Return-Path: Received: from wb1-a.mail.utexas.edu ([128.83.126.134] verified) by logan.com (CommuniGate Pro SMTP 4.2) with ESMTP-TLS id 363775 for flyrotary@lancaironline.net; Fri, 13 Aug 2004 12:04:40 -0400 Received-SPF: none receiver=logan.com; client-ip=128.83.126.134; envelope-from=msteitle@mail.utexas.edu Received: (qmail 34159 invoked from network); 13 Aug 2004 16:04:08 -0000 Received: from dhcp-191-101.per.utexas.edu (HELO benefits3.mail.utexas.edu) (146.6.191.101) by wb1.mail.utexas.edu with RC4-SHA encrypted SMTP; 13 Aug 2004 16:04:08 -0000 Message-Id: <5.1.1.5.2.20040813105511.0260dd58@localhost> X-Sender: msteitle@mail.utexas.edu@localhost X-Mailer: QUALCOMM Windows Eudora Version 5.1.1 Date: Fri, 13 Aug 2004 11:04:02 -0500 To: "Rotary motors in aircraft" From: Mark Steitle Subject: Re: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps In-Reply-To: Mime-Version: 1.0 Content-Type: multipart/alternative; boundary="=====================_14999593==.ALT" --=====================_14999593==.ALT Content-Type: text/plain; charset="iso-8859-1"; format=flowed Content-Transfer-Encoding: quoted-printable David, Thanks for the explanation. I had thought of using an airspeed indicator,= =20 but wasn't sure that it would give me an accurate reading without using a=20 pitot tube. Does the tube size need to be the same as the pitot inlet hole? If I used a water manometer instead, how do you convert inches to mph? It= =20 seems too that it could make a difference as to where on the surface of the= =20 radiator you take the reading. One other thing came to mind after I posted my message. I'm not sure the=20 3-rotor pumps 50% more coolant than the 2-rotor. While the 20B produces=20 50% more heat, I don't know if it pumps 50% more coolant. I do know that=20 the impeller is larger on the 20B. Has anyone measured the flow of the 20B= =20 water pump? Mark (So close, yet so far) At 10:44 AM 8/13/2004 -0500, you wrote: >"How does one determine the airspeed across the radiator?" > >Put a valve that functions like a "selectable" T fitting in your airspeed >pitot line. (same as an "alternate static pressure valve/switch like in a >Cessna?) Run the "new/additional" pitot line to a spot behind your >radiator. Go fly at a specific speed, then switch your airspeed indicator >from "regular pitot line" to the "radiator airspeed pitot line" and write >down the results. Repeat at different speeds (climb, economy cruise, high >cruise, top or race speed). > >Or, can plumb in a simple water manometer. Kevin Horton has posted lots of >stuff on that. I have the e-mails and websites saved if you want to go= that >route. > >David > >----- Original Message ----- >From: "Mark Steitle" >To: "Rotary motors in aircraft" >Sent: Friday, August 13, 2004 10:02 AM >Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT >Coolant was : [FlyRotary] Re: coolant temps > > >Ed, >Wow, that was a great explanation. I think I even understood most of >it. ;-) I'm assuming that for a 3-rotor I would just multiply the 2400 >BTU figure by 1.5 to arrive at a BTU requirement of 3600 for the >coolant? Same for the air requirement (240 * 1.5 =3D 360 lbs, or 5524.75 * >1.5 =3D 8287 CFM). My single radiator core is approx. 18" x 20" or 360 sq. >in. or 2.5 cubic feet. Then 8287/2.5 =3D 3314.8 ft/min (55.25 ft/sec) air >velocity, which equals 37.67 mph. > >Does this look right? I can cool my engine at 37 mph airflow across the >core? How does one determine the airspeed across the radiator? > >Mark >(See what you started???) > >At 10:14 AM 8/13/2004 -0400, you 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. > > > >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 > > > >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. > > > >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. > > > >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 > >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?? > > > >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 =3D 240 lbs/min. Since its > >>temperature is raised 10 degree F we have > >> > >>BTU =3D 240 * 10 * 1 =3D 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 (=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 . > >> > >> > >> > >> > >> > >> > >> > >>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 =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 > >>Q =3D BTU/min of heat removed by coolant mass flow > >> Assuming 30 GPM coolant flow =3D 30*8 =3D 240 lb/min mass flow.= specific > >> heat of water Cp =3D 1.0 > >> 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 > >>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 > >> 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 > > > > >> Homepage: http://www.flyrotary.com/ > >> Archive: http://lancaironline.net/lists/flyrotary/List.html --=====================_14999593==.ALT Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable David,
Thanks for the explanation.  I had thought of using an airspeed indicator, but wasn't sure that it would give me an accurate reading without using a pitot tube.  Does the tube size need to be the same as the pitot inlet hole?

