Return-Path: Sender: (Marvin Kaye) To: flyrotary@lancaironline.net Date: Wed, 04 Feb 2004 12:47:29 -0500 Message-ID: X-Original-Return-Path: Received: from wb2-a.mail.utexas.edu ([128.83.126.136] verified) by logan.com (CommuniGate Pro SMTP 4.1.8) with ESMTP-TLS id 2975210 for flyrotary@lancaironline.net; Wed, 04 Feb 2004 11:18:15 -0500 Received: (qmail 33057 invoked from network); 4 Feb 2004 16:18:13 -0000 Received: from dhcp-191-101.per.utexas.edu (HELO benefits3.mail.utexas.edu) (146.6.191.101) by wb2.mail.utexas.edu with RC4-SHA encrypted SMTP; 4 Feb 2004 16:18:13 -0000 X-Original-Message-Id: <5.1.1.5.2.20040204100848.020cf688@localhost> X-Sender: msteitle@mail.utexas.edu@localhost X-Mailer: QUALCOMM Windows Eudora Version 5.1.1 X-Original-Date: Wed, 04 Feb 2004 10:18:06 -0600 X-Original-To: "Rotary motors in aircraft" From: Mark Steitle Subject: Re: [FlyRotary] air flow, induction air scoop In-Reply-To: Mime-Version: 1.0 Content-Type: multipart/alternative; boundary="=====================_64047015==.ALT" --=====================_64047015==.ALT Content-Type: text/plain; charset="iso-8859-1"; format=flowed Content-Transfer-Encoding: quoted-printable I don't understand how a 13B can flow 277.77 cu ft/min at 6000rpm, while a= =20 20B flows 275 cu ft. min.(Al's numbers), unless the 275 cu ft figure is for= =20 4000rpm instead of 6000 rpm. Using Ed's formula, I get 416.67 cu ft/min at= =20 6000rpm, or 31.66 lbs/minute. Which number is correct? Mark S. At 12:23 PM 2/1/2004 -0800, you wrote: > >Subject: [FlyRotary] Re: air flow > > > Here is a formula 80 CID * RPM/1728. Example for 6000 rpm, 80*6000 >rpm/1728 =3D 277.77 Cubic Feet of air per minute. A cubic foot of air (at= sea >level standard day) is approx 0.076 lbm/cubic foot so 277.77*0.076 =3D= 21.11 >lbm/min air mass at 6000 rpm . Just change the rpm to what ever value you >want. Also note the weight of air of course decreases with altitude so >277.77 CFM provides 21.11 lbm/minute at sea level but less at altitude. > >Note this assumes 100 % Volumetric efficiency. > >Ed Anderson > >Oh; boy, now you ve done it. My beautiful intake scoop is already=20 >installed, and I apparently designed it based on the wrong number. > >You re numbers look right to me. I ll have see if I can figure out what I= =20 >did. > >Thanks. > >Al > >Your Welcome, Al > > Perhaps you used the correct number for the design and just had a=20 > brain fade when you quoted the number for CFM - Hopefully? > >Ed Anderson > >Ed: >Well, it seems that I did use my number (275 cfm) and then computed a=20 >scoop inlet area of 3.9 sq. in. based on an airspeed of 100 mph (steep=20 >climb). When I fabricated it, I made a little larger so it would be less= =20 >restrictive for low speed. Yesterday I made a more accurate measurement=20 >of the opening, and found it to be 5 =BC sq in. Assuming everything is=20 >ideal, this would be about right for zero ram pressure at about 130 mph,=20 >which is close to my typical climb speed. At typical cruise of 200 mph at= =20 >say 8000 ft, I should have same ram pressure for a little better cruise=20 >efficiency. So all that is fine. > >There is always the trade off of a scoop that is not restrictive on the=20 >takeoff roll, and one that gives you some ram air advantage without much=20 >drag at high speed. I am considering a small, weighted trapdoor in the=20 >bottom of the intake duct inside the cowl that will open and allow more=20 >air in when the pressure in the duct is less than ambient, and closes when= =20 >the pressure is more than ambient. > >Al > > > > >> Homepage: http://www.flyrotary.com/ > >> Archive: http://lancaironline.net/lists/flyrotary/List.html --=====================_64047015==.ALT Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable I don't understand how a 13B can flow 277.77 cu ft/min at 6000rpm, while a 20B flows 275 cu ft. min.(Al's numbers), unless the 275 cu ft figure is for 4000rpm instead of 6000 rpm.  Using Ed's formula, I get 416.67 cu ft/min at 6000rpm, or 31.66 lbs/minute.  Which number is correct?
 
Mark S.

At 12:23 PM 2/1/2004 -0800, you wrote:

 
Subject: [FlyRotary] Re: air flow
 
 
 Here is a formula   80 CID * RPM/1728.  Example for 6000 rpm,  80*6000
rpm/1728 =3D 277.77 Cubic Feet of air per minute.  A cubic foot of air (at sea
level standard day) is approx 0.076 lbm/cubic foot   so 277.77*0.076 =3D 21.11
lbm/min air mass at 6000 rpm .  Just change the rpm to what ever value you
want.  Also note the weight of air of course decreases with altitude so
277.77 CFM provides 21.11 lbm/minute at sea level but less at altitude.
 
Note this assumes 100 % Volumetric efficiency.
 
Ed Anderson
 
Oh; boy, now you ve done it.  My beautiful intake scoop is already installed, and I apparently designed it based on the wrong number.
 
You re numbers look right to me.  I ll have see if I can figure out what I did.
 
Thanks.
 
Al
 
Your Welcome, Al
 
    Perhaps you used the correct number for the design and just had a brain fade when you quoted the number for CFM - Hopefully?
 
Ed Anderson
 
Ed:
Well, it seems that I did use my number (275 cfm) and then computed a scoop inlet area of 3.9 sq. in. based on an airspeed of 100 mph (steep climb).  When I fabricated it, I made a little larger so it would be less restrictive for low speed.  Yesterday I made a more accurate measurement of the opening, and found it to be 5 =BC sq in.  Assuming everything is ideal, this would be about right for zero ram pressure at about 130 mph, which is close to my typical climb speed.  At typical cruise of 200 mph at say 8000 ft, I should have same ram pressure for a little better cruise efficiency.  So all that is fine.
 
There is always the trade off of a scoop that is not restrictive on the takeoff roll, and one that gives you some ram air advantage without much drag at high speed.  I am considering a small, weighted trapdoor in the bottom of the intake duct inside the cowl that will open and allow more air in when the pressure in the duct is less than ambient, and closes when the pressure is more than ambient.
 
Al
 
 

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