X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from fed1rmmtao102.cox.net ([68.230.241.44] verified) by logan.com (CommuniGate Pro SMTP 5.3.9) with ESMTP id 4455221 for flyrotary@lancaironline.net; Fri, 03 Sep 2010 19:56:52 -0400 Received-SPF: none receiver=logan.com; client-ip=68.230.241.44; envelope-from=alventures@cox.net Received: from fed1rmimpo02.cox.net ([70.169.32.72]) by fed1rmmtao102.cox.net (InterMail vM.8.00.01.00 201-2244-105-20090324) with ESMTP id <20100903235607.BPLD23088.fed1rmmtao102.cox.net@fed1rmimpo02.cox.net> for ; Fri, 3 Sep 2010 19:56:07 -0400 Received: from BigAl ([72.199.216.236]) by fed1rmimpo02.cox.net with bizsmtp id 2Pw71f00T56cS2o04Pw7ip; Fri, 03 Sep 2010 19:56:07 -0400 X-VR-Score: 0.00 X-Authority-Analysis: v=1.1 cv=RFAs0e+2sU49DU7hBECFSPlG3t5ZI1He37/O2GpqT7s= c=1 sm=1 a=CWPqKAhi16IA:10 a=lN8H/RjlhkCyIsyuOn2r7w==:17 a=Ia-xEzejAAAA:8 a=xwHMckwuqR4CU1P9ETUA:9 a=yPCvCPT1fq9rbwrHpswA:7 a=P9djTBrRkLLBeIndRwqaU5I-RU0A:4 a=CjuIK1q_8ugA:10 a=EzXvWhQp4_cA:10 a=foijEI2K064do9S5:21 a=I5pjWX6KAkAGbpFp:21 a=yMhMjlubAAAA:8 a=_oOmw0CmSGM6Wjb9Tp4A:9 a=_d7dxQn28s6979WbX94A:7 a=ElXN6NLCSNVRKcHRT5to4T81TDoA:4 a=lN8H/RjlhkCyIsyuOn2r7w==:117 X-CM-Score: 0.00 Authentication-Results: cox.net; none From: "Al Gietzen" To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Re: Radiator Math Date: Fri, 3 Sep 2010 16:56:50 -0800 Message-ID: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_000D_01CB4B89.00CD5310" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.6863 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5931 Importance: Normal Thread-Index: ActLvkq9efJt37XjTjOgXabp9p2feAACYR0g In-Reply-To: This is a multi-part message in MIME format. ------=_NextPart_000_000D_01CB4B89.00CD5310 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable There is another point to clarify regarding the inlet scoop area, which = I probably should have made earlier. Since I fly a pusher; my thinking is = in terms free-stream air velocity into the scoop, and that is the ref case = for the 50 sq.in. number. The case for an intake strategically located = behind the prop; that velocity is higher; and considerably higher during = takeoff roll.=20 =20 In the case of Tracy's setup, that is a significant factor. Still; I'd suggest he has an issue making a full power, extended climb on a 90F day without overheating the coolant. I think Tracy will agree (right, = Tracy?) =20 But he doesn't have to. As he says, after he has climbed to maybe = pattern altitude (or whatever) he can start pulling back the power. During that climb he also can take advantage of the heat capacity of the all the = steel, aluminum, and coolant. It takes a while to heat that up. And he has mentioned considering installing a water spray augment system. =20 There is good reason to have the minimum air intake that allows you to complete your climb - lower drag for the entire flight. At cruise, = running at, say, 55% power, the 28 sq in is entirely adequate. =20 Al =20 -----Original Message----- From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of George Lendich Sent: Friday, September 03, 2010 3:18 PM To: Rotary motors in aircraft Subject: [FlyRotary] Re: Radiator Math =20 Chris, From Tracy's response to Steve's request on Tracy's cooling = configuration and performance outcomes, it might reveal that his calculation for = cooling might look like this. =20 3 rotor=3D 270hp x 42.41=3D 11450.7 Btu x 66%(Mistral) =3D 7557.46Btu = water cooling. Al's figures might suggest 63% or 7213.5 Btu=20 =20 If Al's 50 sq" INLET cools for 4,800Btu, then ( extrapolated) 75 sq" = might cool for 7213.5Btu. Tracy's Rad 18x12x2.625 =3D 567cu" ( 216 sq") 25% ( 25% to 40% Rule) of 216 sq" is 54 sq" Tracy's Inlet 6"dia =3D 28.27 sq" or 13% of Rad face and only 38% of my extrapolation of Al's calculations. Tracy may have estimated 2.1 cu" per HP (as in your original = calculations). =20 Remembering that Tracy has been at this for a long time and come to his conclusions through trial and error and you have opted for a 2 pass 5" = thick rad, it's likely that your not comparing apples with apples in your = set-up. =20 However I do find it very interesting when Tracy is doing so well with = what might be considered the minimum requirements if well engineered cooling system. I have included them in my notes. George ( down under) =20 Chris, Sorry 180hp =3D 7633.8 BTU George (down under) =20 Chris, I went overmy notes again after reading other responses. Every hp =3D 42.41 BTU 180 BTU =3D 7633.8 (approx) Mistral suggests 2/3 (66%) heat rejection for water or 5,089.2 BTU Al suggests approx 4,800 or 63% ( I think Al is more accurate from = memory of previous maths examples), however not much in it. Suggested ( my notes) Rad size is 3 cu" per hp =3D 540 cu" Suggested inlet opening is 25 -40% of Rad surface area. If Rad is 2.5" thick =3D 216 sq" (18" x 12"). Then Inlet is 54 sq" - 86.2 sq" Al suggests 50 sq" or 23% approx. Bill suggest 20% min or 43 sq" I've used some rule of thumb figures to get to this ( bottom line) but I believe Al's maths may be more accurate. But from where I'm sitting they are pretty much saying the same thing. George ( down under) =20 Let me start with my old radiator. The core is 8 x 15 x 5. It is a = double pass radiator. My inlet is 36 sq in. I think the radiator is just too thick for good enough air flow at the speeds we are climbing and = cruising, not to mention ground ops. =20 So I geeked out, and did the math that I have found to figure out the correct size. Those who have done this before, please check my math. =20 I found the following requirements/suggestions during my research: =20 1.2 sq in of face per cubic in of displacement 1.2*3*39.9 =3D 143.64 sq in face 2.1 cubic in per HP 2.1*210 =3D 441 cubic in 2.48 cubic in per HP 2.48*210=3D521 cubic in Inlet should be 15% of face 36 sq in inlet =3D 240 sq in face Since my oil cooler is on the side, and works fine, I get to use the = entire bottom of the engine for the radiator. This gives me a max space of 16 = x 18 for 288 sq in face. Using a 2.5 in thick core 288*2.5 =3D 720 cubic in = of radiator. This should do the trick. I could even make it a little = smaller to ensure easy clearance. My inlet may be a little smaller than the radiator can handle, but I don't see how it can hurt to have a slightly larger radiator than the inlet can handle. =20 Along the way I found a reference that said the heat from 13.7 HP is = shed for every 1*C temp differential for every sq ft of intake. Assume = 200F(93C) coolant and a hot day, 90F(30C) I get: (13.7/144) inlet *63 =3D 210 so inlet =3D 35 sq in So up to now I'm feeling pretty good about the math, but please let me = know if I messed it up. =20 I will have to use the wedge shaped duct to move the air through the radiator. So to figure out the height of the duct from the radiator, I again used 15% of the facial area, then divided by the width. 16 x 18 x 15% =3D 43.2 43.2 / 16 =3D 2.7 So Do I really make the front of the wedge only 2.7 inches tall? This = seems pretty small. It would result in a really long, thin triangle. I have = a max of 6.5 inches available, so can easily make it bigger. =20 Thanks for the help. =20 Chris ------=_NextPart_000_000D_01CB4B89.00CD5310 Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable

