Return-Path: Received: from [24.25.9.100] (HELO ms-smtp-01-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.3c1) with ESMTP id 723065 for flyrotary@lancaironline.net; Wed, 09 Feb 2005 21:59:16 -0500 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.100; envelope-from=eanderson@carolina.rr.com Received: from edward2 (cpe-024-074-185-127.carolina.rr.com [24.74.185.127]) by ms-smtp-01-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id j1A2wTbo023335 for ; Wed, 9 Feb 2005 21:58:29 -0500 (EST) Message-ID: <002a01c50f1c$672f6f30$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Same HP = Same Air Mass <> same air Velocity [FlyRotary] Re: Ellison, the missing piece Date: Wed, 9 Feb 2005 21:58:31 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0027_01C50EF2.7E16B8E0" 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_0027_01C50EF2.7E16B8E0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Tom, I have no experience with the Ellison, but the answer may be not in the = total air consumed, but in how it is sucked in. Producing the same HP = requires essentially the same air/fuel regardless - its how it gets = there that may make a difference regarding the Ellison. The aircraft engine gulps in air in large chunks. The four large = cylinders running at say 2800 rpm and only two cylinder "suck" each = revolution. So there air flow characteristic is different than a = rotary. With the rotary you have six faces (piston analogs) of less = displacement rotating (the rotors not the eccentric shaft) at approx = 2000 rpm (for 6000 rpm eccentric shaft). The rotary sips smaller chunks = of air. The total amount of air would have to be the same for both engines (same = HP), however, "Average" covers a multitude of difference in the actual = air flow pattern. I see 2 large masses of air in the intake for the = aircraft engine each revolution. The rotary would have 4 smaller = airmass packages ( Yes, the rotary has six faces but only four have come = around in a 720 deg revolution) in the intake. So the interval between = the center of mass for each package is roughly 1/2 that of the Lycoming. = =20 For a specific example let see what numbers may tell us. Lets take a Lycoming of 360 CID turning at 2800 rpm and a rotary of 80 = CID with the rotors turning at 2100 rpm (6300 E shaft ). This will have = both engines sucking (assuming 100 % Ve for both) approx 291.67 Cubic = Feet/Minute. And assuming the same BSFC they would be producing the = same HP. =20 But, lets see where there are differences. 1st a 360 CID Lycoming at 2800 rpm has a period of revolution of = 2800/60 =3D 46.6666 Revs/Sec or a rotation period of 1/rev-sec =3D 1/ = 46.666 =3D .021428 seconds or 21.428 milliseconds. During that time its = sucking intake air for 2 cylinders in 360 deg of rotation. The rotary = however, has its rotors spinning at 2100 rpm (to draw the same amount of = air) which gives it a rotation period of 2100/60 =3D 33.3333 Revs/Sec or = a period of 1/35 =3D 0.02857 seconds or 28.57 ms. The rotary is also = drawing in two chambers of air in 360 deg of rotation.=20 Here the rotary e shaft is spinning at 6300 rpm to give the rotor a = rotation rate of 2100 rpm 6300/3 =3D 2100. Eshaft rpm =20 Displacement rpm =20 CFM =20 360 2800 291.67 =20 =20 =20 =20 =20 80 6300 291.67=20 =20 =20 =20 =20 =20 6300 40 2100(rotor) 291.67 =20 =20 So right there we have a difference of approx 25% difference in the = rotation time of the pumps pulling in the same average amount of air. = The rotary takes approx 25% more time than the Lycoming to complete a = revolution.. A 360 CID Lycoming (forgetting compression ratios for this discussion) = has 360/4 =3D 90 cid displacement per cylinder or 180 CID for on rev. = The rotary has 40 CID displacement per face and 2 facesx 2 rotors =3D = 4*40 or 160 CID for one rev. So the rotary has 22% less displacement = per revolution and the longer rotation period. 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. The air velocity of the area in the intake for the rotary would appear = to have to be much higher than the Lycoming. If my assumptions and = calculations are correct that would imply (at least to me) that to = minimize air flow restriction a larger opening would be required on the = rotary compared to the same HP Lycoming. Its not that one is taken in = more air its that the rotary has less time and smaller displacement pump = so must take in the air at a higher velocity. The fact that the rotary has no valves to block the flow of air may be = one reason that it can over come what would appear to be less favorable = parameters for sucking air. An additional factor that may play a role is = the fact that air mass pulsation in the rotary intake is less than the = Lycoming. This would mean less starting and stopping of air movement, = so the velocity would seem to remain steadier and on an average higher = than for the air pulses for the Lycoming which if you factor the = start/slowing/start of air flow may lower its overall velocity compared = to the rotary. In summary, while the total air intake in equal for engines producing = equal HP. It is likely that the air flow to the rotary may be = considerably higher in order to ingest the same amount of air over the = same time. This may be why there is a perception that the Ellison = model that may work well for a Lycoming may not work as well for a = rotary. =20 Well, anyhow, that's my best shot - if its incorrect perhaps somebody = can take it from here, but I think the answer lies in the different = pumping configuration of the two engines. Best Regards Ed =20 I'm under the impression I have an answer. Isn't there a law of motor performance that says that two motors = putting out the same horsepower are consuming the same amount of = air&fuel, assuming efficiency differences were not significant? So if you had a 13b and a O-360 putting out the same horsepower for a = single given 1 revolution of the propeller, they should be consuming the = same amount of air and fuel during that 1 propeller revolution. (I THINK = chosing 1 propeller rpm is a correct standard) Bill pointed out that the 13b operates at a higher rpm, and we know = that there's more combustion charges consumed by the 13b to make that 1 = prop rpm. =20 The difference, the missing piece, each 13b combustion charge consumes = a SMALLER amount of fuel/air than the piston powerplants less frequent = combustion charge. ??? So the 13b burns a smaller amount more = frequently. ??? If this is all true, then the Ellison isn't on the trash heap yet.=20 Tom WRJJRS@aol.com wrote: Group, I want to remind everyone about how much a priority the large = volume inlets are to us. I believe Ed Anderson was mentioning in one of = his posts how difficult it can be to get a MAP signal in the airbox of = one of our PP engines. This is a perfect indication of why the smaller = throttle bodies used on some of the slow turning engines will kill our = HP.=20 -------------------------------------------------------------------------= ----- Do you Yahoo!? Yahoo! Search presents - Jib Jab's 'Second Term' ------=_NextPart_000_0027_01C50EF2.7E16B8E0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Tom,
 
