X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from fmailhost01.isp.att.net ([204.127.217.101] verified) by logan.com (CommuniGate Pro SMTP 5.1.12) with ESMTP id 2360847 for flyrotary@lancaironline.net; Sun, 30 Sep 2007 14:12:47 -0400 Received-SPF: pass receiver=logan.com; client-ip=204.127.217.101; envelope-from=ceengland@bellsouth.net Received: from [209.215.63.246] (host-209-215-63-246.jan.bellsouth.net[209.215.63.246]) by bellsouth.net (frfwmhc01) with ESMTP id <20070930181209H0100om04re>; Sun, 30 Sep 2007 18:12:09 +0000 X-Originating-IP: [209.215.63.246] Message-ID: <46FFE6F8.1080905@bellsouth.net> Date: Sun, 30 Sep 2007 13:12:08 -0500 From: Charlie England User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.2) Gecko/20070222 SeaMonkey/1.1.1 MIME-Version: 1.0 To: Rotary motors in aircraft Subject: Re: [FlyRotary] Re: Another cooling question References: In-Reply-To: Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 7bit Bobby J. Hughes wrote: > Snip > That's exactly what I HAD thought, until I was told that the air could > pass through too fast and not pick up as much heat. > > Mark... this is what I was trying to communicate. It could be totally > wrong so let's get more opinions. Ed?? > > *With respect to exit area size only. * > > If the volume of air through the coolant radiator was moving faster than > *optimum* the air delta T would be lower than it would be with slower > air. Slower air at a higher delta T means less air needs to exit the > cowl. What is optimum? Every install is unique so it needs to be viewed > \ identified on each installation. Ed uses a 30% duct air speed as a > reference point. If I understand this correctly the diffuser play a role > in duct airspeed. If the air is not being diffused optimally the > airspeed could be much higher than 30% through the part of the core. > Dennis H. and I observed airspeeds through a test core at 50% + without > a diffuser. > > Your inlets are 72sqin with 306sqin core face for water and 24.75 > sqin with 102 sqin for oil. Both are *competing* for the same exit area. > *IF* this is an exit area problem only then enlarging it should improve > both water and oil proportionally. Water should realize 2/3 of the > improvement and oil 1/3. 306 sqin vs 102 sqin. Your improvements after > opening the exit seem to track this very close. So increasing your exit > area further should show more improvements in both oil and water to a > point. But you do not need more water cooling improvements right? > > Core 306 sqin at an airspeed of 115mph. > > 40% or 46 mph = 8602 cfm 8602 cfm at a 50 deg air delta T = 7741 btu's > \ min > 30% or 34.5 = 6451 cfm 6451 cfm at a 80 deg air delta T = 7741 > btu's \ min > > In this example 2151 cfm less air needs to flow through the water > radiator to produce the same btu rejection. So what effect would 2151 > cfm less air through the water radiator and exit area have on the oil > cooler's ability to flow more air? Would it improve the oil cooler air > flow by 1/3 or 717 cfm? Not sure. > > Can your radiator produce a 80 deg air delta T? It may only produce 60, > 70 or?? > > Am I in left field here? Analysis only valid if cores are identical in construction (resistance to air flow will likely be different with the different cores).