Return-Path: Received: from [24.25.9.101] (HELO ms-smtp-02-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.1.8) with ESMTP id 2899923 for flyrotary@lancaironline.net; Sun, 21 Dec 2003 19:02:03 -0500 Received: from o7y6b5 (clt78-020.carolina.rr.com [24.93.78.20]) by ms-smtp-02-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id hBM01wAc006499 for ; Sun, 21 Dec 2003 19:02:01 -0500 (EST) Message-ID: <001601c3c81e$551c54c0$1702a8c0@WorkGroup> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Dumb intake question Date: Sun, 21 Dec 2003 18:58:29 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0013_01C3C7F4.6BFA0180" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1106 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_0013_01C3C7F4.6BFA0180 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Message=20 ----- Original Message -----=20 From: Russell Duffy=20 To: Rotary motors in aircraft=20 Sent: Sunday, December 21, 2003 11:21 AM Subject: [FlyRotary] Re: Dumb intake question =20 then I'll give it a shot. So based on that - it looks like your length from port to butterfly = inlet of TB ranges from 13.1" for 6500 RPM to 11.5 RPM for 7500 RPM. = This provides time for the pulse to be generated by the closing of the = port and to arrive after that port opens again after bouncing back from = the throttle body once. So from that Length you would need to subtract = your block to port distance of 2.5" and the length of your throttle = body.=20 Excellent! This is just the sort of info I was hoping for, but = there's one thing I still don't understand. When you say the pulse = bounces back from the TB, what is it bouncing off of? I believe the = throat of the TB is straight, with no venturi effect at all, and we'll = assume that the intake runner matches the TB size perfectly. That just = leaves the open butterfly, which has a pretty small profile. Is it = really getting a strong bounce from the open butterfly? What if you had = a slide type opening, like the Ellison TB, where there is nothing left = in the path of the air? Would there just be no bounce? Thanks a bunch, Rusty (special needs student) =20 See, you should have come to class that day{:>), Rusty What happens is when a Finite Amplitude Wave (FAW) hits a change in = cross-sectional area (like an opening from your tube to the atmoshpere = or a plenium - or a decrease to a closed tube) energy is reflected back = down the tube. The amount depends on the extent of cross section change, = but opening to the atmosphere reflects most of the energy. If the = cross sectional area it sees is Larger then a FAW wave of the opposite = type is reflected. If the cross sectional area is smaller, a FAW wave = of the same type is reflected. So in your case, a compression wave is generated when the intake port = closes, the wave travels up the tube until it sees the atmosphere (at = the throat of your throttle body), there a large amount of the energy = (Most of the wave's energy) is reflected back as a expansion wave = (meaning its negative pressure with regard to atmosphere). However, = when it gets to the port the port has closed, so a wave of the same type = (Expansion or negative pressure) is reflected back to the TB throat, = again where it sees a larger cross section. A larger cross section = reflects the opposite type wave, so this time the wave reflected back to = the port is a compression (positive pressure) wave and this when time it = arrives, it finds the intake port now open and helps to shovel more air = into the combustion chamber. Now this is not as desirable as having it = arrive just as the port is closing to overcome reversion - but of some = aid. Ideally, you would want to take the pulse created by an opening port = (in the rotary because that pulse is stronger) and then have it bounce = back just before that port closes. But, that is a duration of approx = 288 degs which would take at 7500 rpm about .0064 Seconds which would = require a length of approx 46 inches for a two bounce system. This (I = think) is one of the reasons that Mazda interconnected the two intake = ports - it reduces the intake length required because the two points of = interest on the different rotors are approx 93 deg apart rather than 288 = deg they are on the same rotor - so the time to travel and therefore the = intake is much shorter. So you say - lets do a four bounce, well the problem there is when the = pulse comes back to the intake port - it finds it open (remember in this = case, it is from Port opening to port closing so the port is open each = time the pulse arrives - the opposite of the short runner case). Since = the port is opened into the chamber in this chase, the FAW sees a = larger cross sectional area and reverses its type which sends an = expansion wave back to the TB. This reflects back a compression wave = which helps shove a bit more air into the chamber but is reflected back = as a expansion wave. This would continue several times (2,4) and each = time some of the pulse's energy would be lost into the atmosphere and = more loss into the chamber. So by the time the port is in its closing = phase and pushing some of its intake charge back out (Reversion) the = pulse is probably too weak to provide much assistance in overcoming the = reversion. FWIW=20 Ed Anderson ------=_NextPart_000_0013_01C3C7F4.6BFA0180 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Message
 
----- Original Message -----
From:=20 Russell=20 Duffy
Sent: Sunday, December 21, 2003 = 11:21=20 AM
Subject: [FlyRotary] Re: Dumb = intake=20 question

 
 then I'll give = it a=20 shot.
 
