X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from web83910.mail.sp1.yahoo.com ([69.147.92.115] verified) by logan.com (CommuniGate Pro SMTP 5.3.9) with SMTP id 4452372 for flyrotary@lancaironline.net; Wed, 01 Sep 2010 15:19:52 -0400 Received-SPF: none receiver=logan.com; client-ip=69.147.92.115; envelope-from=keltro@att.net Received: (qmail 77037 invoked by uid 60001); 1 Sep 2010 19:12:19 -0000 DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=att.net; s=s1024; t=1283368339; bh=MXKq1tCPKjpESHzihgROKsWEXdrj1S0QIiNNRvRR1RQ=; h=Message-ID:X-YMail-OSG:Received:X-Mailer:References:Date:From:Subject:To:In-Reply-To:MIME-Version:Content-Type; b=VMi4q7KQSoDbibUWC2is1yf7emz0wc6CjnOsN9qCsrZn2P0FPFoOaPkp3hDWWaVNUHsB7fjAhWoJkhx+0bU9MHbL5CTaKqwJjC7bEY66yLqfOfXLm8dfY2ECbBT9UFa7/amJji3N0SERjKXJWcDjPNIs0gPheGrMMJI/3mkn5x8= DomainKey-Signature:a=rsa-sha1; q=dns; c=nofws; s=s1024; d=att.net; h=Message-ID:X-YMail-OSG:Received:X-Mailer:References:Date:From:Subject:To:In-Reply-To:MIME-Version:Content-Type; b=3gyCMV8FzyS7uKGuPic8y7GZxKSeHqUK45o9sEIk6/jqdiAH2KVPnUEcsGA8D+HF+JcuAB0Tz4HlirRzc5cca1kdJukXLc0Gp1PvPmo7iJdGNZ9qjhDEyp4NW7DrGikc/OUVEFKIQZf2fKeLGlrrC++CedX3wpZO956RLCMXPHk=; Message-ID: <804134.74978.qm@web83910.mail.sp1.yahoo.com> X-YMail-OSG: 7LDs7.oVM1ld48WUxaI44lR1kblrAAORDilJPlH2l9a1tsl J_TlPby.0IkeCSgRTnxccyBaJh8t7j4dqtVZsgkvYUk2LxGgfW6I19OVoy3k NWb6VU99swgXtB_laxw.2EMVv4b2UJTTLKFms9hpvdnMp4KF0tJlIml0NzSN e9CLygfjCcYresO3QlxQaesMg6OyMsoopGz.RRexjOwMvhTqlXewDxHxSwZl PrnizQCLu34EYt2R8bw0Twl25uu5Ljq3zeKiqIOexDtnSNbZdgNvs3s5gGQ6 xnPmQH0J5.9St2Wq0S3xKEbFKZ0fEmOMCxjFVWW_KCLgjmcfpffCoF4Bq_Om c6ujpKsIrkEbw8iBXzDoMmdHng5Ta81QGHxbcto4- Received: from [208.114.46.32] by web83910.mail.sp1.yahoo.com via HTTP; Wed, 01 Sep 2010 12:12:19 PDT X-Mailer: YahooMailRC/470 YahooMailWebService/0.8.105.279950 References: Date: Wed, 1 Sep 2010 12:12:19 -0700 (PDT) From: Kelly Troyer Subject: Re: Radiator Math To: Rotary motors in aircraft In-Reply-To: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="0-91696725-1283368339=:74978" --0-91696725-1283368339=:74978 Content-Type: text/plain; charset=utf-8 Content-Transfer-Encoding: quoted-printable Chris I would use 25% minimum facial area for this type duct (4.5 inches) t= hen =0Ablock=0Aprogressively during flight tests to find out how small an i= nlet you can live =0Awith and=0Aas Bill Jepson said get as much unobstructe= d=C2=A0 area on outlet side of the =0Aradiator as =0A=0Ayou can............= .FWIW=C2=A0=0AKelly Troyer=0A"Dyke Delta"_13B ROTARY Engine=0A"RWS"_RD1C/EC= 2/EM2=0A"Mistral"_Backplate/Oil Manifold =0A=0A=0A=0A=0A___________________= _____________=0AFrom: "WRJJRS@aol.com" =0ATo: Rotary motors= in aircraft =0ASent: Tue, August 31, 2010 11:= 21:39 PM=0ASubject: [FlyRotary] Re: Radiator Math=0A=0AIn a message dated 8= /31/2010 7:40:36 PM Pacific Daylight Time, =0Acandtmallory@embarqmail.com w= rites:=0ALet me start with my old radiator.=C2=A0 The core is 8 x 15 x 5.= =C2=A0 It is a double pass =0Aradiator.=C2=A0 My inlet is 36 sq in.=C2=A0 I= think the radiator is just too thick for =0Agood enough air flow at the sp= eeds we are climbing and cruising, not to mention =0Aground ops.=0A>=C2=A0= =0A>So I geeked out, and did the math that I have found to figure out the c= orrect =0A>size.=C2=A0 Those who have done this before, please check my mat= h.=0A>=C2=A0=0A>I found the following requirements/suggestions during my re= search:=0A>=C2=A0=0A>1.2 sq in of face per cubic in of displacement=0A>=C2= =A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0= =C2=A0=C2=A0 1.2*3*39.9 =3D 143.64 sq in face=0A>2.1 cubic in per HP=0A>=C2= =A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0= =C2=A0=C2=A0 2.1*210 =3D 441 cubic in=0A>2.48 cubic in per HP=0A>=C2=A0=C2= =A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0= =C2=A0 2.48*210=3D521 cubic in=0A>Inlet should be 15% of face=0A>=C2=A0=C2= =A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0= =C2=A0 36 sq in inlet =3D 240 sq in face=0A>Since my oil cooler is on the s= ide, and works fine, I get to use the entire =0A>bottom of the engine for t= he radiator.=C2=A0 This gives me a max space of 16 x 18 for =0A>288 sq in f= ace.=C2=A0 Using a 2.5 in thick core 288*2.5 =3D 720 cubic in of radiator.= =C2=A0 =0A>This should do the trick.=C2=A0 I could even make it a little sm= aller to ensure easy =0A>clearance.=C2=A0 My inlet may be a little smaller = than the radiator can handle, but I =0A>don=E2=80=99t see how it can hurt t= o have a slightly larger radiator than the inlet can =0A>handle.=0A>=C2=A0= =0A>Along the way I found a reference that said the heat from 13.7 HP is sh= ed for =0A>every 1*C temp differential for every sq ft of intake.=C2=A0 Ass= ume 200F(93C) coolant =0A>and a hot day, 90F(30C) I get:=0A>=C2=A0=C2=A0=C2= =A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0= (13.7/144) inlet *63 =3D 210=C2=A0 so inlet =3D 35 sq in=0A>So up to now I= =E2=80=99m feeling pretty good about the math, but please let me know if I = =0A>messed it up.=0A>=C2=A0=0A>I will have to use the wedge shaped duct to = move the air through the radiator.=C2=A0 =0A>So to figure out the height of= the duct from the radiator, I again used 15% of =0A>the facial area, then = divided by the width.=0A>=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0= =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 16 x 18 x 15% =3D 43.2=0A>=C2=A0= =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2= =A0=C2=A0 43.2 / 16 =3D 2.7=0A>So Do I really make the front of the wedge o= nly 2.7 inches tall?=C2=A0 This seems =0A>pretty small.=C2=A0 It would resu= lt in a really long, thin triangle.=C2=A0 I have a max of =0A>6.5 inches av= ailable, so can easily make it bigger.=0A>=C2=A0=0A>Thanks for the help.=0A= >=C2=A0=0A>Chris=0AChris,=0AIf I recall correctly the wedge duct requires a= bit more inlet than that about =0A20% is minimum. More important than the = inlet, the area behind the radiator is =0Acritical. You need 4 inches of un= obstructed area if using a wedge duct and more =0Ais better.=0ABill Jepson --0-91696725-1283368339=:74978 Content-Type: text/html; charset=utf-8 Content-Transfer-Encoding: quoted-printable
=0A
Chris I would use 25% minimum facial area for this = type duct (4.5 inches) then block
=0A
progressively during flight = tests to find out how small an inlet you can live with and
=0A
as = Bill Jepson said get as much unobstructed  area on outlet side of the = radiator as
=0A
you can.............FWIW 
=0A
 = ;
Kelly Troyer
"Dyke Delta"_13B ROTARY Engine
"RWS"_RD1C/EC2/EM2=
"Mistral"_Backplate/Oil Manifold =0A

