Return-Path: Sender: (Marvin Kaye) To: flyrotary Date: Thu, 27 Mar 2003 02:51:25 -0500 Message-ID: X-Original-Return-Path: Received: from cpimssmtpu12.email.msn.com ([207.46.181.87] verified) by logan.com (CommuniGate Pro SMTP 4.1b1) with ESMTP id 2087858 for flyrotary@lancaironline.net; Wed, 26 Mar 2003 20:42:33 -0500 X-Originating-IP: 68.7.218.110 Received: from BigAl ([68.7.218.110]) by cpimssmtpu12.email.msn.com with Microsoft SMTPSVC(5.0.2195.4905); Wed, 26 Mar 2003 17:41:59 -0800 Reply-To: From: "Al Gietzen" X-Original-To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Re: Thick vs Thin X-Original-Date: Wed, 26 Mar 2003 17:42:56 -0800 Organization: ALVentures X-Original-Message-ID: <000c01c2f402$33c2c6b0$6400a8c0@BigAl> MIME-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.4024 Importance: Normal In-Reply-To: X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 X-Original-Return-Path: alventures@email.msn.com X-OriginalArrivalTime: 27 Mar 2003 01:41:59.0555 (UTC) FILETIME=[0ED1FD30:01C2F402] David said: > Yes, except. The forward energy used to slow down that incoming air is > offset by any thrust you may be able to obtain from the exiting air (or, > if you prefer, there is also significant drag at the exit point, this > drag is inversely proportional to the speed of the exiting air). Like > in a jet engine, what matters is the difference between the intake and > exhaust velocity. And this is where I disagree with your first point, > the rad is the main determinant of this difference in velocity between > intake and exhaust (given optimum ducting- which I'll admit is darn near > impossible). The more energy you loose passing air through the rad, the > less you will be able to speed it for smooth exit. And it is velocity > through the rad (because of its cubed effect) that is the primary > determinant of energy loss there. Tracy again: I promise not to risk 2 hours of typing this time :-) Correct, much of the debate hinges on this point. It is my contention that 90+ percent of the energy represented by the high speed air used to cool the engine is lost in either thick or thin radiator. There is no significant energy to recover. This is easily verified by measuring the velocity of the air coming out of the rad. Typical values are on the order of 20 mph or less. To verify, Calculate the velocity of 3000 cfm (typical value for a 180 HP engine) going through an 18"x 20" rad. Since energy is equal to (1/2 Mass) x (Velocity *squared*) is is obvious that recovered energy is a lost cause. But let's calculate it for fun. Set mass arbitrarily at 2. First calculate energy at 200 mph (2/2) * (200 * 200) = 40,000. Now do it for 20 mph (2/2) * (20 * 20) = 400 This shows that 99% of the energy in that 200 mph air is lost going through the rad (only 1% is left). Tracy But wait - aren't you forgetting that you still have pressure left to accelerate the air in the exit duct? You converted dynamic pressure to static pressure in the inlet duct, now you convert the static pressure back to dynamic pressure at the exit. The pressure drop through the radiator determines how much energy you have left to accelerate the air. Al