X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from mail03.syd.optusnet.com.au ([211.29.132.184] verified) by logan.com (CommuniGate Pro SMTP 5.1.11) with ESMTPS id 2212336 for flyrotary@lancaironline.net; Sun, 29 Jul 2007 23:56:04 -0400 Received-SPF: none receiver=logan.com; client-ip=211.29.132.184; envelope-from=lendich@optusnet.com.au Received: from george (d211-31-248-169.dsl.nsw.optusnet.com.au [211.31.248.169]) by mail03.syd.optusnet.com.au (8.13.1/8.13.1) with SMTP id l6U3tHpT008160 for ; Mon, 30 Jul 2007 13:55:18 +1000 Message-ID: <004801c7d25d$750127f0$a9f81fd3@george> From: "George Lendich" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Fw: [FlyRotary] Air/fuel flow Date: Mon, 30 Jul 2007 13:55:22 +1000 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0045_01C7D2B1.45E505F0" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.2180 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.2180 X-Antivirus: avast! (VPS 0657-0, 12/12/2006), Outbound message X-Antivirus-Status: Clean This is a multi-part message in MIME format. ------=_NextPart_000_0045_01C7D2B1.45E505F0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Well I'll be Ed, 2.4137sq" =3D (.877x.877xPi =3D 2.416), therefore .877 Radius or 1.754 = Dia or 44.55 mm for the carb opening. I'm working on the rest! George ( down under) =20 George, I've done a bit of this type of calculation, perhaps Lynn = won't mind if I give it a stab 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_0045_01C7D2B1.45E505F0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Well I'll be Ed,
2.4137sq" =3D (.877x.877xPi =3D 2.416), = therefore .877=20 Radius or 1.754 Dia or 44.55 mm for the carb opening.
I'm working on the rest!
George ( down under)
  
George, I've done a bit of this type of = calculation,=20 perhaps Lynn won't mind if I give it a stab
 
 To get an airflow in CFM to a fuel Flow = in Gallons=20 per minute

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

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

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

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

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

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

For inlet, you take the = CFM and=20 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=20 you wanted slower air flow then the area of the opening would = increase, faster=20 and it would decrease  up to a choke point where the air flow = would flow=20 no faster.

Hope that helped

Ed NOT LYnn
----- = Original=20 Message -----
From: George Lendich
To: Rotary motors in = aircraft=20
Sent: 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=20 x 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 = under)=20
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