There is another point to clarify regarding the inlet scoop area, which I probably should have made = earlier.  Since I fly a pusher; my thinking is in terms free-stream air velocity = into the scoop, and that is the ref case for the 50 sq.in. number. The case for = an intake strategically located behind the prop; that velocity is higher; = and considerably higher during takeoff roll.

 

In the case of = Tracy’s setup, that is a significant factor.  Still; I’d suggest he = has an issue making a full power, extended climb on a 90F day without = overheating the coolant. I think Tracy will = agree (right, Tracy?)

 

But he doesn’t have = to.  As he says, after he has climbed to maybe pattern altitude (or whatever) he = can start pulling back the power.  During that climb he also can take advantage of the heat capacity of the all the steel, aluminum, and = coolant.  It takes a while to heat that up. And he has mentioned considering = installing a water spray augment system.

 

There is good reason to have the = minimum air intake that allows you to complete your climb – lower drag for = the entire flight.  At cruise, running at, say, 55% power, the 28 sq in = is entirely adequate.

 

Al

 

-----Original = Message-----
From: Rotary motors in = aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of George Lendich
Sent: Friday, September = 03, 2010 3:18 PM
To: Rotary motors in = aircraft
Subject: [FlyRotary] Re: = Radiator Math

 

Chris,

From Tracy's response to = Steve's request on Tracy's cooling configuration and performance outcomes, it might reveal that his calculation for cooling might look like = this.

