X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from ms-smtp-01.southeast.rr.com ([24.25.9.100] verified) by logan.com (CommuniGate Pro SMTP 5.1.9) with ESMTP id 2079010 for flyrotary@lancaironline.net; Fri, 01 Jun 2007 15:02:00 -0400 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.100; envelope-from=eanderson@carolina.rr.com Received: from edward2 (cpe-024-074-103-061.carolina.res.rr.com [24.74.103.61]) by ms-smtp-01.southeast.rr.com (8.13.6/8.13.6) with SMTP id l51J19YX016346 for ; Fri, 1 Jun 2007 15:01:09 -0400 (EDT) Message-ID: <002001c7a47f$3f74f5a0$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: PP Ve??? was Re: Intake CFM air flow Date: Fri, 1 Jun 2007 15:01:23 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_001D_01C7A45D.B8149960" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3028 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_001D_01C7A45D.B8149960 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable 12A Pports can do 310 HP at 10,700 RPM at .667 . Lynn E. Hanover Thanks for that Data point, Lynn. I see if I can use it to get a = gestimate on PP Ve. So if a 12A is 73 cubic inches then at 10700 rpm and 100%Ve it would = flow 73*10700/(1728) =3D 452 CFM air flow.=20 So if at this 100% Ve we get less power than 310HP then we can assume = the Ve of the 12A must be greater than 100%Ve to give us more power. At sea level standard day 1 cubic foot of air =3D 0.076 lbms. So for = that flow we would have 0.076 * 452 =3D 34.35 lbs/min. Now I don't = have any idea what Air Fuel ratio a rotary racer uses but I best power = is reportedly to close to 12:1 =20 Assuming a race air fuel ratio of around 12:1 then the fuel needed = for that ratio at that airflow. Then the fuel needed would be 34.35 = /12 :=3D 2.8624 lb of gasoline per minute. =20 A lb of gasoline has 19000 BTU depending on octane. Higher octane = has less so assuming 19000 BTU/Lbm gasoline, we can next calculate the = power being produced in the engine. Converting 2.8624lb/min of gasoline into lb/sec we have 2.8624/60 =3D = 0.047708 lb/sec. To find the BTU we have .048*19000 =3D 912 BTU/sec. IF ALL this energy = were converted to torque it would give 912 *778 =3D 709536 ft-lbs of = torque. or divide by 550 =3D 1290 HP!!! Unfortunately, we know approx 50% goes out the tail pipe as heat and = another 25% (more or less) is Waste heat rejected by our coolers leaving = us somewhere around 25-30% depending on whose estimate you use for = efficiency of a rotary engine of 1290 *.25 =3D 322 HP or using 30% 1290 = *.30 =3D 387 HP Hummm, since the 12A PP is producing 310, but the calculations shows = it should be getting closer to 322 HP that would suggest a PP port 12A = flows less than 100% Ve. If fact, it would suggest that the Ve of the = 12A at 10700 rpm is closer to 310/322 *100 =3D 96.27%Ve. But, this is instantaneous BHP, I have not subtracted for mechanical = or other inefficiencies so taking a guess that amounts to around 5% of = the total. Then to get a dyno of 310 HP the engine would need to = produce 310*1.05 =3D 325 HP. So here we would get 325/322 =3D 101% Ve = for the 12A at 10700 rpm. Given we know that some racers are restricted by the size of the = intakes permitted (is this true for the PP, Lynn?) perhaps that is why = the VE seems a bit on the low side. But, that's just a guess. Well, = that was my best crack at trying to determine the efficiency of a PP. =20 So anybody else having an idea or source of information or opinion - = jump in. Ed =20 ------=_NextPart_000_001D_01C7A45D.B8149960 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
 
12A  Pports can do 310 HP at = 10,700 RPM at=20 .667 .
 
Lynn E. Hanover
 
 
Thanks for that Data point, Lynn.  I see if I can use it to = get a=20 gestimate on PP Ve.

 So if a 12A is 73 cubic inches = then=20 at 10700 rpm and 100%Ve it would flow
 
73*10700/(1728) =3D   452 CFM air flow. 
 
 So if at this 100% Ve we get less power than = 310HP then=20 we can assume the Ve of the 12A must be greater than 100%Ve to give us = more=20 power.
 
At sea level standard day 1 cubic foot of air =3D 0.076 = lbms.  So for=20 that flow we would have 0.076 * 452 =3D  34.35 lbs/min.  =  Now I=20 don't have any idea what Air Fuel ratio a rotary racer uses but I best = power=20 is reportedly to  close to 12:1 
 
 Assuming a race air fuel ratio of around 12:1 then the fuel = needed=20 for that ratio at that airflow.  Then  the fuel needed would = be=20 34.35 /12 :=3D  2.8624 lb of gasoline per minute.   =
 
  A lb of gasoline has 19000 BTU depending on=20 octane.  Higher octane has less  so assuming 19000 = BTU/Lbm=20 gasoline, we can next calculate the power being produced in the = engine.
 
Converting 2.8624lb/min of gasoline into lb/sec we have = 2.8624/60 =3D=20 0.047708 lb/sec.
 
To find the BTU we have .048*19000 =3D 912 BTU/sec. IF ALL = this energy=20 were converted to torque it would give  912 *778 =3D 709536 = ft-lbs of=20 torque.  or divide by 550 =3D 1290 HP!!!
 
Unfortunately, we know approx 50% goes out the tail pipe as heat = and=20 another 25% (more or less) is Waste heat rejected by our coolers = leaving us=20 somewhere around 25-30% depending on whose estimate you use=20 for efficiency of a rotary engine of  1290 *.25 = =3D 322 HP=20 or using 30% 1290 *.30 =3D 387 HP
 
 
Hummm, since the 12A PP is producing 310, = but the calculations=20 shows it should be getting closer to 322 HP that would suggest a PP = port 12A=20 flows less than 100% Ve.  If fact, it would suggest that the Ve = of the=20 12A at 10700 rpm is closer to 310/322 *100 = =3D 96.27%Ve.
 
But, this is instantaneous BHP,  I have not subtracted for=20 mechanical or other inefficiencies so taking a guess that amounts = to=20 around 5% of the total.  Then to get a dyno of 310 HP = the=20 engine would need to produce 310*1.05 =3D 325 HP.  So here we = would get=20 325/322 =3D  101% Ve for the 12A at 10700 rpm.
 
Given we know that some racers are restricted by the size of = the=20 intakes permitted (is this true for the PP, Lynn?) perhaps that is why = the VE=20 seems a bit on the low side.  But, that's just a guess.  = Well, that=20 was my best crack at trying to determine the efficiency of a PP.  =
 
So anybody else having an idea or source of information or = opinion - jump=20 in.
 
Ed
 
 
 
 
 
 
 
 
 
 
 
 
 
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