X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from web81309.mail.mud.yahoo.com ([68.142.199.125] verified) by logan.com (CommuniGate Pro SMTP 5.2c1) with SMTP id 2498784 for flyrotary@lancaironline.net; Wed, 21 Nov 2007 01:53:42 -0500 Received-SPF: none receiver=logan.com; client-ip=68.142.199.125; envelope-from=ron2369@sbcglobal.net Received: (qmail 22415 invoked by uid 60001); 21 Nov 2007 06:53:04 -0000 DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=s1024; d=sbcglobal.net; h=X-YMail-OSG:Received:Date:From:Subject:To:In-Reply-To:MIME-Version:Content-Type:Content-Transfer-Encoding:Message-ID; b=vLjvleLnzZrSla0kRTZoLiT00RwCYGIVdmS0Dy8+VHzFkDIv5rrLZbokBaG4GaHwFp7x6c3wSvMeFdrm1roUUeWqiV5EfMsVUeNAfS0GEkdiJB7X/fRHgkfNN0+nnJgsuztIeumnfAnQZvh/1nXV3zcGisGlyIrpN7EAOv4y2/g=; X-YMail-OSG: hAIy01UVM1nF4TiQoRCy5EAhOROXhe9GtWYcT7ZuMlpXE1dv_nxGUOwJm6g0HfWqVQ-- Received: from [71.142.232.57] by web81309.mail.mud.yahoo.com via HTTP; Tue, 20 Nov 2007 22:53:04 PST Date: Tue, 20 Nov 2007 22:53:04 -0800 (PST) From: Ron Springer Subject: Re: [FlyRotary] Re: Thick vs Thin was : Diffuser Configuration Comparison To: Rotary motors in aircraft In-Reply-To: MIME-Version: 1.0 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: 8bit Message-ID: <182393.9340.qm@web81309.mail.mud.yahoo.com> I have to agree with you argument below. It has been argued that since so little thrust can be recovered downstream of the radiator, that it should be assumed that nothing can be recovered. So, in that case, the lowest mass flow is the lowest drag. Well, that would be true in that case, but the inital assumption is a bit extreme. The only way to achieve it is to put a black hole just downstream of the radiator! All the mass flow that enters the inlet will exit the outlet. The effect of the exit flow is significant and I don't think it can be ignored. [The next part of this email is optional reading material!] Conservation of mass says that the mass flow rate entering the inlet will equal the mass flow rate exiting the outlet. Mass flow rate = density x area x velocity Consider an example with the inlet and exit areas the same. Then, density times velocity is the same at the inlet and the outlet. The density at the outlet will be lower than at the inlet because the air has been heated. So, the velocity must be higher at the outlet than at the inlet. This is true for this case regardless of the drag created by the radiator, or the efficiency of the system, as long as inlet amd outlet areas match. (Actually, if drag is increased too much, the inlet will spill air and the mass flow rate will be lowered to below that of a full flowing inlet. A smaller streamtube of air would be captured, there would be external diffusion prior to the lip of the inlet, and the effective inlet area (or capture area) would need to be used instead, which is less than the physical area of the inlet.) More sophisticated examples can be done for non-equal areas, but then it becomes more involved. The point is that the air exits with significant velocity as long as there is significant mass flow and the effects can't be ignored. Ron --- David Leonard wrote: > I do not believe that useful pressure recovery is a > pipe-dream. Some report > that the P-51 even gained thrust from the cooling > exit (I don't think they > mean net thrust from the system thought). Also, > both Bill Eslick and I have > the same experience that our cowl flaps want to pull > themselves wide open by > the pressure available at the cooling exit (well > into the slip stream). > > Oops, sorry, couldn't help myself. > > -- > David Leonard > > Turbo Rotary RV-6 N4VY > http://N4VY.RotaryRoster.net > http://RotaryRoster.net >