X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from qw-out-2122.google.com ([74.125.92.24] verified) by logan.com (CommuniGate Pro SMTP 5.3c4) with ESMTP id 4034094 for flyrotary@lancaironline.net; Mon, 21 Dec 2009 15:40:08 -0500 Received-SPF: pass receiver=logan.com; client-ip=74.125.92.24; envelope-from=rwstracy@gmail.com Received: by qw-out-2122.google.com with SMTP id 9so1147266qwb.25 for ; Mon, 21 Dec 2009 12:39:35 -0800 (PST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=gamma; h=domainkey-signature:mime-version:sender:received:in-reply-to :references:date:x-google-sender-auth:message-id:subject:from:to :content-type; bh=/xerhxp11WIpuU7k+Fb7DpjF++L6QkLAlGd/d5rZF6c=; b=FbdMA/zKEHj3UBL96z44rAiuk8CxXg6A7ZcCjkr/cBLua+YV9eGF7A9qn2L5qii5IW qHf/uENclCGWT3PorJroO2pLg23Kp1ASxZ3tw9NuBSPkLp00hWlUC7r1bQxTjyKtCbtW z8mWd/GZftDZfdecpL2+nyoxecAhaHSL4+Vys= DomainKey-Signature: a=rsa-sha1; c=nofws; d=gmail.com; s=gamma; h=mime-version:sender:in-reply-to:references:date :x-google-sender-auth:message-id:subject:from:to:content-type; b=ACvhZBuLQ7q7+i+AP5UtI7cHmBv4z5ULoICtzU9yC9F78AYlW24qcPRwUEXpNoN22M CPYAASVnl7JlAv//04xxWVqghGMpkp7WHryPFeuuYaM3+duu7HT/dRMTWZjBcToqUqYZ NmKIyWrd/G2MEzY62LNXit9Sa3BROArq/K8/I= MIME-Version: 1.0 Sender: rwstracy@gmail.com Received: by 10.224.114.9 with SMTP id c9mr4048202qaq.281.1261427974986; Mon, 21 Dec 2009 12:39:34 -0800 (PST) In-Reply-To: References: Date: Mon, 21 Dec 2009 15:39:34 -0500 X-Google-Sender-Auth: a7168b83d20abde1 Message-ID: <1b4b137c0912211239t24262bafud40f83a4bec1cdc5@mail.gmail.com> Subject: Re: [FlyRotary] Air Flow Question From: Tracy Crook To: Rotary motors in aircraft Content-Type: multipart/alternative; boundary=000feaea1ad15bdeaf047b431691 --000feaea1ad15bdeaf047b431691 Content-Type: text/plain; charset=ISO-8859-1 I assume you mean 400% of the inlet cross sectional *area*. CFM air in = CFM air out so 45 kts is about right given your numbers. But I think the weak point in your assumptions is that the air entering the inlet is going at 180 kts. (or whatever your TAS is). There will likely be some external diffusion going on so that column of air velocity may not = TAS. The pressure required to force air through a typical oil cooler at 45 kts is probably a lot higher than you are likely to achieve. Just a gut feel, I could be wrong. Just as a reference point, my oil cooler face area to inlet area ratio is 650 %. Inlet is ~16 sq.in. The radiator face to inlet ratio is 560 %. Are these the right numbers to use? Don't know yet but today's RV-8 flight gave me excellent oil & water temps (167 / 157) on a 55 deg day. Still have lots of cooling outlet cleanup work to do and still flying without all my intersection fairings in place so I expect cooling to only get better. I cleaned the exterior of my O2 sensor with brake parts cleaner for today's flight but it was still dead. Didn't try removing it to clean the hot end though. Tracy (Whooo Hooo! no smoke today. I'm closer to taking this thing somewhere you can see it Ed! ) On Mon, Dec 21, 2009 at 2:03 PM, Thomas Mann wrote: > If I have a volume of air entering my scoop at 180 kts and expand the > volume of the chamber by 400% can I expect the speed of the airflow to drop > to 45 kts at that point? > > > > T Mann > > --000feaea1ad15bdeaf047b431691 Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable I assume you mean 400% of the inlet cross sectional area.=A0=A0=A0 C= FM air in =3D CFM air out so 45 kts is about right given your numbers.=A0 B= ut I think the weak point in your assumptions is that the air entering the = inlet is going at 180 kts. (or whatever your TAS is). There will likely be = some external diffusion going on so that column of air velocity may not=A0 = =3D TAS.=A0=A0 The pressure required to force air through a typical oil coo= ler at 45 kts is probably a lot higher than you are likely to achieve.=A0 J= ust a gut feel, I could be wrong.

Just as a reference point, my oil cooler face area to inlet area ratio = is 650 %.=A0 Inlet is ~16 sq.in. The radiator = face to inlet ratio is 560 %.=A0=A0=A0 Are these the right numbers to use?= =A0 Don't know yet but today's RV-8 flight gave me excellent oil &a= mp; water temps (167 / 157) on a 55 deg day.=A0 Still have lots of cooling = outlet cleanup work to do and still flying without all my intersection fair= ings in place so I expect cooling to only get better.=A0

I cleaned the exterior of my O2 sensor with brake parts cleaner for tod= ay's flight but it was still dead.=A0 Didn't try removing it to cle= an the hot end though.

Tracy (Whooo Hooo!=A0 no smoke today. =A0 I&#= 39;m closer to taking this thing somewhere you can see it Ed!=A0 )

On Mon, Dec 21, 2009 at 2:03 PM, Thomas Mann= <tmann@n200lz.com= > wrote:

If I have a volume of air entering my scoop at 180 kts and expand the volume of the chamber by 400% can I expect the speed of the airf= low to drop to 45 kts at that point?

=A0

T Mann


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