X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from bay0-omc2-s17.bay0.hotmail.com ([65.54.190.92] verified) by logan.com (CommuniGate Pro SMTP 5.3.9) with ESMTP id 4451966 for flyrotary@lancaironline.net; Wed, 01 Sep 2010 10:15:32 -0400 Received-SPF: pass receiver=logan.com; client-ip=65.54.190.92; envelope-from=stol83001@live.com Received: from BAY143-W24 ([65.54.190.125]) by bay0-omc2-s17.bay0.hotmail.com with Microsoft SMTPSVC(6.0.3790.4675); Wed, 1 Sep 2010 07:14:56 -0700 Message-ID: Return-Path: stol83001@live.com Content-Type: multipart/alternative; boundary="_7c134d5a-7b90-423e-9d89-8b5185f34b0a_" X-Originating-IP: [75.174.174.132] From: ben haas To: Subject: RE: [FlyRotary] Radiator Math Date: Wed, 1 Sep 2010 08:14:56 -0600 Importance: Normal In-Reply-To: References: MIME-Version: 1.0 X-OriginalArrivalTime: 01 Sep 2010 14:14:56.0610 (UTC) FILETIME=[0DF8C420:01CB49E0] --_7c134d5a-7b90-423e-9d89-8b5185f34b0a_ Content-Type: text/plain; charset="Windows-1252" Content-Transfer-Encoding: quoted-printable Most cooling issues are outlet related=2C not inlet sizing...... =20 Think of air as a chain.. You can pull it alot easier that you can push it. =20 Your set up is even more critical as the thick core can only breath so much= through its dense fins. You need a large pressure drop on the discharge si= de to draw air through it... =20 IMHO Ben Haas www.haaspowerair.com =20 To: flyrotary@lancaironline.net From: candtmallory@embarqmail.com Date: Tue=2C 31 Aug 2010 22:39:52 -0400 Subject: [FlyRotary] Radiator Math 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 th= ick for good enough air flow at the speeds we are climbing and cruising=2C = not to mention ground ops. =20 So I geeked out=2C and did the math that I have found to figure out the cor= rect size. Those who have done this before=2C 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=2C and works fine=2C I get to use the en= tire bottom of the engine for the radiator. This gives me a max space of 1= 6 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 th= e radiator can handle=2C but I don=92t see how it can hurt to have a slight= ly 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 f= or every 1*C temp differential for every sq ft of intake. Assume 200F(93C)= coolant and a hot day=2C 90F(30C) I get: (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=2C 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 radiat= or. So to figure out the height of the duct from the radiator=2C I again u= sed 15% of the facial area=2C 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 seem= s pretty small. It would result in a really long=2C thin triangle. I have= a max of 6.5 inches available=2C so can easily make it bigger. =20 Thanks for the help. =20 Chris = --_7c134d5a-7b90-423e-9d89-8b5185f34b0a_ Content-Type: text/html; charset="Windows-1252" Content-Transfer-Encoding: quoted-printable Most cooling issues are outlet related=2C not inlet sizing......
 =3B
Think of air as a chain.. You can pull it alot easier that you can push it.=
 =3B
Your set up is even more critical as the thick core can only breath so much= through its dense fins. You need a large pressure drop on the discharge si= de to draw air through it...
 =3B
IMHO

Ben Haas
www.haaspowera= ir.com



 =3B
To: flyrotary@lancaironline.net
From: candtmallory@embarqmail.com
Dat= e: Tue=2C 31 Aug 2010 22:39:52 -0400
Subject: [FlyRotary] Radiator Math<= BR>

Let me start with my old radiator. =3B The core= is 8 x 15 x 5. =3B It is a double pass radiator. =3B My inlet is 3= 6 sq in. =3B I think the radiator is just too thick for good enough air= flow at the speeds we are climbing and cruising=2C not to mention ground o= ps.

 =3B

So I geeked out=2C and did the math that I have fou= nd to figure out the correct size. =3B Those who have done this before= =2C please check my math.

 =3B

I found the following requirements/suggestions duri= ng my research:

 =3B

1.2 sq in of face per cubic in of displacement

 =3B =3B =3B =3B =3B =3B&nb= sp=3B =3B =3B =3B =3B =3B =3B =3B =3B 1.2*3= *39.9 =3D 143.64 sq in face

2.1 cubic in per HP

 =3B =3B =3B =3B =3B =3B&nb= sp=3B =3B =3B =3B =3B =3B =3B =3B =3B 2.1*2= 10 =3D 441 cubic in

2.48 cubic in per HP

 =3B =3B =3B =3B =3B =3B&nb= sp=3B =3B =3B =3B =3B =3B =3B =3B =3B 2.48*= 210=3D521 cubic in

Inlet should be 15% of face

 =3B =3B =3B =3B =3B =3B&nb= sp=3B =3B =3B =3B =3B =3B =3B =3B =3B 36 sq= in inlet =3D 240 sq in face

Since my oil cooler is on the side=2C and works fin= e=2C I get to use the entire bottom of the engine for the radiator. =3B= This gives me a max space of 16 x 18 for 288 sq in face. =3B Using a 2= .5 in thick core 288*2.5 =3D 720 cubic in of radiator. =3B This should = do the trick. =3B I could even make it a little smaller to ensure easy = clearance. =3B My inlet may be a little smaller than the radiator can h= andle=2C but I don=92t see how it can hurt to have a slightly larger radiat= or than the inlet can handle.

 =3B

Along the way I found a reference that said the hea= t from 13.7 HP is shed for every 1*C temp differential for every sq ft of i= ntake. =3B Assume 200F(93C) coolant and a hot day=2C 90F(30C) I get:

 =3B =3B =3B =3B =3B =3B&nb= sp=3B =3B =3B =3B =3B =3B =3B =3B =3B (13.7= /144) inlet *63 =3D 210 =3B so inlet =3D 35 sq in

So up to now I=92m feeling pretty good about the ma= th=2C but please let me know if I messed it up.

 =3B

I will have to use the wedge shaped duct to move th= e air through the radiator. =3B So to figure out the height of the duct= from the radiator=2C I again used 15% of the facial area=2C then divided b= y the width.

 =3B =3B =3B =3B =3B =3B&nb= sp=3B =3B =3B =3B =3B =3B =3B =3B =3B 16 x = 18 x 15% =3D 43.2

 =3B =3B =3B =3B =3B =3B&nb= sp=3B =3B =3B =3B =3B =3B =3B =3B =3B 43.2 = / 16 =3D 2.7

So Do I really make the front of the wedge only 2.7= inches tall? =3B This seems pretty small. =3B It would result in a= really long=2C thin triangle. =3B I have a max of 6.5 inches available= =2C so can easily make it bigger.

 =3B

Thanks for the help.

 =3B

Chris

= --_7c134d5a-7b90-423e-9d89-8b5185f34b0a_--