I have no experience with the Ellison, = but the=20 answer may be  not in the total air consumed, but in how it is = sucked=20 in.  Producing the same HP requires essentially the same air/fuel=20 regardless - its how it gets there that may make a difference regarding = the=20 Ellison.
 
 
  The aircraft engine gulps in air = in=20  large chunks.  The four large cylinders running at say 2800 = rpm and=20 only two cylinder "suck" each revolution.  So there air flow = characteristic=20 is different than a rotary.  With the rotary you have six faces = (piston=20 analogs) of less displacement rotating (the rotors not the eccentric = shaft) at=20 approx 2000 rpm (for 6000 rpm eccentric shaft).  The rotary sips = smaller=20 chunks of air.
 
The total amount of air would have to = be the same=20 for both engines (same HP), however, "Average" covers a multitude of = difference=20 in the actual air flow pattern.  I see 2 large masses of air in the = intake=20 for the aircraft engine each revolution.  The rotary would have 4 = smaller=20 airmass packages ( Yes, the rotary has six faces but only four have come = around=20 in a 720 deg revolution) in the intake. So the interval between the = center of=20 mass for each package is roughly 1/2 that of the Lycoming.  =
 
For a specific example let see what = numbers may=20 tell us.
 
Lets take a Lycoming of 360 CID turning = at 2800 rpm=20 and a rotary of 80 CID with the rotors turning at 2100 rpm (6300 E shaft = ).  This will have both engines sucking (assuming 100 % Ve for = both) approx=20 291.67 Cubic Feet/Minute.  And assuming = the same=20 BSFC they would be producing the same HP. 
 
But, lets see where there are=20 differences.
 
1st a 360 CID Lycoming at =  2800 rpm has a=20 period of revolution of 2800/60 =3D 46.6666 Revs/Sec or a rotation = period of=20 1/rev-sec =3D 1/ 46.666 =3D .021428 seconds or 21.428 = milliseconds.  During=20 that time its  sucking intake air  for 2 = cylinders in=20 360 deg of rotation.  The rotary however, has its rotors spinning = at 2100=20 rpm (to draw the same amount of air) which gives it a rotation period of = 2100/60=20 =3D 33.3333 Revs/Sec or a period of 1/35 =3D 0.02857 seconds or 28.57 = ms. The=20 rotary is also drawing in two chambers of air in 360 deg of=20 rotation. 
 