=0A

=0A
= =0A
=0AFrom: "WRJJRS@aol.com" <WRJJRS@aol.com>
To: Rotary motors in aircraft &l= t;flyrotary@lancaironline.net>
S= ent: Tue, August 31, 2010 11:21:39 PM
Subject: [FlyRotary] Re: Radiator Math
<= BR>=0A
= =0A
In a message dated 8/31/2010 7:40:36 PM Pacific Daylight Time, cand= tmallory@embarqmail.com writes:
=0A
=0A
=0A

Let me start with my old radiator.&= nbsp; 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 en= ough air flow at the speeds we are climbing and cruising, not to mention gr= ound ops.

=0A

 

=0A

So I= geeked out, and did the math that I have found to figure out the correct s= ize.  Those who have done this before, please check my math.

=0A

 

=0A

I found the following r= equirements/suggestions during my research:

=0A

&nbs= p;

=0A

1.2 sq in of face per cubic in of displacemen= t

=0A

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

=0A

2.1 cubic in per HP

=0A

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

=0A

2.48 = cubic in per HP

=0A

     &n= bsp;          2.48*210=3D521 c= ubic in

=0A

Inlet should be 15% of face

=0A

          =       36 sq in inlet =3D 240 sq in face

=0A

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 m= e a max space of 16 x 18 for 288 sq in face.  Using a 2.5 in thick cor= e 288*2.5 =3D 720 cubic in of radiator.  This should do the trick.&nbs= p; 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 inl= et can handle.

=0A

 

=0A

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 200= F(93C) coolant and a hot day, 90F(30C) I get:

=0A

&n= bsp;            = ;   (13.7/144) inlet *63 =3D 210  so inlet =3D 35 sq in

= =0A

So up to now I=E2=80=99m feeling pretty good about = the math, but please let me know if I messed it up.

=0A

 

=0A

I will have to use the wedge shaped d= uct 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 d= ivided by the width.

=0A

    &nb= sp;           16 x 18 x 1= 5% =3D 43.2

=0A

      =           43.2 / 16 =3D 2.7=0A

So Do I really make the front of the wedge only 2.= 7 inches tall?  This seems pretty small.  It would result in a re= ally long, thin triangle.  I have a max of 6.5 inches available, so ca= n easily make it bigger.

=0A

 

=0A

Thanks for the help.

=0A

 

=0A<= P class=3DMsoNormal>Chris

=0A
= =0A
Chris,
=0A
If I recall correctly the wedge duct requires a= bit more inlet than that about 20% is minimum. More important than the inl= et, the area behind the radiator is critical. You need 4 inches of unobs= tructed area if using a wedge duct and more is better.
=0A
Bil= l Jepson
--0-91696725-1283368339=:74978--