X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from fed1rmmtao107.cox.net ([68.230.241.39] verified) by logan.com (CommuniGate Pro SMTP 5.3.9) with ESMTP id 4452070 for flyrotary@lancaironline.net; Wed, 01 Sep 2010 11:17:18 -0400 Received-SPF: none receiver=logan.com; client-ip=68.230.241.39; envelope-from=alventures@cox.net Received: from fed1rmimpo03.cox.net ([70.169.32.75]) by fed1rmmtao107.cox.net (InterMail vM.8.00.01.00 201-2244-105-20090324) with ESMTP id <20100901151641.WDNQ3990.fed1rmmtao107.cox.net@fed1rmimpo03.cox.net> for ; Wed, 1 Sep 2010 11:16:41 -0400 Received: from BigAl ([72.199.216.236]) by fed1rmimpo03.cox.net with bizsmtp id 1TGg1f00X56cS2o04TGgZr; Wed, 01 Sep 2010 11:16:41 -0400 X-VR-Score: 0.00 X-Authority-Analysis: v=1.1 cv=e4n1ZjYY/MidH6WTRS5qGD+cSaIt2UqZNViAts1RHs8= c=1 sm=1 a=Bu6Sa6AsMZ4A:10 a=lN8H/RjlhkCyIsyuOn2r7w==:17 a=Ia-xEzejAAAA:8 a=k7PCEpV72_DKBWIWr9EA:9 a=B3z0WBLlw17V3zjC6nMA:7 a=yzeNd0JuuxF3VWWbBs4cen3SJ7UA:4 a=wPNLvfGTeEIA:10 a=EzXvWhQp4_cA:10 a=kDj8S-WyppzUBPpn:21 a=oyp9ORgP1ENq9ofQ:21 a=yMhMjlubAAAA:8 a=ez5uSvjFezxBBMak_yUA:9 a=LJvEv-sTGY5L0ryrvPcA:7 a=o-mlurwcyehgvZpWTkgirr1oXo4A:4 a=lN8H/RjlhkCyIsyuOn2r7w==:117 X-CM-Score: 0.00 Authentication-Results: cox.net; none From: "Al Gietzen" To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Radiator Math Date: Wed, 1 Sep 2010 08:17:18 -0800 Message-ID: <627CE15F511845CAB4770EAC4073CFFF@BigAl> MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0000_01CB49AE.184DF350" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.6863 Importance: Normal Thread-Index: ActJfw234xoPlB+6QgeI/Oh/SAvzTgAbatMg In-Reply-To: X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5931 This is a multi-part message in MIME format. ------=_NextPart_000_0000_01CB49AE.184DF350 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Here=92s a brief summary of cooling air intake size; a.k.a., air flow requirements. The inlet area requirement is not determined by first choosing the = radiator area. It is computed from the engine power, and speed of the aircraft. = The engine power determines the amount of heat to be rejected, and the speed determines how much air can flow into the scoop. =20 Let=92s assume your engine output in full power climb (which is the = highest demand on the cooling system) is 180 hp. (Skipping a couple of steps) = The amount of heat transferred to the coolant, and hence to the air, is = about 4800 BTU/min. Generally we=92d like to design for an air temp rise = through the rad of 50 =96 80 F. Knowing the specific heat of air we can compute the = air flow. Assuming a 70F DT; it is about 3500 cfm. =20 Now let=92s assume your climb speed is 90 kts. Knowing the volume of = air, and the speed it is coming into the scoop, we can compute the inlet area of about 80 sq. in. =96 assuming a very effective scoop. That=92s a good = starting point; but you can scale that to some different assumption; e.g., you = might consider that your actual hp for climb-out is something less due to altitude/temp .etc. The inlet area will scale directly with power or aircraft speed. =20 Now you can consider how much you need to expand (thereby slowing) the = air to ensure that you have sufficient static pressure to overcome the = pressure drop of the core. For a typical rad of maybe 2 =BD=94 thickness and 16 = fins/in a ratio of roughly 1:4 has been shown to be pretty good. For a thicker = rad you may need more pressure recovery (more expansion) to overcome a = bigger pressure drop; a thinner rad can take a lower ratio. So here you see a dilemma =96 larger ratio, bigger rad face area =96 hum-m-m, leads to = thinner rad for same volume. All this assumes that the pressure at the rad exit is ambient or lower. =20 I skipped all the formulas cuz this is all the time I have for this = right now. But maybe gives different insight. =20 Al G =20 -----Original Message----- From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Chris and Terria Sent: Tuesday, August 31, 2010 6:40 PM To: Rotary motors in aircraft Subject: [FlyRotary] Radiator Math =20 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=92t 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=92m 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_0000_01CB49AE.184DF350 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable

Here’s a brief summary of = cooling air intake size; a.k.a., air flow requirements.

The inlet area requirement is not determined by first choosing the radiator area. =A0It is computed from = the engine power, and speed of the aircraft.=A0 The engine power determines the = amount of heat to be rejected, and the speed determines how much air can flow into = the scoop.

 

Let’s assume your engine = output in full power climb (which is the highest demand on the cooling system) is = 180 hp. =A0(Skipping a couple of steps) The amount of heat transferred to the = coolant, and hence to the air, is about 4800 BTU/min. Generally we’d like = to design for an air temp rise through the rad of 50 – 80 F. Knowing = the specific heat of air we can compute the air flow.=A0 Assuming a 70F = DT; it is about 3500 cfm.

 

Now let’s assume your climb = speed is 90 kts. Knowing the volume of air, and the speed it is coming into = the scoop, we can compute the inlet area of about 80 sq. in. – = assuming a very effective scoop.=A0 That’s a good starting point; but you can = scale that to some different assumption; e.g., you might consider that your = actual hp for climb-out is something less due to altitude/temp .etc.=A0 The inlet = area will scale directly with power or aircraft speed.

 

Now you can consider how much you = need to expand (thereby slowing) the air to ensure that you have sufficient = static pressure to overcome the pressure drop of the core. =A0For a typical rad = of maybe 2 =BD” thickness and 16 fins/in a ratio of roughly 1:4 has been = shown to be pretty good.=A0 For a thicker rad you may need more pressure recovery = (more expansion) to overcome a bigger pressure drop; a thinner rad can take a = lower ratio.=A0 So here you see a dilemma – larger ratio, bigger rad = face area – hum-m-m, leads to thinner rad for same = volume.=A0 All this assumes that the pressure at the rad exit is ambient or = lower.

 

I skipped all the formulas cuz = this is all the time I have for this right now. But maybe gives different = insight.

 

Al G

 

-----Original = Message-----
From: Rotary motors in = aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Chris and Terria
Sent: Tuesday, August 31, = 2010 6:40 PM
To: Rotary motors in = aircraft
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 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

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

2.1 cubic in per HP

       &nbs= p;        2.1*210 =3D 441 cubic in

2.48 cubic in per HP

       &nbs= p;        2.48*210=3D521 cubic in

Inlet should be 15% of face

       &nbs= p;        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:

       &nbs= p;        (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.

       &nbs= p;        16 x 18 x 15% =3D 43.2

       &nbs= p;        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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