X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from mailrelay.embarq.synacor.com ([208.47.184.3] verified) by logan.com (CommuniGate Pro SMTP 5.3.9) with ESMTP id 4451399 for flyrotary@lancaironline.net; Tue, 31 Aug 2010 22:40:31 -0400 Received-SPF: pass receiver=logan.com; client-ip=208.47.184.3; envelope-from=candtmallory@embarqmail.com Return-Path: DKIM-Signature: v=1; a=rsa-sha1; d=embarqmail.com; s=s012408; c=relaxed/simple; q=dns/txt; i=@embarqmail.com; t=1283308796; h=From:Subject:Date:To:MIME-Version:Content-Type; bh=1ocrmoxzmq9khMvwawhzlJPteVo=; b=KvYdi3rZvxTbJFunwylnTXVifwFN4o+mO3z25/wvkFVPzZAOJulPRPnC0nmFs/PN c/4TbCnRAd/nNhIPCFUZWnqHZKzekmTIsyQ37QqyAV73srqyrVHH8Ys8KMG/A0VE; X-BINDING: X-Spam-Rating: None X_CMAE_Category: 0,0 Undefined,Undefined X-CNFS-Analysis: v=1.1 cv=5UfXMjiySVA2/TZBxb2XJXmM2GV5+Di6hbsGaPh/FUE= c=1 sm=0 a=veypUrOAYd02yeQPBQVYUA==:17 a=7KbRZ2D8I6najw4jbF4A:9 a=e8ollgF-v3-98x0msNoA:7 a=aL0CUduYkpTfnr95iwckMDwrWrUA:4 a=CjuIK1q_8ugA:10 a=ocQleZY4KRtNrMYq:21 a=zd4cdW5T35VOAh8b:21 a=yMhMjlubAAAA:8 a=SSmOFEACAAAA:8 a=YIv5yD4nKJzdKgCE7YQA:9 a=49DHyN7uFpRbsr608QsA:7 a=p_uRSL4wLutRk_208-BGbNc39V4A:4 a=veypUrOAYd02yeQPBQVYUA==:117 X-CM-Score: 0 X-Scanned-by: Cloudmark Authority Engine Authentication-Results: smtp02.embarq.synacor.com smtp.user=candtmallory@embarqmail.com; auth=pass (LOGIN) Received: from [71.0.26.31] ([71.0.26.31:16643] helo=AcerPC) by mailrelay.embarq.synacor.com (envelope-from ) (ecelerity 2.2.2.40 r(29895/29896)) with ESMTPA id 0C/77-01937-CFCBD7C4; Tue, 31 Aug 2010 22:39:56 -0400 From: "Chris and Terria" To: "'Rotary motors in aircraft'" Subject: Radiator Math Date: Tue, 31 Aug 2010 22:39:52 -0400 Message-ID: <000601cb497e$f47498a0$dd5dc9e0$@com> MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0007_01CB495D.6D62F8A0" X-Mailer: Microsoft Office Outlook 12.0 Thread-Index: ActJfvQSMjCzu9PaTaiKJiDqBbyjTQ== Content-Language: en-us This is a multi-part message in MIME format. ------=_NextPart_000_0007_01CB495D.6D62F8A0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit 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 1.2*3*39.9 = 143.64 sq in face 2.1 cubic in per HP 2.1*210 = 441 cubic in 2.48 cubic in per HP 2.48*210=521 cubic in Inlet should be 15% of face 36 sq in inlet = 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 = 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: (13.7/144) inlet *63 = 210 so inlet = 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. 16 x 18 x 15% = 43.2 43.2 / 16 = 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_0007_01CB495D.6D62F8A0 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

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

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

2.1 cubic in per HP

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

2.48 cubic in per HP

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

Inlet should be 15% of face

         &= nbsp;      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:

         &= nbsp;      (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.

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

         &= nbsp;      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

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