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 4454080 for flyrotary@lancaironline.net; Thu, 02 Sep 2010 22:02:39 -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 71FFB7D4AB for ; Fri, 3 Sep 2010 10:02:02 +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 DC78BC59AE for ; Fri, 3 Sep 2010 10:02:01 +0800 (WST) Message-ID: <9237713658D144A68793A9BA827B0D9E@ownerf1fc517b8> From: "George Lendich" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Radiator Math Date: Fri, 3 Sep 2010 12:02:05 +1000 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0027_01CB4B5F.D3650360" 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_0027_01CB4B5F.D3650360 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Chris, I went over my 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_0027_01CB4B5F.D3650360 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Chris,
I went over my 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 water=20 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 =3D=20 540 cu"
Suggested inlet opening is 25 -40% of = Rad surface=20 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%=20 min or 43 sq"
I've used some rule of thumb figures to = get to this=20 ( bottom line) but I believe Al's maths may be more = 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 x 15=20 x 5.  It is a double pass radiator.  My inlet is 36 sq = in.  I=20 think the radiator is just too thick for good enough air flow at the = speeds we=20 are climbing and cruising, not to mention ground 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 my=20 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 get=20 to use the entire bottom of the engine for the radiator.  This = gives me a=20 max space of 16 x 18 for 288 sq in face.  Using a 2.5 in thick = core=20 288*2.5 =3D 720 cubic in of radiator.  This should do the = trick.  I=20 could even make it a little smaller to ensure easy clearance.  My = inlet=20 may be a little smaller than the radiator can handle, but I don=92t = see how it=20 can hurt to have a slightly larger radiator than the inlet can=20 handle.

 

Along the way I found a reference that said the = heat from=20 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, but=20 please let me know if I messed it up.

 

I will have to use the wedge shaped duct to move = the air=20 through the radiator.  So to figure out the height of the duct = from the=20 radiator, I again used 15% of the facial area, then divided by the=20 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 inches=20 tall?  This seems pretty small.  It would result in a really = long,=20 thin triangle.  I have a max of 6.5 inches available, so can = easily make=20 it bigger.

 

Thanks for the help.

 

Chris

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