X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from poplet2.per.eftel.com ([203.24.100.45] verified) by logan.com (CommuniGate Pro SMTP 5.3.9) with ESMTP id 4455184 for flyrotary@lancaironline.net; Fri, 03 Sep 2010 19:18:10 -0400 Received-SPF: none receiver=logan.com; client-ip=203.24.100.45; envelope-from=lendich@aanet.com.au Received: from sv1-1.aanet.com.au (mail.aanet.com.au [203.24.100.34]) by poplet2.per.eftel.com (Postfix) with ESMTP id 86E917D235 for ; Sat, 4 Sep 2010 07:17:35 +0800 (WST) Received: from ownerf1fc517b8 (203.171.92.134.static.rev.aanet.com.au [203.171.92.134]) by sv1-1.aanet.com.au (Postfix) with SMTP id E4798C5973 for ; Sat, 4 Sep 2010 07:17:34 +0800 (WST) Message-ID: <9C46032DEE8D4D94B4E775921AC999E7@ownerf1fc517b8> From: "George Lendich" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Radiator Math Date: Sat, 4 Sep 2010 09:17:38 +1000 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0057_01CB4C12.04D74450" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.5931 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5931 X-Antivirus: avast! (VPS 100830-1, 08/30/2010), Outbound message X-Antivirus-Status: Clean This is a multi-part message in MIME format. ------=_NextPart_000_0057_01CB4C12.04D74450 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable 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=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). 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. =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_0057_01CB4C12.04D74450 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Chris,
From Tracy's response to Steve's = request on Tracy's=20 cooling configuration and performance outcomes, it might reveal = that his=20 calculation for cooling might look like this.
 
3 rotor=3D 270hp x 42.41=3D 11450.7 Btu = x 66%(Mistral)=20 =3D 7557.46Btu water cooling. Al's figures might suggest 63% or 7213.5=20 Btu 
 
If Al's 50 sq" INLET cools for = 4,800Btu, then (=20 extrapolated) 75 sq" might cool for  7213.5Btu.
Tracy's Rad 18x12x2.625 =3D 567cu" ( = 216=20 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=20 and only 38% of my extrapolation of Al's calculations.
Tracy may have estimated 2.1 cu" per HP = (as in your=20 original calculations).
 
Remembering that Tracy has been at this = for a long=20 time and come to his conclusions through trial and error and you have = opted for=20 a 2 pass 5" thick rad, it's likely that your not comparing apples with = apples in=20 your set-up.
 
However I do find it very interesting = when Tracy is=20 doing so well with what might be considered the minimum = requirements if=20 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=20 responses.
Every hp =3D 42.41 BTU
180 BTU =3D 7633.8 = (approx)
Mistral suggests 2/3 (66%) heat = rejection for=20 water or 5,089.2 BTU
Al suggests approx 4,800 or 63% ( I = think Al is=20 more accurate from memory of previous maths examples), however not = much in=20 it.
Suggested ( my notes) Rad size is 3 = cu" per hp=20 =3D 540 cu"
Suggested inlet opening is 25 -40% = of Rad=20 surface area.
If Rad is 2.5" thick =3D 216 sq" = (18" x=20 12").
Then Inlet is 54 sq" - 86.2 = sq"
Al suggests 50 sq" or 23% approx. = Bill suggest=20 20% min or 43 sq"
I've used some rule of thumb = figures to get to=20 this ( bottom line) but I believe Al's maths may be more=20 accurate.
But from where I'm sitting they are = pretty much=20 saying the same thing.
George ( down under)
 

Let me start with my old radiator.  The = core is 8=20 x 15 x 5.  It is a double pass radiator.  My inlet is 36 = sq=20 in.  I think the radiator is just too thick for good enough = air flow=20 at the speeds we are climbing and cruising, not to mention ground=20 ops.

 

So I geeked out, and did the math that I have = found to=20 figure out the correct size.  Those who have done this = before, please=20 check my math.

 

I found the following = requirements/suggestions during=20 my research:

 

1.2 sq in of face per cubic in of=20 displacement

         &= nbsp;     =20 1.2*3*39.9 =3D 143.64 sq in face

2.1 cubic in per HP

         &= nbsp;     =20 2.1*210 =3D 441 cubic in

2.48 cubic in per HP

         &= nbsp;     =20 2.48*210=3D521 cubic in

Inlet should be 15% of face

         &= nbsp;     =20 36 sq in inlet =3D 240 sq in face

Since my oil cooler is on the side, and works = fine, I=20 get to use the entire bottom of the engine for the radiator.  = This=20 gives me a max space of 16 x 18 for 288 sq in face.  Using a = 2.5 in=20 thick core 288*2.5 =3D 720 cubic in of radiator.  This should = do the=20 trick.  I could even make it a little smaller to ensure easy=20 clearance.  My inlet may be a little smaller than the = radiator can=20 handle, but I don=92t see how it can hurt to have a slightly = larger radiator=20 than the inlet can handle.

 

Along the way I found a reference that said = the heat=20 from 13.7 HP is shed for every 1*C temp differential for every sq = ft of=20 intake.  Assume 200F(93C) coolant and a hot day, 90F(30C) I=20 get:

         &= nbsp;     =20 (13.7/144) inlet *63 =3D 210  so inlet =3D 35 sq = in

So up to now I=92m feeling pretty good about = the math,=20 but please let me know if I messed it up.

 

I will have to use the wedge shaped duct to = move the=20 air through the radiator.  So to figure out the height of the = duct=20 from the radiator, I again used 15% of the facial area, then = divided by=20 the width.

         &= nbsp;     =20 16 x 18 x 15% =3D 43.2

         &= nbsp;     =20 43.2 / 16 =3D 2.7

So Do I really make the front of the wedge = only 2.7=20 inches tall?  This seems pretty small.  It would result = in a=20 really long, thin triangle.  I have a max of 6.5 inches = available, so=20 can easily make it bigger.

 

Thanks for the help.

 

Chris

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