X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from cdptpa-omtalb.mail.rr.com ([75.180.132.121] verified) by logan.com (CommuniGate Pro SMTP 5.1.12) with ESMTP id 2360786 for flyrotary@lancaironline.net; Sun, 30 Sep 2007 13:18:35 -0400 Received-SPF: pass receiver=logan.com; client-ip=75.180.132.121; envelope-from=eanderson@carolina.rr.com Received: from edward2 ([24.74.103.61]) by cdptpa-omta02.mail.rr.com with SMTP id <20070930171756.WLKS3965.cdptpa-omta02.mail.rr.com@edward2> for ; Sun, 30 Sep 2007 17:17:56 +0000 Message-ID: <001901c80385$d9fc3f60$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Another cooling question Date: Sun, 30 Sep 2007 13:17:59 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0016_01C80364.52B892E0" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3138 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 This is a multi-part message in MIME format. ------=_NextPart_000_0016_01C80364.52B892E0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Rather than me state my oft express viewpoint about slow moving fluid = and its cooling effectiveness, here are some folks with a world of = experience and knowledge and what they say about it. http://www.stewartcomponents.com/tech_tips/Tech_Tips_3.htm Q =3D M*Cp*Delta T. There is no doubt that if you move air slower through your core, you are = going to increase its temperature more than faster moving air through = the core (everything else being the same). However, that does not necessarily mean you are removing more heat from = the total system. Let me try to show the relative effects of M and = Delta T of air flow through a 1 sq ft core Velocity(ft/sec) Mass Flow(lbm/min) Delta T(F) Q(BTU/Min) 10 45.9 50 = 573.75 (Baseline) Lets increase delta T to the maximum that I have heard reported (100F) 10 45.9 100 = 1147.5 So doubling the delta T doubles the heat rejected - no surprise there, = but this is the largest delta T that I have heard reported. Now lets vary mass flow by varying velocity 20 91.8 50 = 1147.5 40 183.6 25 = 1147.5 So here, I can reduce the delta T by as much as 75 deg (100-75 =3D 25) = but by modestly increasing the duct velocity from 10 ft/sec ( 6 mph) to = 40 ft/sec (27 Mph) I can reject the same amount of heat at only a = 25F increase in air temps. So my point is that we appear to hit the practical limit on hear = transfer increase by increasing delta T much sooner than we hit a limit = by increasing mass flow. =20 Now IF you have a very efficient core design and can get a 100 Delta T = when the air velocity is 40 ft/sec then you are really cooking as that = would give you=20 40 183.6 100 = 4590 BTU/Min So how do you get maximum heat transfer - I imagine it is a combination = of duct and core design optimized for that particular set of parameter = values - like smaller holes, more fins, wavy fins, thicker core, etc. = But alas those would only be optimum for one set of values and because = we want to cool from sitting on the ground to going 200 MPH, we must = accept a compromise that accommodates both of those regimes of = operation. Cooling is sort of like life - a continuing series of compromises {:>) My 0.02 Ed =20 ----- Original Message -----=20 From: Bobby J. Hughes=20 To: Rotary motors in aircraft=20 Sent: Sunday, September 30, 2007 1:32 PM Subject: [FlyRotary] Re: Another cooling question Snip That's exactly what I HAD thought, until I was told that the air could = pass through too fast and not pick up as much heat. =20 Mark... this is what I was trying to communicate. It could be totally = wrong so let's get more opinions. Ed?? With respect to exit area size only. =20 If the volume of air through the coolant radiator was moving faster = than optimum the air delta T would be lower than it would be with slower = air. Slower air at a higher delta T means less air needs to exit the = cowl. What is optimum? Every install is unique so it needs to be viewed = \ identified on each installation. Ed uses a 30% duct air speed as a = reference point. If I understand this correctly the diffuser play a role = in duct airspeed. If the air is not being diffused optimally the = airspeed could be much higher than 30% through the part of the core. = Dennis H. and I observed airspeeds through a test core at 50% + without = a diffuser.=20 Your inlets are 72sqin with 306sqin core face for water and 24.75 sqin = with 102 sqin for oil. Both are competing for the same exit area. IF = this is an exit area problem only then enlarging it should improve both = water and oil proportionally. Water should realize 2/3 of the = improvement and oil 1/3. 306 sqin vs 102 sqin. Your improvements after = opening the exit seem to track this very close. So increasing your exit = area further should show more improvements in both oil and water to a = point. But you do not need more water cooling improvements right? Core 306 sqin at an airspeed of 115mph. 40% or 46 mph =3D 8602 cfm 8602 cfm at a 50 deg air delta T =3D 7741 = btu's \ min 30% or 34.5 =3D 6451 cfm 6451 cfm at a 80 deg air delta T =3D = 7741 btu's \ min In this example 2151 cfm less air needs to flow through the water = radiator to produce the same btu rejection. So what effect would 2151 = cfm less air through the water radiator and exit area have on the oil = cooler's ability to flow more air? Would it improve the oil cooler air = flow by 1/3 or 717 cfm? Not sure. =20 Can your radiator produce a 80 deg air delta T? It may only produce = 60, 70 or?? Am I in left field here? Bobby (flow testing new radiator ducts prototype today)=20 -------------------------------------------------------------------------= ----- From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] = On Behalf Of Mark Steitle Sent: Sunday, September 30, 2007 4:28 AM To: Rotary motors in aircraft Subject: [FlyRotary] Another cooling question ED wrote: Mark, if you really had excess air flowing through your radiators the = coolant would drop more than 4 Deg F. In fact, the more air flow the = more coolant Delta T you would drop through the radiator.=20 That's exactly what I HAD thought, until I was told that the air could = pass through too fast and not pick up as much heat. This didn't make = sense to me. Maybe I wasn't listening closely and missed the point = altogether (wouldn't be the first time). =20 What I DO know is that the air is flowing faster through the water = radiator than the oil radiator. (I'm not sure I have the ASI's hooked = up correctly, but they're both hooked up the same). I have a pitot = behind each radiator hooked up to two separate ASI's. In slow cruise, = say 125-130 kts, the water radiator ASI will read about 110knts and the = oil ASI will read about 90 kts. The way it was behaving before I opened = up the exit, it appeared that the air from the water radiator was trying = to exit backwards through the oil inlet. I say this because of how high = the oil temps were reading. I enlarged the cowl exit, and both the = water and oil temps dropped significantly. =20 The ASI's are referencing the static port for these readings; should = they be referencing cowl or cabin pressure instead? Airspeeds readings = seem awfully high to me. =20 Mark (Going to the airport today to recalibrate temp sensors) ------=_NextPart_000_0016_01C80364.52B892E0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Rather than me state my oft express viewpoint = about slow=20 moving fluid and its cooling effectiveness, here are some folks with a = world of=20 experience and knowledge and what they say about it.
 
