X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from ms-smtp-04.southeast.rr.com ([24.25.9.103] verified) by logan.com (CommuniGate Pro SMTP 5.1.9) with ESMTP id 2079214 for flyrotary@lancaironline.net; Fri, 01 Jun 2007 17:07:06 -0400 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.103; 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-04.southeast.rr.com (8.13.6/8.13.6) with SMTP id l51L6EIC016126 for ; Fri, 1 Jun 2007 17:06:14 -0400 (EDT) Message-ID: <002501c7a490$b8e0e1e0$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: BSHP Approach was [FlyRotary] Re: PP Ve??? was Re: Intake CFM air flow Date: Fri, 1 Jun 2007 17:06:28 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0022_01C7A46F.318C6C80" 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_0022_01C7A46F.318C6C80 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Thanks, Blake. I missed that point.=20 I did find one minor error in your transcription of your example to e = mail - a digit was dropped. I also used 10700 rpm as Lynn reported vice = the 10000 rpm in you example. With that using your approach (which is certainly simpler and more = accurate than the one I used) I came up with 110.19% VE which would = appear to give a PP a VE in the vicinity of 110% Ve. But, again, it would be nice to know if there were any restrictors or = if that was clean and clear intake. =20 Lynn? Ed ----- Original Message -----=20 From: "Blake Lewis" To: "Rotary motors in aircraft" Sent: Friday, June 01, 2007 4:35 PM Subject: [FlyRotary] Re: PP Ve??? was Re: Intake CFM air flow > 12A Pports can do 310 HP at 10,700 RPM at .667 >=20 > If the .667 is BSFC then start with that. > density of air at sea level 15C =3D .0765 lbs/cf > air/fuel =3D 12 >=20 > .667 bsfc x 310hp is 206.77 lbs/hr fuel >=20 > 206.77lbs x 12 =3D 2481.24 lbs/hr air >=20 > 2481.24 lbs/hr / 60min/hr =3D 41.354 lbs/min >=20 > (41.354 lbs/min) / (.0765 lbs/cf) * (1728 ci/cf) =3D 93411 ci /min = =3D 934113. Dropped a digit in transcription to e mail. >=20 > (93411ci/min) / (10000 rev/min) =3D 93.41 ci/rev =20 (934113 ci/min)/(10700 rev/min) =3D 87.300 ci/rev, I used the 10700 = rpm Lynn cited in his example >=20 > 1300cc / (2.54cm * 2.54cm * 2.54cm)/ci =3D 79.33 ci volume >=20 > 93.41ci / 79.33ci =3D 1.1774 =3D 117.74 % VE using the Ci/rev I got with 10700 rpm and 310 HP I arrived at 87.33 ci/79.22 ci =3D 110.19 % VE Good calculation approach, I'll save it for later use >=20 > Blake >=20 >=20 > On 6/1/07, Ed Anderson wrote: >> >> >> >> >> 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. >> >> 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 >> >> 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. >> >> 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. >> >> So anybody else having an idea or source of information or opinion - = jump >> in. >> >> Ed >=20 > -- > Homepage: http://www.flyrotary.com/ > Archive and UnSub: = http://mail.lancaironline.net:81/lists/flyrotary/List.html ------=_NextPart_000_0022_01C7A46F.318C6C80 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Thanks, Blake.  I missed that=20 point. 
 
 I did find one minor error in your = transcription of=20 your example to e mail - a digit was dropped. I  also used = 10700 rpm=20 as Lynn reported vice the 10000 rpm in you example.
 
With that using your approach (which is = certainly simpler=20 and more accurate than the one I used) I came up with 110.19% VE  = which=20 would appear to give a PP a VE in the vicinity of  110% = Ve.
 
  But, again, it would be nice to know if = there were=20 any restrictors or if that was clean and clear intake.  =
 
Lynn?
 
Ed
 
 
 
 
----- Original Message -----
From: "Blake Lewis" <blake.lewis@gmail.com>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Sent: Friday, June 01, 2007 4:35 PM
Subject: [FlyRotary] Re: PP Ve??? was Re: Intake = CFM air=20 flow

> 12A  Pports can do 310 HP at 10,700 = RPM at=20 .667
>
> If the .667 is BSFC then start with that.
> = density=20 of air at sea level 15C =3D .0765 lbs/cf
> air/fuel =3D 12
> =
>=20 .667 bsfc x 310hp is 206.77 lbs/hr fuel
>
> 206.77lbs x 12 = =3D=20 2481.24 lbs/hr air
>
> 2481.24 lbs/hr  /  = 60min/hr =3D=20 41.354 lbs/min
>
> (41.354 lbs/min)  / (.0765 lbs/cf) = * (1728=20 ci/cf) =3D 93411 ci /min  =3D  934113.  Dropped a digit in transcription to e=20 mail.
>
> (93411ci/min) / (10000 rev/min) = =3D 93.41=20 ci/rev 
 
