X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from fed1rmmtao104.cox.net ([68.230.241.42] verified) by logan.com (CommuniGate Pro SMTP 5.3c4) with ESMTP id 4036631 for flyrotary@lancaironline.net; Wed, 23 Dec 2009 11:20:33 -0500 Received-SPF: none receiver=logan.com; client-ip=68.230.241.42; envelope-from=alventures@cox.net Received: from fed1rmimpo03.cox.net ([70.169.32.75]) by fed1rmmtao104.cox.net (InterMail vM.8.00.01.00 201-2244-105-20090324) with ESMTP id <20091223161957.YHMR16123.fed1rmmtao104.cox.net@fed1rmimpo03.cox.net> for ; Wed, 23 Dec 2009 11:19:57 -0500 Received: from BigAl ([72.192.128.205]) by fed1rmimpo03.cox.net with bizsmtp id LgKw1d0074S1t5C04gKwrm; Wed, 23 Dec 2009 11:19:56 -0500 X-VR-Score: -100.00 X-Authority-Analysis: v=1.1 cv=hxRkenI+XOLoHPjz51PhBd6+pm8nOe1puOOeXS7yE/Y= c=1 sm=1 a=Vegc0WxVmH5BHtpNDyThtA==:17 a=NB-te26q-JLC-l5d9ZIA:9 a=JB2MfBS6ycegNrDKPYEA:7 a=p6_tBt7TYfltX4a331Lyw_fqjLkA:4 a=KVoZRe9gm5njMNXRwRoA:9 a=fDsu-mntF0Bna0lL7pcA:7 a=Va6SwSTfVgvxAv3D1VVnpwDkRa8A:4 a=Vegc0WxVmH5BHtpNDyThtA==:117 X-CM-Score: 0.00 From: "Al Gietzen" To: "'Rotary motors in aircraft'" Subject: Air Flow Question Date: Wed, 23 Dec 2009 08:20:37 -0800 Message-ID: <90EA3D29E8584A35A91FAB668D262B71@BigAl> MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0016_01CA83A8.CE881690" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.6856 In-Reply-To: Thread-Index: AcqD1MEh6Zvp/kMAThGQQWx8GolPOAAEM1Ug X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5579 Importance: Normal This is a multi-part message in MIME format. ------=_NextPart_000_0016_01CA83A8.CE881690 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: quoted-printable 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 ------=_NextPart_000_0016_01CA83A8.CE881690 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

Tracy wrote:

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

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%.

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.

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

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

Merry Christmas and Happy New = Year to all.

Al G

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