If I used a water manometer instead, how do you convert inches to mph?  It seems too that it could make a difference as to where on the surface of the radiator you take the reading. 

One other thing came to mind after I posted my message.  I'm not sure the 3-rotor pumps 50% more coolant than the 2-rotor.  While the 20B produces 50% more heat, I don't know if it pumps 50% more coolant.  I do know that the impeller is larger on the 20B.  Has anyone measured the flow of the 20B water pump? 
 
Mark
(So close, yet so far)


 At 10:44 AM 8/13/2004 -0500, you wrote:
"How does one determine the airspeed across the radiator?"

Put a valve that functions like a "selectable" T fitting in your airspeed
pitot line. (same as an "alternate static pressure valve/switch like in a
Cessna?)   Run the "new/additional" pitot line to a spot behind your
radiator.  Go fly at a specific speed, then switch your airspeed indicator
from "regular pitot line" to the "radiator airspeed pitot line" and write
down the results.  Repeat at different speeds (climb, economy cruise, high
cruise, top or race speed).

Or, can plumb in a simple water manometer.  Kevin Horton has posted lots of
stuff on that.  I have the e-mails and websites saved if you want to go that
route.

David

----- Original Message -----
From: "Mark Steitle" <msteitle@mail.utexas.edu>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Sent: Friday, August 13, 2004 10:02 AM
Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re: DeltaT
Coolant was : [FlyRotary] Re: coolant temps


Ed,
Wow, that was a great explanation.  I think I even understood most of
it.  ;-)   I'm assuming that for a 3-rotor I would just multiply the 2400
BTU figure by 1.5 to arrive at a BTU requirement of 3600 for the
coolant?  Same for the air requirement (240 * 1.5 =3D 360 lbs, or 5524.75 *
1.5 =3D 8287 CFM).  My single radiator core is approx. 18" x 20" or 360 sq.
in. or 2.5 cubic feet.  Then 8287/2.5 =3D  3314.8 ft/min (55.25 ft/sec) air
velocity, which equals 37.67 mph.

Does this look right?  I can cool my engine at 37 mph airflow across the
core?  How does one determine the airspeed across the radiator?

Mark
(See what you started???)

At 10:14 AM 8/13/2004 -0400, you 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.
>
>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
>
>Q =3D W*deltaT*cp, we have Q =3D 240*10*1 =3D 2400 BTU/minute is require= d 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.
>
>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.
>
>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
>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??
>
>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: <mailto:ms= teitle@mail.utexas.edu>Mark Steitle
>To: <mailto= :flyrotary@lancaironline.net>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 =3D 240 lbs/min.  Since its
>>temperature is raised 10 degree F we have
>>
>>BTU =3D 240 * 10 * 1 =3D 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
>><http://www.bartleby.com/65/en/Englsh-u.html>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 .
>>
>>
>>
>>
>>
>>
>>
>>Ed Anderson
>>RV-6A N494BW Rotary Powered
>>Matthews, NC
>>----- Original Message -----
>>From: <mailto:daveleonard@cox.net>DaveLeonard
>>To: <mailto= :flyrotary@lancaironline.net>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 =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
>>Q =3D BTU/min of heat removed by coolant mass flow
>>  Assuming 30 GPM coolant flow =3D 30*8 =3D 240 lb/min mass flow. specific
>> heat of water  Cp =3D 1.0
>>  Solving for DeltaT =3D Q/(W*Cp) =3D 2400/(240*1)  =3D&nb= sp; 2400/240 =3D 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) =3D 2400/168 =3D 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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