   = (934113=20 ci/min)/(10700 rev/min) =3D 87.300=20 ci/rev,  I used the 10700 rpm Lynn cited in his=20 example
>
> = 1300cc /=20 (2.54cm * 2.54cm * 2.54cm)/ci =3D 79.33 ci volume
>
> = 93.41ci /=20 79.33ci =3D 1.1774 =3D 117.74 % VE
 
using the Ci/rev I got = with 10700=20 rpm and 310 HP I arrived at
 
87.33 ci/79.22 ci = =3D 110.19 %=20 VE
 
Good calculation = approach,=20 I'll save it for later use

>
> Blake
>
>
> On 6/1/07, Ed = Anderson=20 <eanderson@carolina.rr.com> = wrote:
>>
>>
>>
>>
>> = 12A =20 Pports can do 310 HP at 10,700 RPM at .667 .
>>
>> = Lynn E.=20 Hanover
>>
>>
>> Thanks for that Data point,=20 Lynn.  I see if I can use it to get a gestimate
>> on PP=20 Ve.
>>
>>  So if a 12A is 73 cubic inches then at = 10700=20 rpm and 100%Ve it would flow
>>
>> 73*10700/(1728)=20 =3D   452 CFM air flow.
>>
>>  So if at = this 100%=20 Ve we get less power than 310HP then we can assume the
>> Ve of = the 12A=20 must be greater than 100%Ve to give us more = power.
>>
>> At=20 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=20 lbs/min.   Now I don't have any idea
>> what Air Fuel = ratio a=20 rotary racer uses but I best power is reportedly to
>> close to = 12:1
>>
>>  Assuming a race air fuel ratio of = around 12:1=20 then the fuel needed for that
>> ratio at that airflow. =20 Then  the fuel needed would be 34.35 /12 :=3D  = 2.8624
>> lb of=20 gasoline per minute.
>>
>>   A lb of = gasoline has=20 19000 BTU depending on octane.  Higher octane has
>> = less  so=20 assuming 19000 BTU/Lbm gasoline, we can next calculate the = power
>>=20 being produced in the engine.
>>
>> Converting = 2.8624lb/min of=20 gasoline into lb/sec we have 2.8624/60 =3D 0.047708
>>=20 lb/sec.
>>
>> To find the BTU we have .048*19000 =3D = 912=20 BTU/sec. IF ALL this energy were
>> converted to torque it = would=20 give  912 *778 =3D 709536 ft-lbs of torque.  or
>> = divide by=20 550 =3D 1290 HP!!!
>>
>> Unfortunately, we know approx = 50% goes=20 out the tail pipe as heat and another
>> 25% (more or less) is = Waste=20 heat rejected by our coolers leaving us
>> somewhere around = 25-30%=20 depending on whose estimate you use for efficiency
>> of a = rotary=20 engine of  1290 *.25 =3D 322 HP or using 30% 1290 *.30 =3D 387=20 HP
>>
>>
>> Hummm, since the 12A PP is = producing 310,=20 but the calculations shows it
>> should be getting closer to = 322 HP=20 that would suggest a PP port 12A flows
>> less than 100% = Ve.  If=20 fact, it would suggest that the Ve of the 12A at
>> 10700 rpm = is closer=20 to 310/322 *100 =3D 96.27%Ve.
>>
>> But, this is = instantaneous=20 BHP,  I have not subtracted for mechanical or
>> other=20 inefficiencies so taking a guess that amounts to around 5% of = the
>>=20 total.  Then to get a dyno of 310 HP the engine would need to=20 produce
>> 310*1.05 =3D 325 HP.  So here we would get = 325/322=20 =3D  101% Ve for the 12A at
>> 10700 = rpm.
>>
>>=20 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=20 why the VE seems
>> a bit on the low side.  But, that's = just a=20 guess.  Well, that was my best
>> crack at trying to = determine the=20 efficiency of a PP.
>>
>> So anybody else having an = idea or=20 source of information or opinion - jump
>> = in.
>>
>>=20 Ed
>
> --
> Homepage:  http://www.flyrotary.com/
> Archive=20 and UnSub:   http://mail.lancaironline.net:81/lists/flyrotary/List.html
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