Return-Path: Received: from [65.173.216.66] (account ) by logan.com (CommuniGate Pro WebUser 4.0b8) with HTTP id 1770291 for ; Tue, 24 Sep 2002 11:09:17 -0400 From: "Marvin Kaye" Subject: Re: [LML] Re: RAM-air induction To: lml X-Mailer: CommuniGate Pro Web Mailer v.4.0b8 Date: Tue, 24 Sep 2002 11:09:17 -0400 Message-ID: In-Reply-To: MIME-Version: 1.0 Content-Type: text/plain; charset="ISO-8859-1"; format="flowed" Content-Transfer-Encoding: 8bit Posted for "Christopher Zavatson" : Villi, If the air intake were sized to keep area x cruise speed equal to engine volume demand, you would not be able to recover any ram effect. You could still expand the cross sectional area and slow down the flow to reduce losses through a filter etc, but you would have no net gain in MP. To gain the most ram, you really need your induction system to act like a pitot tube and not allow any flow into the engine. Once you start allowing air into the engine, you begin losing pressure. In our case the amount lost is, however, very small. The volume of air ingested by the engine is less than that which could be taken in by our 3 inch diameter intake. The result is a reduction in velocity and a pressure rise even if you never expand beyond the 3 inch diameter cross section. This ratio of actual inlet cross section to engine volume demand/cruise speed sets the maximum ram pressure that can be recovered. Flow losses in the induction system must now be subtracted from this maximum theoretically possible value. Since these losses are a function of velocity, slowing the flow in the induction system down helps minimize these losses. Here is another way to look at this. Close up the throttle completely and fly at 200 kts. The induction system is acting like a giant pitot tube and is seeing 100% of the stagnation pressure. If you start to open the throttle, letting some air to pass out the back of your induction system, the pressure inside will begin to drop. Now since you are trying to supply it faster in one end than it can be consumed at the other, you still have a net gain of pressure. I calculate the ratio to be about 3.5 in this case which leads to about 90% of the stagnation pressure being theoretically recoverable(before induction losses are included). Chris Zavatson N91CZ 360std BTW it looks like you got off by an order of magnitude in calculating area x speed (7 x 218740) s/b 1.4 mil in^3. not 14 mil in^3. Excess available volume is close to 3.5x instead of 64x.