 So based on that - it looks = like your=20 length from port to butterfly inlet of TB ranges from 13.1" for 6500 = RPM to=20 11.5 RPM for 7500 RPM.  This provides time for the pulse to be = generated=20 by the closing of the port and to arrive after that port opens again = after=20 bouncing back from the throttle body once.  So from that Length = you would=20 need to subtract your block to port distance of 2.5" and the length of = your=20 throttle body. 
 
Excellent!  = This is=20 just the sort of info I was hoping for, but there's one thing I=20 still don't understand.  When you say the pulse bounces = back=20 from the TB, what is it bouncing off of?   I believe = the throat=20 of the TB is straight, with no venturi effect at all, and we'll assume = that=20 the intake runner matches the TB size perfectly.  That just = leaves=20 the open butterfly, which has a = pretty small=20 profile.   Is it really getting a strong bounce from the = open=20 butterfly?  What if you had a slide type opening, like the = Ellison TB,=20 where there is nothing left in the path of the air?  Would there = just be=20 no bounce?
 
Thanks a=20 bunch,
Rusty (special needs=20 student)
 
See, you should have come to class = that=20 day{:>), Rusty
 
  What happens is when a Finite = Amplitude=20 Wave (FAW) hits a change in cross-sectional area (like an opening from = your=20 tube to the atmoshpere or a plenium - or a decrease to a closed tube) = energy=20 is reflected back down the tube. The amount depends on the extent of = cross=20 section change, but opening to the atmosphere reflects most of the = energy.=20   If the cross sectional area it sees is Larger then a FAW wave = of the=20 opposite type is reflected.  If the cross sectional area is = smaller, a=20 FAW wave of the same type is reflected.
 
So in your case, a compression wave = is generated=20 when the intake port closes, the wave travels up the tube until = it sees=20 the atmosphere (at the throat of your throttle body), there a large = amount of=20 the energy (Most of the wave's energy) is reflected back as a = expansion wave=20 (meaning its negative pressure with regard to atmosphere).  = However, when=20 it gets to the port the port has closed, so a wave of the same type = (Expansion=20 or negative pressure) is reflected back to the TB throat, again where = it sees=20 a larger cross section.  A larger cross section reflects the = opposite=20 type wave, so this time the wave reflected back to the port is a=20 compression (positive pressure) wave and this when time it = arrives,=20 it finds the intake port now open and helps to shovel = more air=20 into the combustion chamber.  Now this is not as desirable as = having it=20 arrive just as the port is closing to overcome reversion - but of some = aid.
 
Ideally, you would want to take the = pulse created=20 by an opening port (in the rotary because that pulse is stronger) and = then=20 have it bounce back just before that port closes.  But, that is a = duration of approx 288 degs which would take at 7500 rpm about .0064 = Seconds=20 which would require a length of approx 46 inches for a two bounce=20 system.  This (I think) is one of the reasons that Mazda = interconnected=20 the two intake ports - it reduces the intake length required because = the two=20 points of interest on the different rotors are approx 93 deg apart = rather than=20 288 deg they are on the same rotor - so the time to travel and = therefore=20 the intake is much shorter.
 
So you say - lets do a four bounce, = well the=20 problem there is when the pulse comes back to the intake port - it = finds it=20 open (remember in this case, it is from Port opening to port = closing so=20 the port is open each time the pulse arrives - the opposite = of the=20 short runner case).  Since the  port is opened into the = chamber in=20 this chase, the FAW sees a larger cross sectional area and reverses = its type=20 which sends an expansion wave back to the TB.  This reflects back = a=20 compression wave which helps shove a bit more air into the chamber but = is=20 reflected back as a expansion wave.  This would continue several = times=20 (2,4) and each time some of the pulse's energy would be lost into the=20 atmosphere and more loss into the chamber.  So by the time the = port is in=20 its closing phase and pushing some of its intake charge back out = (Reversion)=20 the pulse is probably too weak to provide much assistance in = overcoming the=20 reversion.
 
FWIW
 
Ed = Anderson
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