Here the rotary e shaft is spinning at = 6300 rpm to=20 give the rotor a rotation rate of 2100 rpm 6300/3 =3D 2100.

Eshaft rpm   =

   Displacement

     rpm   =20

CFM

360

2800

291.67

80

6300

291.67

6300

40

 2100(rotor)

291.67

 
 
 So right there we have a = difference of approx=20 25% difference in the rotation time of the pumps pulling in the same = average=20 amount of air. The rotary takes  approx = 25% more=20 time than the Lycoming to complete a revolution..
 
 A 360 CID Lycoming (forgetting = compression=20 ratios for this discussion) has 360/4 =3D 90 cid displacement per = cylinder or 180=20 CID for on rev.  The rotary has 40 CID displacement per face = and 2=20 facesx 2 rotors =3D  4*40 or 160 CID for one rev.  So the = rotary=20 has 22%  less displacement per revolution and the  longer = rotation period.
 
So if the rotary has less displacement = of the=20 sucking component and must take 25% longer for each revolution.  = Therefore=20 the only way it can obtain an equal amount of air is for the intake air = to have=20 a higher velocity than the Lycoming does.
 
The air velocity of the area in the = intake for the=20 rotary would appear to have to be much higher than the Lycoming.  = If my=20 assumptions and calculations are correct that would imply (at least to = me) that=20 to minimize air flow restriction a larger opening would be required = on the=20 rotary compared to the same HP Lycoming.  Its not that one is = taken in=20 more air its that the rotary has less time and smaller displacement pump = so must=20 take in the air at a higher velocity.
 
  The fact that the rotary has no = valves to=20 block the flow of air may be one reason that it can over come what would = appear=20 to be  less favorable parameters for sucking air. An additional = factor that=20 may play a role is the fact that air mass pulsation in the rotary intake =  is less than the Lycoming.  This would mean less starting and = stopping of air movement, so the velocity would seem to remain steadier = and on=20 an average higher than for the air pulses for the Lycoming which if you = factor=20 the start/slowing/start of air flow may lower its overall velocity = compared to=20 the rotary.
 
In summary, while the total air intake = in equal for=20 engines producing equal HP. It is likely that the air flow to the=20 rotary may be considerably higher in order to ingest the same = amount of air=20 over the same time.  This may be why  there is a = perception that=20 the Ellison model  that may work well for a Lycoming may not work = as well=20 for a rotary. 
 
Well, anyhow, that's my best shot - if = its=20 incorrect perhaps somebody can take it from here, but I think the answer = lies in=20 the different pumping configuration of the two engines.
 
Best Regards
 
Ed
 
I'm under the impression I have an answer.
 
Isn't there a law of motor performance that says that two motors = putting=20 out the same horsepower are consuming the same amount of air&fuel, = assuming efficiency differences were not significant?
 
So if you had a 13b and a O-360 putting out the same horsepower = for a=20 single given 1 revolution of the propeller, they should be consuming = the same=20 amount of air and fuel during that 1 propeller revolution. (I THINK = chosing 1=20 propeller rpm is a correct standard)
 
Bill pointed out that the 13b operates at a higher rpm, and we = know that=20 there's more combustion charges consumed by the 13b to make that 1 = prop=20 rpm. 
 
The difference, the missing piece, each 13b combustion charge = consumes a=20 SMALLER amount of fuel/air than the piston powerplants less frequent=20 combustion charge.   ???   So the 13b burns a = smaller=20 amount more frequently.   ???
 
If this is all true, then the Ellison isn't on the trash heap = yet.
 
Tom

WRJJRS@aol.com wrote:
Group,
 I want to remind everyone about how much a priority the = large=20 volume inlets are to us. I believe Ed Anderson was mentioning in one = of his=20 posts how difficult it can be to get a MAP signal in the airbox of = one of=20 our PP engines. This is a perfect indication of why the smaller = throttle=20 bodies used on some of the slow turning engines will kill our HP.=20

Do you Yahoo!?
Yahoo! Search presents - Jib=20 Jab's 'Second Term' ------=_NextPart_000_0027_01C50EF2.7E16B8E0--