X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from ms-smtp-02.southeast.rr.com ([24.25.9.101] verified) by logan.com (CommuniGate Pro SMTP 5.1.11) with ESMTP id 2212201 for flyrotary@lancaironline.net; Sun, 29 Jul 2007 21:30:41 -0400 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.101; 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-02.southeast.rr.com (8.13.6/8.13.6) with SMTP id l6U1U5sA014772 for ; Sun, 29 Jul 2007 21:30:05 -0400 (EDT) Message-ID: <000001c7d249$3e51eaf0$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Air/fuel flow Date: Sun, 29 Jul 2007 20:21:49 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0006_01C7D21E.17FF07F0" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3138 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_0006_01C7D21E.17FF07F0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable George, I've done a bit of this. To get an airflow in CFM to a fuel = Flow in Gallons per minute I start with your airflow of 173.6 CFM and convert that to mass air = flow rather than volume because the air/fuel ratio is based on mass = ratio not volume ratio. =20 The reason for this is that a cubic foot of fuel will weigh the same = everywhere but a cubic foot of air will weigh differently at different = air densities (altitudes). I'll use the notation of Ft^3 (foot cube) = for CF and Ft^2 (foot squared) for Sq ft as it makes it easier to show = where units cancel out. At sea level the mass/weigth of 1 cubic foot of air is around 0.076 = lbm/Ft^3 (of air). So with 173.6 Ft^3/Min * 0.076 lbm/Ft^3 =3D = 13.1936 lbm/minute of air(the two CFM or Ft^3 factors cancel out leaving = you with units of pounds mass (lbm) per minute). Then taking your = air/fuel ratio value of 14.7 we have 13.1936 /14.7 =3D 0.8975 lbm/minute = of fuel. Various figures are used for weight of gasoline depending in part on its = octane. The most common use figure is 6 lbs/ gallon, so I'll use that = figure. So 1 minute at your flow rate we have 0.8975 Lbm/Min / 6 lb/gal =3D = 0.149587 Gal/Min or in more conventional terms 0.149587 gal/min * 60 = min/hour =3D 8.975 Gal/hour fuel flow for that air flow and air/fuel = ratio. There are other approaches, but unless I've screwed up this should be = pretty close to the answer. For inlet, you take the CFM and you need one other factor - what = velocity do you want through your inlet? Lets say you want a higher = velocity of around 176 feet/sec (120 MPH) then we know that Volume =3D = Area * length. If were want 176 feet/sec velocity from 176 CFM air flow = then coverting CFM to cubic feet per second were have 176Ft^3/min / 60 = Second/Minute =3D 29.333 Ft^3/ Second So solving for Area (of opening) =3D Volume/Length =3D 2.9333Ft^3/sec/ = 175 Ft/Sec =3D .01676 Ft^2of area which is 0.01676Ft^2 * 144 = inch^2/Ft^2 =3D 2.4137 inch^2 of opening for that assumed velocity. Again, if I haven't screwed up for an air velocity through the opening = of 176 feet/sec that would be the opening size. If you wanted slower = air flow then the area of the opening would increase, faster and it = would decrease up to a choke point where the air flow would flow no = faster. Hope that helped Ed NOT LYnn ----- Original Message -----=20 From: George Lendich=20 To: Rotary motors in aircraft=20 Sent: Sunday, July 29, 2007 7:10 PM Subject: [FlyRotary] Air/fuel flow Lynn, Following from previous post from your self, I'm looking at AIR FLOW and = FUEL FLOW for my single rotor. EG [40 cu" per face x 3 ( 7,500/3)]/1728 =3D 173.6 CFM( Air)/ 14.7 ( = Ratio)=3D 11.8 cfm ( Fuel) 40x3x2,000/1728=3D138CFM( Air)/14.7( ratio)=3D9.44cfm( fuel) Question, 1. how does one convert cfm to Gallons per minute? 2. how does one convert CFM to diameter of inlet? George ( down under) ------=_NextPart_000_0006_01C7D21E.17FF07F0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable George, I've done a bit of = this.  To get an=20 airflow in CFM to a fuel Flow in Gallons per minute

I  start = with=20 your airflow of 173.6 CFM and convert that to mass air flow rather than = volume=20 because the air/fuel ratio is based on mass ratio not volume = ratio. =20

The reason for this is that a cubic foot of fuel will weigh the = same=20 everywhere but a  cubic foot of air will weigh differently at = different air=20 densities (altitudes).  I'll use the notation of Ft^3 (foot cube) = for CF=20 and Ft^2 (foot squared) for Sq ft as it makes it easier to show where = units=20 cancel out.

At sea level the mass/weigth of 1 cubic foot of air = is around=20 0.076 lbm/Ft^3 (of air).  So with 173.6 Ft^3/Min  * 0.076=20 lbm/Ft^3  =3D 13.1936 lbm/minute of air(the two CFM or Ft^3 factors = cancel=20 out leaving you with units of pounds mass (lbm) per minute).  Then = taking=20 your air/fuel ratio value of 14.7 we have 13.1936 /14.7 =3D 0.8975 = lbm/minute of=20 fuel.

Various figures are used for weight of gasoline depending = in part=20 on its octane.  The most common use figure is 6 lbs/ gallon, so = I'll use=20 that figure.

So 1 minute at your flow rate we have  0.8975 = Lbm/Min /=20 6 lb/gal =3D 0.149587 Gal/Min or in more conventional terms 0.149587 = gal/min * 60=20 min/hour =3D 8.975 Gal/hour fuel flow for that air flow = and=20 air/fuel ratio.

There are other approaches, but unless I've = screwed up=20 this should be pretty close to the answer.

For inlet, you take = the CFM=20 and you need one other factor - what velocity do you want through your=20 inlet?  Lets say you want a higher velocity of around 176 feet/sec = (120=20 MPH)  then we know that Volume =3D Area * length.  If were = want 176=20 feet/sec velocity from 176 CFM air flow then coverting CFM to cubic feet = per=20 second were have 176Ft^3/min / 60 Second/Minute =3D 29.333 Ft^3/=20 Second

So solving for Area (of opening) = =3D=20 Volume/Length =3D 2.9333Ft^3/sec/ 175 Ft/Sec =3D .01676 Ft^2of = area which=20 is 0.01676Ft^2  * 144 inch^2/Ft^2 =3D 2.4137 inch^2 of opening for = that=20 assumed velocity.

Again, if I haven't screwed up for an air = velocity=20 through the opening of 176 feet/sec that would be the opening = size.  If you=20 wanted slower air flow then the area of the opening would increase, = faster and=20 it would decrease  up to a choke point where the air flow would = flow no=20 faster.

Hope that helped

Ed NOT LYnn
----- Original = Message=20 -----
From: George Lendich
To: Rotary motors in aircraft =
Sent:=20 Sunday, July 29, 2007 7:10 PM
Subject: [FlyRotary] Air/fuel=20 flow



Lynn,
Following from previous post from your = self, I'm=20 looking at AIR FLOW and FUEL FLOW for my single rotor.
EG [40 cu" per = face x=20 3 ( 7,500/3)]/1728 =3D 173.6 CFM( Air)/ 14.7 ( Ratio)=3D 11.8 cfm (=20 Fuel)
40x3x2,000/1728=3D138CFM( Air)/14.7( ratio)=3D9.44cfm(=20 fuel)
Question,
1. how does one convert cfm to Gallons per = minute?
2.=20 how does one convert CFM to diameter of inlet?
George ( down=20 under) ------=_NextPart_000_0006_01C7D21E.17FF07F0--