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 363895 for flyrotary@lancaironline.net; Fri, 13 Aug 2004 13:54:56 -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 i7DHsLPg007369 for ; Fri, 13 Aug 2004 13:54:22 -0400 (EDT) Message-ID: <003a01c4815e$93a74b60$2402a8c0@EDWARD> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Marks Question Answer to when is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps Date: Fri, 13 Aug 2004 13:54:27 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0037_01C4813D.0C496020" 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_0037_01C4813D.0C496020 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Wait!!! Mark, remember this was all for the example of Tracy Crook = burning 7.0 gallons/hour of fuel. Now, that is near the low end of what = a two rotor will see and as Tracy stated that was for a Cruise condition = - which is the regime of flight that normally is the least demanding so = far as cooling requirements. You really need to look at the most severe = regime which is generally take off and initial climb. There you are = putting out full power and the air speed is slowest (therefore you = cooling mass flow is lowest). But, yes what ever the two rotor BTU cooling needs are (all other things = being equal) for a specific power output then the 3 rotor's needs = would be approx 1.5 times that. Remember this cooling stuff is far from as simple as you might have = surmised from my calculations. These equations will give you the = coolant and airflow capable of removing your waste heat - however, = whether you particular system provides that air mass flow through your = radiators and heat transfer is a different problem. Your ducting and = type diffuser and cooler core are significant factors as to whether 1. you get your minimal airflow required and=20 2. That your cores effectively transfer the heat to that cooling air = mass. For instance you could hang your radiators out in the slip stream and = possible get the required minimum airflow through them (if you fly fast = enough) but the cooling drag would be extremely high - not desirable. = To get maximum cooling with minimum drag takes a bit of more. Its generally accepted that you should ideally design your cooling = system to either=20 1. Provide adequate cooling with an airflow velocity around 10% of = CRUISE velocity 2 Or 30% of your climb velocity In some cases the two rules produce approx the same air velocity = figure for the cooler. =20 So if your climb velocity was say 120 MPH IAS then your "Optimum" = airflow through your radiator would be approx 120 *.3 =3D 36 MPH If your cruise airspeed was 200 mph then your design optimum (assuming = you chose cruise as you design point) would be approx 20 MPH. So first you need to decide on which regime you want to design for and = the trade offs involved. That establishes your airspeed and power = parameters. I choose climb cruise which leaves me with a slight cooling = deficit until I hit approx 120 mph in climb. That means my temperature = has continue to climb until I hit my design airspeed. But, I accept I = will have increased temps for a short periods I spent very little time = in take off and initial climb (to pattern altitude or maybe 2500 AGL. Then you need to estimate your power produced during that phase of = operation, in my cause I use 170 HP - I may not make that but if its = less then my cooling problem is less severe {:>). Then you figure your = waste BTU you need to get rid of. The spreadsheet I sent will do that = for you for either a 2 rotor or 3 rotor - but you have to change the = value from 2 to 3 on the spread sheet. =20 So you find the rpm you are operating at and the power/BTU rejection you = must achieve. Figure that the number of BTU you must reject through oil = and coolant to be the same amount of BTU as the HP produced (that's what = the spreadsheet does). It breaks out that into the coolant and oil BTU = rejection requirements. Then once you have the BTUs (for that set of operating parameters) you = can start getting a handle on your cooling challenge. Since your total surface area is appox 165 sq inches greater that 360 vs = 195 (for 2 gm cores used on the two rotors), I would say you have the = basic radiator needed to cooling your 3 rotor provided you get the = minimum required airflow. Lets see if I use 170 HP for my climb power figure lets assume you will = use 1.5 times that or 255 HP for your 3 rotor. That would mean you need = to reject approx 7834 BTU (not the 2400 BTU of the two rotor at lean = cruise conditions) with your radiator (and another 3917 BTU from your = oil). So we need an airmass flow through your 360 sq inch to carry away 7834 = BTU/min. =20 So again turning to our heat transfer equation we have Q =3D W DeltaT = cp. Lets assume the deltaT (increased in air temp caused by dumping the = heat into it) is around 20 - 50F(it will be more when you are operating = at a higher temps of full power situations, so lets pick 30F). The air = mass flow required W =3D Q/(DeltaT *0.25) W =3D 7834 /(50*.25) =3D 626 lbms/min. 626 lbm/0.076 =3D 8236.8 CFM = (already factored in the 0.25 cp). So changing that to cubic ft/sec we = divide by 60 , 8237/60 =3D 137 cubic ft/sec . Taking your radiator core = surface are of 360/144 =3D 2.5 sq ft and we have ( 137 cubic ft/sec)/(2.5 sq ft) =3D 54.31 ft/sec air velocity required. = =3D 37.4 mph air flow. Using the 30% climb rule this means you would = need to climb (while producing 255 hp) at an airspeed of around 37/0.3 = =3D 124 mph to maintain cooling at that power setting and your current = radiator size. Now I fudged a little bit by assuming a DeltaT of 50F. =20 I don't know what delta T your system will produce having no data points = on your radiator core. I purposely picked 50F as I knew it would = produce an airflow close to what you had calculated. (See you can make = the figures provide any answer you want provided you get to make the = assumptions {:>)). However, lets say your core's delta T is closer to the delta Ts reported = by GM core users which is generally 20-30F. Generally the GM core users have been reporting a 20 -31F rise in air = temps at high power settings. Lets say your radiator produced the same = increase in temp of 30F and not the 50F I earlier assumed. Then W =3D 7834/(30*.25) =3D 1045 lbm/min of air flow needed. = 1045/0.076 =3D 137434 Cubic ft/Min divide by 60 gives 229 cubic ft/Sec taking your 2.5 sq ft V =3D 229/2.5 =3D 92 ft/sec of air velocity =3D = 62.7 MPH through your radiator. So this is approx double the required = airflow over the case where the air temp increase was assumed to be 50F. = Still all in all you radiator should have no problem getting rid of = the heat - although your cooling drag may be a bit higher than optimum. So just trying to point out that while these calculations can give you a = starting point, just some small differences in assumptions (as I did = with the deltaT ) can make the difference between success and not. But = since your radiator is sufficiently large, the worst penalty you are = likely to paid is a bit more drag. Assuming that your ducting and = diffuser provides airflow reasonably close to that calculated. Hope this helps 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 11:02 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. =20 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? =20 Mark (See what you started???) Ed Anderson=20 RV-6A N494BW Rotary Powered=20 Matthews, NC ------=_NextPart_000_0037_01C4813D.0C496020 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Wait!!!  Mark, remember this was = all for the=20 example of Tracy Crook burning 7.0 gallons/hour of fuel.  Now, that = is near=20 the low end of what a two rotor will see and as Tracy stated that was = for a=20 Cruise condition - which is the regime of flight that normally is the = least=20 demanding so far as cooling requirements.  You really need to look = at the=20 most severe regime which is generally take off and initial climb.  = There=20 you are putting out full power and the air speed is slowest (therefore = you=20 cooling mass flow is lowest).
 
