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 5.3c4) with ESMTP id 4037025 for flyrotary@lancaironline.net; Wed, 23 Dec 2009 17:05:23 -0500 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 55E38173870 for ; Thu, 24 Dec 2009 06:04:49 +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 97182BEC018 for ; Thu, 24 Dec 2009 06:04:48 +0800 (WST) Message-ID: <368C0DB379DA47D9A1C450745C1D7EC4@ownerf1fc517b8> From: "George Lendich" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Air Flow Question Date: Thu, 24 Dec 2009 08:04:48 +1000 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0062_01CA846F.C31C2B80" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.5843 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5579 X-Antivirus: avast! (VPS 091223-0, 12/23/2009), Outbound message X-Antivirus-Status: Clean This is a multi-part message in MIME format. ------=_NextPart_000_0062_01CA846F.C31C2B80 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Ed, That's how I would see it as well. Just using the rule of thumb rules, I = figure he's using about 250hp max that's 250x3 =3D 750cu" Rad. Inlet 5x5xPi =3D 78.5 sq". Considering a rad 2.75 thick 750/2.75=3D 272 = sq", therefore 78.5/272x100 =3D 28.86% inlet. Now that sounds about right to me, however Tracy could be running less = hp. If it 5" diameter, I will have to rethink 'rules of thumb'. If I remember correctly some use 2cu" per hp for air cooled engines with = a much higher Delta T and much wider tolerances to allow them to run at = those higher temperatures. George ( down under) Al, looking at photos of Tracy's inlets, I would hazard a guess that = the inlet may be more than 5" in diameter. If Tracy misspoke and meant = 5" radius then that would give 10" diameter or approx 78 sq inch =3D = 0.547 sq ft which then taking =20 4000 CFM /0.547 =3D 7322 ft/min =3D 83 MPH somewhat more realistic =20 But, then Tracy may mean 5" diameter in which case I am going to fly = down, pour him full of beer and badger the secret out of him {:>) =20 Ed =20 Ed Anderson Rv-6A N494BW Rotary Powered Matthews, NC eanderson@carolina.rr.com http://www.andersonee.com http://www.dmack.net/mazda/index.html http://www.flyrotary.com/ http://members.cox.net/rogersda/rotary/configs.htm#N494BW http://www.rotaryaviation.com/Rotorhead%20Truth.htm -------------------------------------------------------------------------= ----- From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] = On Behalf Of Al Gietzen Sent: Wednesday, December 23, 2009 11:21 AM To: Rotary motors in aircraft Subject: [FlyRotary] Air Flow Question =20 Tracy wrote: My 5" round inlet for the radiator looks ridiculously small compared = to yours but so far it is cooling the 20B OK. =20 =20 Now that sort of boggles my mind as it seems to violate the laws of = physics. Let's just take a modest climb power of, say; 225 hp. At that = power, the energy going into the coolant is about 6000 Btu/min. In = order to remove that amount of heat, at a typical air temp increase of = 75 degrees; takes about 4000 cfm air flow. A 5" dia inlet is 0.14 sq = feet, meaning an average inlet velocity about 29,000 ft/min, or 330 mph. = Even at 100 air temp increase (unlikely on a 90F day) it's 250 mph. = And I'm guessing your climb speed is half that. Similar math suggests = you'd be limited to a steady state (cruise) power of about 50%. =20 Of course, being a pusher driver, I think of inlet air speeds in terms = of the speed of the airplane. So does the fact that the inlet is behind = prop give a much higher effective inlet velocity? I've been thinking = that the turbulence in the prop wash would negate a good portion of the = extra mean air velocity because of reduced inlet effectiveness. =20 It will be interesting to know how it works out on a hot day.=20 =20 Great that you have your 20B in the air and working well. =20 Merry Christmas and Happy New Year to all. =20 Al G =20 =20 =20 __________ Information from ESET NOD32 Antivirus, version of virus = signature database 3267 (20080714) __________ The message was checked by ESET NOD32 Antivirus. http://www.eset.com ------=_NextPart_000_0062_01CA846F.C31C2B80 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Ed,
That's how I would see it as well. Just = using the=20 rule of thumb rules, I figure he's using about 250hp max that's 250x3 = =3D 750cu"=20 Rad.
Inlet 5x5xPi =3D 78.5 sq". Considering = a rad 2.75=20 thick 750/2.75=3D 272 sq", therefore 78.5/272x100 =3D=20 28.86% inlet.
Now that sounds about right to me, = however Tracy=20 could be running less hp.
If it 5" diameter, I will have to = rethink 'rules of=20 thumb'.
 
