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 4451602 for flyrotary@lancaironline.net; Wed, 01 Sep 2010 03:37:59 -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 2322A7D17A for ; Wed, 1 Sep 2010 15:37:22 +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 83E04C5B11 for ; Wed, 1 Sep 2010 15:37:21 +0800 (WST) Message-ID: From: "George Lendich" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Radiator Math Date: Wed, 1 Sep 2010 17:37:23 +1000 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0017_01CB49FC.5644B240" 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_0017_01CB49FC.5644B240 Content-Type: text/plain; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Chris, My notes going back some time suggest, 30 to 35% of frontal area (of = the rad surface area) are the inlet requirements. Also 3 to 3.5 cu" for each HP - I believe that Tracy runs with less. There are different inlet requirements for air cooled and water cooled = as the air cooled runs at a higher Delta T. George ( down under) In a message dated 8/31/2010 7:40:36 PM Pacific Daylight Time, = candtmallory@embarqmail.com writes: 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=E2=80=99t 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=E2=80=99m 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 Chris, If I recall correctly the wedge duct requires a bit more inlet than = that about 20% is minimum. More important than the inlet, the area = behind the radiator is critical. You need 4 inches of unobstructed area = if using a wedge duct and more is better. Bill Jepson ------=_NextPart_000_0017_01CB49FC.5644B240 Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable =EF=BB=BF
 
 Chris,
My notes going back some time suggest, = 30 to 35%=20 of frontal area (of the rad surface area) are the inlet = requirements.
Also 3 to 3.5 cu" for each HP - I = believe that=20 Tracy runs with less.
There are different inlet requirements = for air=20 cooled and water cooled as the air cooled runs at a higher Delta = T.
George ( down under)
 
In a message dated 8/31/2010 7:40:36 PM Pacific Daylight Time, candtmallory@embarqmail.com=20 writes:

Let me start with my old radiator.  The = core is 8 x=20 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 at=20 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 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=20 a 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=E2=80=99t 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=E2=80=99m 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=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

Chris,
If I recall correctly the wedge duct requires a bit more inlet = than that=20 about 20% is minimum. More important than the inlet, the area behind = the=20 radiator is critical. You need 4 inches of unobstructed area if = using a=20 wedge duct and more is better.
Bill Jepson
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