But, yes what ever the two rotor BTU = cooling needs=20 are (all other things being equal)  for a specific power = output  then=20 the 3 rotor's needs would be approx 1.5 times that.
 
 
 Remember this cooling stuff is = far from as=20 simple as you might have surmised from my calculations.  These = equations=20 will give you the coolant and airflow capable of removing your waste = heat -=20 however, whether you particular system provides that air mass flow = through your=20 radiators and heat transfer is a different problem.  Your ducting = and type=20 diffuser and cooler core are  significant factors as to=20 whether
 
1.  you get your minimal airflow = required and=20
 
2. That your cores effectively transfer = the heat to=20 that cooling air mass.
 
For instance you could hang your = radiators out in=20 the slip stream and possible get the required minimum airflow through = them (if=20 you fly fast enough) but the cooling drag would be extremely high - not=20 desirable.  To get maximum cooling with minimum drag takes a bit of = more.
Its generally accepted that you should = ideally=20 design your cooling system to either
 
1.  Provide adequate cooling with = an airflow=20 velocity around 10% of CRUISE velocity
 
2  Or 30% of your climb = velocity
 
In some cases  the two  rules = produce=20 approx the same air velocity figure for the cooler. 
 
So if your climb velocity was say 120 = MPH IAS then=20 your "Optimum" airflow through your radiator would be approx 120 *.3 =3D = 36=20 MPH
 
If your cruise airspeed was 200 mph = then your=20 design optimum (assuming you chose cruise as you design = point) would be=20 approx 20 MPH.
 
So first you need to decide on which = regime you=20 want to design for and the trade offs involved.  That establishes = your=20 airspeed and power parameters.  I choose climb cruise which leaves = me with=20 a slight cooling deficit until I hit approx 120 mph in climb.  That = means=20 my temperature has continue to climb until I hit my design = airspeed.  But,=20 I accept I will have increased temps for a short periods I spent very = little=20 time in take off and initial climb (to pattern altitude or maybe 2500=20 AGL.
 
Then you need to estimate your power = produced=20 during that phase of operation, in my cause I use 170 HP - I may not = make that=20 but if its less then my cooling problem is less severe {:>).  = Then you=20 figure your waste BTU you need to get rid of.  The spreadsheet I = sent will=20 do that for you for either a 2 rotor or 3 rotor - but you have to change = the=20 value from 2 to 3 on the spread sheet. 
 
