Received-SPF: pass
 receiver=logan.com; client-ip=64.102.122.148; envelope-from=echristl@cisco.com
Message-ID: <420B9462.8000000@cisco.com>
Date: Thu, 10 Feb 2005 12:05:38 -0500
From: Ernest Christley <echristl@cisco.com>
User-Agent: Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.7.3) Gecko/20040929
MIME-Version: 1.0
To: Rotary motors in aircraft <flyrotary@lancaironline.net>
Subject: Re: [FlyRotary] Rotary AirFlow Equation? Help?
References: <list-723645@logan.com>
In-Reply-To: <list-723645@logan.com>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Content-Transfer-Encoding: 7bit

Ed Anderson wrote:

> In a previous posting, I alluded to my suspicions that a commonly 
> accepted formula for calculating airflow for the rotary engine might 
> be flawed. I was asked off-line to expound on that suspicion.
>  
>  Well, I can think of no better audience to present my suspicions to 
> and see what you think.  Last time I tried this on another list I was 
> berated, beat upon and chastised for daring to raise the question - 
> but, never one to be daunted for long and knowing this list has some 
> experienced and clear thinkers, here goes again.
>  
> *1.   The racing crowd was thrown into a turmoil when the rotary 
> appeared on the scene - what displacement class should it be place in??*
>  
> Well eventually the following was apparently (If not agreed to was at 
> least accepted) that the 1300 cc 13B rotary was equivalent to a 160 
> CID 4 stroke engine.  Then the standard air flow formula could be 
> applied - which is
>  
> Air Flow (CFM) = Displacement*rpm/(2 *1728)*Ve  (Volumetric efficiency 
> (Ve) is generally assume to be 100% or 1, so we can drop it for this 
> discussion as well as the effects of compression ratios)
>  
> How did they arrive at 160 CID displacement, well they apparently 
> decided to reference the rotary  to the standard 720 Deg 4 stroke 
> rotation cycle even though for all six faces of the two rotors to go 
> through their cycle requires 1080 deg of e shaft rotation - well that 
> is 360 deg more than the standard of 720Deg.  So if you limit 
> consideration to only 720deg of rotation only 4 of the six rotor faces 
> have occurred, so since the displacement by a single face is 40 CID 
> then 4* 40 = 160 CID.  So that appears to be how the decision of 160 
> CID equivalent displacement was arrived at. No problem with that part 
> - but, continuing on:
>  
> So taking the formula and assuming 6000 rpm, we have Air Flow = 
> 160*6000/(2*1728) = 277.77 CFM (assuming 100% Ve).
>  
> *2.  So what's the problem?  Well, unfortunately, I like to fully 
> understand how and why something works rather than just plugging in 
> numbers to a formula (although, heaven knows I do enough of that). *
>  
>  So here is how I started sliding down the slipper slope {:>).
>  
> We know that the rotors turn at a speed 3 times less than the E 
> shaft.  So if the e shaft is turning at 6000 rpm then the rotors are 
> turning at 6000/3 = 2000 rpm.  Its the rotor of course that actually 
> suck in the air - not the eccentric shaft.  So we should be able to 
> arrive at the same answer from either reference point (rotor or e 
> shaft).  Well the formula above uses the e shaft reference.  So I 
> looked at what the rotor reference would produce (expecting the same 
> answer).
>  
> If the rotors are spinning at 2000 rpm, we know that 2 of the rotor 
> faces have gone through their cycle for each 360 deg revolution of the 
> e shaft or 4 will have for 720 deg.  Looking at it from the rotor 
> reference if the e shaft has turned 720 deg then the rotor has turned 
> 720/3 = 240 degrees.  240/360 = 0.6666 * 6 faces = 4 faces.  So that 
> confirms that 4 faces go through their cycle in 720 degs of e shaft 
> rotation.  So if 4 faces of the two rotors complete their cycle in 
> 720Deg eshaft then we have
>  
> Air Flow = 4(number of faces) *40 (displacement for each face) * rpm 
> (rotor rpm)/(1728)    note: we drop the division by 2 which in the 
> original 4 stroke reciprocating engine equation takes into account 
> that only 1/2 of the cylinders are sucking air on each revolution. 
>  
>
Ed, I think you've blinded yourself with science 8*)

Each face of the rotor will cycle 40cid of air on each revolution.  You 
can ignore the eshaft.  Every time the rotor turns around, each face 
will have processed 40cid. If a rotor is turning at 2000 RPM, then three 
faces will each cycle 2000*40cid of air.  You bring in referencing off 
the eshaft, and all of a sudden only 4 faces are processing air.  I 
think that is the mistake.

You reference 720 degrees of eshaft rotation.  But all that means is 
that the rotors haven't completed their cycle.  You're just ignoring 
every third rotor face.