http:= //www.stewartcomponents.com/tech_tips/Tech_Tips_3.htm
 
Q =3D M*Cp*Delta T.
 
There is no doubt that if you move air slower = through your=20 core, you are going to increase its temperature more than faster moving = air=20 through the core (everything else being the same).
However, that does not necessarily mean you are = removing=20 more heat from the total system.  Let me try to show the relative = effects=20 of M and Delta T of air flow  through a 1 sq ft core
 
Velocity(ft/sec)   Mass=20 Flow(lbm/min)   Delta T(F)    =20 Q(BTU/Min)
 
10        =    =20             =    =20 45.9            =    =20             = 50   =20             =    =20     573.75  (Baseline)
 
Lets increase delta T to the maximum that I have = heard=20 reported (100F)
 
10        =    =20             =    =20 45.9            =    =20             = 100   =20             =    =20     1147.5
 
So doubling the delta T doubles the heat = rejected - no=20 surprise there, but this is the largest delta T that I have heard=20 reported.
 
Now lets vary mass flow by varying = velocity
 
20         &nbs= p;            = ;       =20   91.8            =             =    =20 50            =    =20          1147.5
 
40         &nbs= p;            = ;         183.6  &= nbsp;=20             =    =20     25        =    =20             =    =20 1147.5
 
 
So here,  I can reduce the delta T by as = much as 75=20 deg (100-75 =3D 25) but by modestly increasing the duct velocity from 10 = ft/sec=20 (  6 mph) to   40 ft/sec (27 Mph)   I = can=20 reject the same amount of heat at only a 25F increase in air = temps.
 
So my point is that we appear to hit the = practical limit=20 on hear transfer increase by increasing delta T much sooner than we hit = a=20 limit by  increasing mass flow. 
 
Now IF you have a very efficient core design and = can get a=20 100 Delta T when the air velocity is 40 ft/sec then you are really = cooking as=20 that would give you
 
40        =20             =    =20     183.6        =    =20             = 100   =20        =20          4590 = BTU/Min
 
So how do you get maximum heat transfer - I = imagine=20 it is a combination of duct and core design optimized for that = particular set of=20 parameter values - like smaller holes, more fins, wavy fins, thicker = core,=20 etc.  But alas those would only be optimum for one set of values = and=20 because we want to cool from sitting on the ground to going 200 = MPH,=20 we must accept a compromise that accommodates both of those regimes = of=20 operation.
 