So you find the rpm you are operating = at and the=20 power/BTU rejection you must achieve.  Figure that the number of = BTU you=20 must reject through oil and coolant to be the same amount of BTU as the = HP=20 produced (that's what the spreadsheet does).  It breaks out that = into the=20 coolant and oil BTU rejection requirements.
 
Then once you have the BTUs (for that = set of=20 operating parameters) you can start getting a handle on your cooling=20 challenge.
 
Since your total surface area is appox = 165 sq=20 inches greater that 360 vs 195 (for 2 gm cores used on the two rotors), = I would=20 say you have the basic radiator needed to cooling your 3 rotor provided = you get=20 the minimum required airflow.
 
Lets see if I use 170 HP for my climb = power figure=20 lets assume you will use 1.5 times that or 255 HP for your 3 = rotor.  That=20 would mean you need to reject approx 7834 BTU (not the 2400 BTU of the = two rotor=20 at lean cruise conditions) with your radiator (and another 3917 BTU from = your=20 oil).
 
So we need an airmass flow through your = 360 sq inch=20 to carry away 7834 BTU/min. 
 
So again turning to our heat transfer = equation we=20 have  Q =3D W DeltaT cp.  Lets assume the deltaT (increased in = air temp=20 caused by dumping the heat into it) is around 20 - 50F(it will be = more when=20 you are operating at a higher temps of full power situations, so lets = pick=20 30F).   The air mass flow required W =3D Q/(DeltaT = *0.25)
 
W =3D 7834 /(50*.25) =3D 626 = lbms/min.  626=20 lbm/0.076 =3D 8236.8  CFM (already factored in the 0.25 = cp).  So=20 changing that to cubic ft/sec we divide by 60 , 8237/60 =3D 137 = cubic ft/sec=20 .  Taking your radiator core surface are of 360/144 =3D 2.5 sq ft = and we=20 have
 
( 137 cubic ft/sec)/(2.5 sq ft) = =3D 54.31=20 ft/sec air velocity required.  =3D 37.4 mph air flow.  = Using the=20 30% climb rule this means you would need to climb (while producing 255 = hp) at an=20 airspeed of around 37/0.3 =3D 124  mph to maintain cooling at = that power=20 setting and your current radiator size. Now I fudged a little bit by = assuming a=20 DeltaT of 50F. 
 
I don't know what delta T your system = will produce=20 having no data points on your radiator core.  I purposely picked = 50F as I=20 knew it would produce an airflow close to what you had calculated.  = (See=20 you can make the figures provide any answer you want provided you get to = make=20 the assumptions {:>)).
 
However, lets say your core's delta T = is=20 closer to the delta Ts reported by GM core users which is = generally=20 20-30F.
 
Generally the GM core users have been = reporting a=20 20 -31F rise in air temps at high power settings.  Lets say your = radiator=20 produced the same increase in temp of 30F and not the 50F I earlier=20 assumed.
 
Then W =3D 7834/(30*.25) =3D 1045 = lbm/min  of air=20 flow needed.  1045/0.076 =3D 137434 Cubic ft/Min divide by 60 gives = 229 cubic=20 ft/Sec
 taking your 2.5 sq ft V =3D = 229/2.5 =3D  92=20 ft/sec of air velocity =3D 62.7 MPH through your radiator.  So this = is approx=20 double the required airflow over the case where the air temp increase = was=20 assumed to be 50F.   Still all in all you radiator should have = no=20 problem getting rid of the heat - although your cooling drag may be a = bit higher=20 than optimum.
 
So just trying to point out that while = these=20 calculations can give you a starting point, just some small differences = in=20 assumptions (as I did with the deltaT ) can make the difference between = success=20 and not. But since your radiator is sufficiently large, the worst = penalty you=20 are likely to paid is a bit more drag.  Assuming that your ducting = and=20 diffuser provides airflow reasonably close to that = calculated.
 
 
Hope this helps
 
Ed
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
From:=20 Mark=20 Steitle
Sent: Friday, August 13, 2004 = 11:02=20 AM
Subject: [FlyRotary] Re: Answer = to when=20 is 2 gallons enough?Re: DeltaT Coolant was : [FlyRotary] Re: coolant=20 temps

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

Does this=20 look right?  I can cool my engine at 37 mph airflow across the=20 core?  How does one determine the airspeed across the=20 radiator?   

Mark
(See what you=20 started???)
Ed Anderson=20
RV-6A N494BW Rotary Powered=20
Matthews, NC =
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