If I remember correctly some use 2cu" = per hp for=20 air cooled engines with a much higher Delta T and much wider tolerances = to allow=20 them to run at those higher temperatures.
George ( down under)

Al, looking = at photos=20 of Tracy=92s=20 inlets, I would hazard a guess that the inlet may be  more than = 5=94 in=20 diameter.  If Tracy misspoke and meant 5=94=20 radius=20 then that = would give=20 10=94 diameter or approx 78 sq inch =3D 0.547 sq ft which then=20 taking

 

4000 CFM = /0.547 =3D=20 7322 ft/min =3D 83 MPH somewhat more = realistic

 

But, then = Tracy may = mean 5=94=20 diameter in which case I am going to fly down, pour him full of beer = and=20 badger the secret out of him {:>)

 

Ed

 

Ed=20 Anderson

Rv-6A = N494BW Rotary=20 Powered

Matthews,=20 NC

eanderson@carolina.rr.com

http://www.andersonee.com

http://www.dmack.net/mazda/index.html

http://www.flyrotary.com/

http://members.cox.net/rogersda/rotary/configs.htm#N494BW

http://www.r= otaryaviation.com/Rotorhead%20Truth.htm

From:=20 Rotary motors in aircraft = [mailto:flyrotary@lancaironline.net] On=20 Behalf Of Al Gietzen
Sent: Wednesday, December 23, = 2009 11:21=20 AM
To: = Rotary motors in aircraft
Subject: [FlyRotary] Air Flow=20 Question

 

Tracy=20 wrote:

My 5" round inlet for the = radiator looks=20 ridiculously small compared to yours but so far it is cooling the 20B=20 OK. =20 =

 

Now that = sort of=20 boggles my mind as it seems to violate the laws of physics. Let=92s = just take a=20 modest climb power of, say; 225 hp.  At that power, the energy = going into=20 the coolant is about 6000 Btu/min.  In order to remove that = amount of=20 heat, at a typical air temp increase of 75 degrees; takes about 4000 = cfm air=20 flow.  A 5=94 dia inlet is 0.14 sq feet, meaning an average inlet = velocity=20 about 29,000 ft/min, or 330 mph.  Even at 100 air temp increase = (unlikely=20 on a 90F day) it=92s 250 mph.  And I=92m guessing your climb = speed is half=20 that.  Similar math suggests you=92d be limited to a steady state = (cruise)=20 power of about 50%.

 

Of = course, being a=20 pusher driver, I think of inlet air speeds in terms of the speed of = the=20 airplane.  So does the fact that the inlet is behind prop give a = much=20 higher effective inlet velocity?  I=92ve been thinking that the = turbulence=20 in the prop wash would negate a good portion of the extra mean air = velocity=20 because of reduced inlet effectiveness.

 

It will = be=20 interesting to know how it works out on a hot day.=20

 

Great = that you have=20 your 20B in the air and working well.

 

Merry = Christmas and=20 Happy New Year to all.

 

Al=20 G

 

 

 



__________ Information from ESET = NOD32=20 Antivirus, version of virus signature database 3267 (20080714)=20 __________

The message was checked by ESET NOD32 = Antivirus.

http://www.eset.com

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