X-Virus-Scanned: clean according to Sophos on Logan.com X-SpamCatcher-Score: 50 [XX] (67%) RECEIVED: IP not found on home country list (33%) HTML: title tag is empty Return-Path: Received: from [201.225.225.167] (HELO cwpanama.net) by logan.com (CommuniGate Pro SMTP 5.1.7) with ESMTP id 1876098 for flyrotary@lancaironline.net; Thu, 01 Mar 2007 09:10:19 -0500 Received-SPF: none receiver=logan.com; client-ip=201.225.225.167; envelope-from=rijakits@cwpanama.net Received: from [201.224.94.164] (HELO usuario5ebe209) by frontend1.cwpanama.net (CommuniGate Pro SMTP 4.2.10) with SMTP id 104407792 for flyrotary@lancaironline.net; Thu, 01 Mar 2007 09:17:45 -0500 Message-ID: <000f01c75c0b$2faa3cf0$a45ee0c9@usuario5ebe209> From: "Thomas y Reina Jakits" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Pinched ducts was : [FlyRotary] Re: cowl openings for water radiators Date: Thu, 1 Mar 2007 09:09:11 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_000C_01C75BE1.468578B0" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3028 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028 This is a multi-part message in MIME format. ------=_NextPart_000_000C_01C75BE1.468578B0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Thanx Ed! Now for your next experiment: Buy a couple of lipstick cameras, drill some holes into your ducts, but = some strings in there and go fly! :))) Let's see what the airflow does:)) TJ Here is an example from K&W chapter 12 The heat flow coefficient Kp = for turbulent flow in smooth passages (radiator core passages) Kp =3D = 1/2*L/D* 0.326/(Re)^-4 .=20 If Kp is a measure of goodness, then clearly if L increases and D = gets smaller Kp increases. Or if the Reynolds number Re gets smaller Kp = goes up. So what does this mean? It basically shows that for the heat = transfer to be large, the Reynolds number should be low (I.e. the = airflow through the core should be slow), the core should be deep(large = L) and the hole's hydraulic diameter (D) should be small.=20 This makes sense as the thicker the core the more heat transfer = (although the further into the core the less efficient the heat = transfer), the holes should be smaller (area exposed area - with large = holes some of the cooling air in the center will simply not have as much = contact with the hot metal of the core) and the air velocity should be = slow (dwell time adds to heat transferred to the unit volume of air per = unit time). =20 However, if you make the core too thick or the holes too small or = slow the air too much - then your KP factor may be high - but your over = all cooling will suck because you have too little mass flow through a = too restrictive core. This is just one example of where optimizing on = one set of factors can play havoc with the overall system function. = One way of looking at it is that you have to suboptimize a lot of = factors in order to get an optimum system {:>)=20 My 0.02 Ed ------=_NextPart_000_000C_01C75BE1.468578B0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable

Thanx Ed!

Now for your next = experiment:

Buy a couple of lipstick cameras, drill = some holes=20 into your ducts, but some strings in there and go fly! :)))

Let's see what the airflow = does:))

Here is an example from K&W chapter = 12 The=20 heat flow coefficient Kp for turbulent flow in smooth passages=20 (radiator core passages) Kp =3D 1/2*L/D* = 0.326/(Re)^-4 =20 .

If Kp is a measure of goodness, then = clearly if=20 L increases and D gets smaller Kp increases. Or if the = Reynolds number=20 Re gets smaller Kp goes up. So what does this mean? It = basically=20 shows that for the heat transfer to be large, the Reynolds number = should be=20 low (I.e. the airflow through the core should be slow), the core = should be=20 deep(large L) and the hole's hydraulic diameter (D) should be=20 small.

This makes sense as the thicker the = core the=20 more heat transfer (although the further into the core the less = efficient=20 the heat transfer), the holes should be smaller (area exposed = area -=20 with large holes some of the cooling air in the center will simply = not have=20 as much contact with the hot metal of the core) and the air = velocity=20 should be slow (dwell time adds to heat transferred to the unit = volume of=20 air per unit time).

However, if you make the core too thick or = the holes=20 too small or slow the air too much - then your KP factor may = be high -=20 but your over all cooling will suck because you have too little mass = flow=20 through a too restrictive core. This is just one example of = where=20 optimizing on one set of factors can play havoc with the overall = system=20 function. One way of looking at it is that you have to=20 suboptimize a lot of factors in order to get an optimum system = {:>)=20

My 0.02

Ed

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