should be:
My calculations show that at 7 gph fuel burn you need to get rid of 2369 BTU/Min through your coolant/radiators.  I rounded it off to 2400 BTU/min.
Q = W*DeltaT*Cp  Basic Heat/Mass Flow equation  With water as the mass with a weight of 8 lbs/ gallon and a specific heat of 1.0
Q = BTU/min of heat removed by coolant mass flow
 Assuming 30 GPM coolant flow = 30*8 = 240 lb/min mass flow. specific heat of water  Cp = 1.0
 Solving for DeltaT = Q/(W*Cp) = 2400/(240*1)  =  2400/240 = 10 or  your delta T for the parameters specified should be around 10F
Assuming a 50/50 coolant mix with a Cp  of 0.7 you would have approx 2400/(240 *0.7) = 2400/168 = 14.2F so I would say you do not fly with
 a 50/50 coolant mix but something closer to pure water.  But in any case, certainly in the ball park.
You reported 10-12F under those conditions, so I would say condition is 4. Normal operation
Ed
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC

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