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From: "Ed Anderson" <eanderson@carolina.rr.com>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
References: <list-723450@logan.com>
Subject: Re: [FlyRotary] Re: : Same HP = Same Air Mass <> same air Velocity II [FlyRotary] Re: Ellison, the missing piece
Date: Thu, 10 Feb 2005 10:04:14 -0500
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Hi Eric

Thanks for the comments

In this case, The assumption was made that engines that consume the same
amount of air/flue produce essential equivalent HP.  True,  the different
BSFC  for different engines which will result in a different HP output - but
I was trying to determine if the physical configuration of the pumping might
make a difference.

However, I believe that local inlet velocity does matter  - as a narrower
intake/Higher velocity can result in the "choking" of flow at a certain
velocities (there are formulas for determining at what velocity with what
opening that choke point occurs at - but I'll leave that to someone else)
and there is the increased friction loss (how significant I don't know) with
higher velocity and smaller opening.
Average velocity and maximum velocity can be considerably different in an
intake. According to the theory I have read the  Maximum instantaneous
velocity in your intake should not exceed mach 0.6 .  But, the average
velocity could still  could be in the 150-300 ft/sec range.

I agree that Volume of air = velocity * Carb area, however, I think the
dependent and independent variables would cause the equation to be stated
Velocity = Volume of air/Carb area.  Since the volume of a positive
displacement pump is independent of the air velocity it would be the
independent variable in the equation.  However, to be perfectly correct we
should really talk about air mass   flow and not just the velocity as the
density of the intake charge will vary - but, the volume (with a positive
displacement pump) will not.  You get the same volumetric displacement from
the engine whether at idle or WOT - its, of course, the density of the air
in the intake that makes the difference in the power produced.

In any case, like I said the Ellison rotary story could be just that - a
story without factual data or perhaps there is something to it.  Just trying
to see what might make logical sense in looking at it. Not trying to
discourage anyone from using an Ellison, I would be just as interested as
anyone to see what the real world comparison would be.

Best Regards

Ed


----- Original Message -----
From: "Eric Ruttan" <ericruttan@chartermi.net>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Sent: Thursday, February 10, 2005 9:18 AM
Subject: [FlyRotary] Re: : Same HP = Same Air Mass <> same air Velocity II
[FlyRotary] Re: Ellison, the missing piece


> Warning top poster, who cuts the post size down.
>
> A hopothises for your examination.
>
> A 360 Lyc does not produce the same power as a rotary.
>
> If true, then the Ellison card may not get enough air.
>
> If not true, then there is no real reason why the Ellison cannot feed a
> rotary.
>
> Ed, I understand your math, but even if the local inlet velocity is much
> higher, we dont care.  the velocities adverage out to the same, as the
> volume of air = velocity * carb area.
>
> If the velocities are higher, the rotary consumes more air, and makes more
> power.
>
> Eric
>
> ----- Original Message -----
> From: "Ed Anderson" <eanderson@carolina.rr.com>
> To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
> Sent: Thursday, February 10, 2005 8:31 AM
> Subject: [FlyRotary] : Same HP = Same Air Mass <> same air Velocity II
> [FlyRotary] Re: Ellison, the missing piece
>
>
> Good question, Tom.
>
> That interpretation did occur to me.  I think the answer depends on your
> assumptions, IF using commonly accepted formulas for calculating air flow
vs
> rpm and displacement (and considering both are positive displacement
> pumps) - then the 360 CID lycoming turning 2800 rpm and the rotors in the
> rotary turning 2100 rpm (6300 rpm E shaft) ingest the same total quantity
of
> air in one minute - approx 291 CFM.  In comparing the two engines, its
> accepted that you compare them over the standard 720deg 4 stroke cycle -
> that means that 4 of the rotary faces have gone through their cycle in the
> same 720 deg of rotation.
>
> But, assuming the formulas are correct, then they both end up with the
same
> amount of air in the engine to create the same HP.  I think my math is
> correct on the smaller/unit displacement and longer period of rotation for
> the rotary for the same intake of air.  However, in both cases the air
flow
> is pulsating and pulsating differently.  So if the total displacement for
> the rotary over that 720 deg is less than the Lycoming and the time it
takes
> to complete that rotation is slower AND you still ingest the same amount
of
> total Air then the only way I can see that happening is the velocity of
the
> air in the rotary's intake has to be considerably higher than in the
> Lycoming.
>
>  The only other alternative answer I see if that the commonly accepted
> formula for comparing the rotary to the reciprocating 4 stroke is
incorrect
> (I got beat about the head mercilessly by a number of respected rotary
> experts  challenging that formula , so I wont' go there again (at least
not
> now {:>)).
>
>  Air Flow = Total Displacement * RPM/(2 - accounting for only every other
> cylinder sucking on each rev * 1728 (conversion of cubic inches to cubic
> feet)  = TD*RPM/(2*1728)
>
> For the 360 CID Lycoming at 2800 rpm, Air Flow = 360*2800/(2*1728) =
291.66
> CFM
>
> Using the commonly accepted notion that a rotary is equivalent to a 160
CID
> 4 stroke reciprocating engine because of the 4 faces of 40 CID that
complete
> there cycle in 720 deg.
>
> For the 160 CID Rotary at 6000 rpm, Air Flow = 160 * 6300/(2*1728) =
291.66
> CFM
>
> So if both ingest the 291 CFM and the rotary has less total displacement
> (over 720 deg) then disregarding any of my math on rotation period
> differences you still have to account for why the rotary can ingest the
same
> amount of air with less displacement.  (Now I must admit I have my
> suspicions about the commonly accepted (racing approved) formula for the
> rotary.  However, if my suspicions about the rotary formula are correct,
it
> would make the rotary even more efficient at ingesting air - so I won't go
> there {:>)).
>
> If my logic and calculations are correct then this implies the Ve of the
> rotary is considerably better than the Lycoming and is great than 100%. I
> mentioned a few of the reasons why the Ve of the rotary may indeed be
better
> in the previous message.
>
> Now, its possible that the stories about the Ellison not working well on
the
> rotary is just that - a story OR there could be a plausible physical
reason
> as I have poorly attempted to present.
>
>
> Ed
>
>
> ----- Original Message -----
>   From: Tom
>   To: Rotary motors in aircraft
>   Sent: Wednesday, February 09, 2005 11:54 PM
>   Subject: [FlyRotary] Re: Same HP = Same Air Mass <> same air Velocity
> [FlyRotary] Re: Ellison, the missing piece
>
>
>   Ed,
>
>   >The rotary has 40 CID displacement per face and 2 facesx 2 rotors =
4*40
> or 160 CID for one rev.  So the rotary has 22% less displacement per
> revolution and the longer rotation period.<
>
>   and
>
>   >So if the rotary has less displacement of the sucking component and
must
> take 25% longer for each revolution.  Therefore the only way it can obtain
> an equal amount of air is for the intake air to have a higher velocity
than
> the Lycoming does.<
>
>   Isn't 'displacement' equal to the amount of air needing to be ingested?
> So 22% less displacement equates to 22% less air and the rotarys longer
> rotation period gives it more time for air to push in?    And then the
> intake air velocity should be lower?
>
>
>
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