Cooling is sort of like life - a continuing = series of=20 compromises {:>)
 
My 0.02
 
Ed         &nbs= p; =20
 
 
----- Original Message -----
From:=20 Bobby J. = Hughes=20
Sent: Sunday, September 30, = 2007 1:32=20 PM
Subject: [FlyRotary] Re: = Another cooling=20 question

Snip
That's exactly what I HAD thought, until I = was told=20 that the air could pass through too fast and not pick up as much = heat. =20
 
Mark... this is what I was trying to = communicate. It=20 could be totally wrong so let's get more opinions. =20 Ed??
 
With respect to exit area size = only. =20
 
If the volume of air through the coolant = radiator was=20 moving faster than optimum the air delta T would be = lower=20 than it would be with slower air. Slower air at a higher delta T means = less=20 air needs to exit the cowl.  What is optimum? Every install = is=20 unique so it needs to be viewed \ identified on each=20 installation. Ed uses a 30% duct air speed as a reference point. = If I=20 understand this correctly the diffuser play a role in duct airspeed. = If the=20 air is not being diffused optimally the airspeed could be much higher = than 30%=20 through the part of the core. Dennis H. and I observed airspeeds = through a=20 test core at 50% + without a diffuser.
 
Your inlets are 72sqin with 306sqin = core face=20 for water and 24.75 sqin with 102 sqin for oil. Both are=20 competing for the same exit area. IF = this is=20 an exit area problem only then enlarging it should improve both water = and oil=20 proportionally. Water should realize 2/3 of the = improvement and=20 oil 1/3.  306 sqin vs 102 sqin. Your improvements after = opening the=20 exit seem to track this very close. So increasing your exit area = further=20 should show more improvements in both oil and water to a point. But = you do not=20 need more water cooling improvements right?
 
Core 306 sqin at an airspeed of=20 115mph.
 
40% or 46 mph =3D 8602 cfm   8602 = cfm at a 50 deg=20 air delta T =3D 7741 btu's \ min
30% or 34.5 =3D 6451=20 cfm        6451 cfm at a 80 deg air = delta T=20 =3D 7741 btu's \ min
 
In this example 2151 cfm less air needs to = flow through=20 the water radiator to produce the same btu rejection. So what effect = would=20 2151 cfm less air through the water radiator and exit area have on the = oil=20 cooler's ability to flow more air? Would it improve the oil cooler air = flow by=20 1/3 or 717 cfm? Not sure.  
 
Can your radiator produce a 80 deg air delta = T? It may=20 only produce 60, 70 or??
 
Am I in left field here?
 
 
 
Bobby
(flow testing new radiator = ducts prototype=20 today) 
 
 
 
 
 
 
 
 
 
 
 
From: Rotary motors in aircraft=20 [mailto:flyrotary@lancaironline.net] On Behalf Of Mark=20 Steitle
Sent: Sunday, September 30, 2007 4:28 = AM
To:=20 Rotary motors in aircraft
Subject: [FlyRotary] Another = cooling=20 question

ED wrote:
<snip>
 Mark, if you really had excess air flowing through your = radiators=20 the coolant would drop more than 4 Deg F.  In fact, the more air = flow the=20 more coolant Delta T you would drop through the radiator. 
<snip>
 
That's exactly what I HAD thought, until I was told that the air = could=20 pass through too fast and not pick up as much heat.  This didn't = make=20 sense to me.  Maybe I wasn't listening closely and missed the = point=20 altogether (wouldn't be the first time). 
 
What I DO know is that the air is flowing faster through the = water=20 radiator than the oil radiator.  (I'm not sure I have the ASI's = hooked up=20 correctly, but they're both hooked up the same).  I have a pitot = behind=20 each radiator hooked up to two separate ASI's.  In slow cruise, = say=20 125-130 kts, the water radiator ASI will read about 110knts and the = oil ASI=20 will read about 90 kts.  The way it was behaving before I opened = up the=20 exit, it appeared that the air from the water radiator was trying = to exit=20 backwards through the oil inlet.  I say this because of how = high the=20 oil temps were reading.  I enlarged the cowl exit, and both=20 the water and oil temps dropped significantly. 
 
The ASI's are referencing the static port for these readings; = should they=20 be referencing cowl or cabin pressure instead?  Airspeeds = readings seem=20 awfully high to me. 
 
Mark
(Going to the airport today to recalibrate temp sensors)
 
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