 

3 rotor=3D 270hp x 42.41=3D = 11450.7 Btu x 66%(Mistral) =3D 7557.46Btu water cooling. Al's figures might suggest = 63% or 7213.5 Btu 

 

If Al's 50 sq" INLET = cools for 4,800Btu, then ( extrapolated) 75 sq" might cool for  = 7213.5Btu.

Tracy's Rad 18x12x2.625 =3D 567cu" ( 216 sq")

25% ( 25% to 40% Rule) of = 216 sq" is 54 sq"

Tracy's Inlet 6"dia = =3D 28.27 sq" or 13% of Rad face and only 38% of my extrapolation of Al's calculations.

Tracy may have estimated = 2.1 cu" per HP (as in your original calculations).

 

Remembering that Tracy has = been at this for a long time and come to his conclusions through trial and error = and you have opted for a 2 pass 5" thick rad, it's likely that your not comparing apples with apples in your set-up.

 

However I do find it very interesting when Tracy is doing so well with what might be = considered the minimum requirements if well engineered cooling = system.

I have included them in my = notes.

George ( down = under)

=

 

Chris,

Sorry

180hp =3D 7633.8 = BTU

George (down = under)

=

 

Chris,

I went overmy notes again = after reading other responses.

Every hp =3D 42.41 = BTU

180 BTU =3D 7633.8 = (approx)

Mistral suggests 2/3 (66%) = heat rejection for water or 5,089.2 BTU

Al suggests approx 4,800 or = 63% ( I think Al is more accurate from memory of previous maths examples), = however not much in it.

Suggested ( my notes) Rad = size is 3 cu" per hp =3D 540 cu"

Suggested inlet opening is = 25 -40% of Rad surface area.

If Rad is 2.5" thick = =3D 216 sq" (18" x 12").

Then Inlet is 54 sq" - = 86.2 sq"

Al suggests 50 sq" or = 23% approx. Bill suggest 20% min or 43 sq"

I've used some rule of = thumb figures to get to this ( bottom line) but I believe Al's maths may be more = accurate.

But from where I'm sitting = they are pretty much saying the same thing.

George ( down = under)

 

=

Let me start with my old radiator.  The = core is 8 x 15 x 5.  It is a double pass radiator.  My inlet is 36 sq = in.  I think the radiator is just too thick for good enough air flow at the = speeds we are climbing and cruising, not to mention ground = ops.

 

So I geeked out, and did the math that I have = found to figure out the correct size.  Those who have done this before, = please check my math.

 

I found the following = requirements/suggestions during my research:

 

1.2 sq in of face per cubic in of = displacement

       &nbs= p;        1.2*3*39.9 =3D 143.64 sq in face

2.1 cubic in per HP

       &nbs= p;        2.1*210 =3D 441 cubic in

2.48 cubic in per HP

       &nbs= p;        2.48*210=3D521 cubic in

Inlet should be 15% of face

       &nbs= p;        36 sq in inlet =3D 240 sq in face

Since my oil cooler is on the side, and works = fine, I get to use the entire bottom of the engine for the radiator.  This = gives me a max space of 16 x 18 for 288 sq in face.  Using a 2.5 in thick = core 288*2.5 =3D 720 cubic in of radiator.  This should do the = trick.  I could even make it a little smaller to ensure easy clearance.  My = inlet may be a little smaller than the radiator can handle, but I don’t = see how it can hurt to have a slightly larger radiator than the inlet can = handle.

 

Along the way I found a reference that said = the heat from 13.7 HP is shed for every 1*C temp differential for every sq ft of intake.  Assume 200F(93C) coolant and a hot day, 90F(30C) I = get:

       &nbs= p;        (13.7/144) inlet *63 =3D 210  so inlet =3D 35 sq = in

So up to now I’m feeling pretty good = about the math, but please let me know if I messed it up.

 

I will have to use the wedge shaped duct to = move the air through the radiator.  So to figure out the height of the duct = from the radiator, I again used 15% of the facial area, then divided by the = width.

       &nbs= p;        16 x 18 x 15% =3D 43.2

       &nbs= p;        43.2 / 16 =3D 2.7

So Do I really make the front of the wedge = only 2.7 inches tall?  This seems pretty small.  It would result in a = really long, thin triangle.  I have a max of 6.5 inches available, so can = easily make it bigger.

 

Thanks for the help.

 

Chris

------=_NextPart_000_000D_01CB4B89.00CD5310--