X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from mail08.syd.optusnet.com.au ([211.29.132.189] verified) by logan.com (CommuniGate Pro SMTP 5.1.11) with ESMTPS id 2212290 for flyrotary@lancaironline.net; Sun, 29 Jul 2007 23:36:57 -0400 Received-SPF: none receiver=logan.com; client-ip=211.29.132.189; envelope-from=lendich@optusnet.com.au Received: from george (d211-31-248-169.dsl.nsw.optusnet.com.au [211.31.248.169]) by mail08.syd.optusnet.com.au (8.13.1/8.13.1) with SMTP id l6U3aE8C024864 for ; Mon, 30 Jul 2007 13:36:15 +1000 Message-ID: <001801c7d25a$cbbded60$a9f81fd3@george> From: "George Lendich" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Air/fuel flow Date: Mon, 30 Jul 2007 13:36:19 +1000 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0015_01C7D2AE.9C8EB890" 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_0015_01C7D2AE.9C8EB890 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Thanks Ed and Lynn, I will go over those figures again, so I'm familiar with the process. What is the rule of thumb fuel burn for a 2 rotor at 6,000 RPM cruise as = a quick comparison? George ( down under) ----- Original Message -----=20 From: Ed Anderson=20 To: Rotary motors in aircraft=20 Sent: Monday, July 30, 2007 10:21 AM Subject: [FlyRotary] Re: Air/fuel flow 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_0015_01C7D2AE.9C8EB890 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Thanks Ed and Lynn,
I will go over those figures again, so = I'm familiar=20 with the process.
What is the rule of thumb fuel burn for = a 2=20 rotor at 6,000 RPM cruise as a quick comparison?
George ( down under)
 
 
 
 
 
 
----- Original Message -----
From:=20 Ed=20 Anderson
Sent: Monday, July 30, 2007 = 10:21=20 AM
Subject: [FlyRotary] Re: = Air/fuel=20 flow

George, I've done a bit of = this.  To get=20 an airflow in CFM to a fuel Flow in Gallons per minute

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

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

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

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

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

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

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

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

Again, if I haven't screwed up for an air velocity = through=20 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=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 ------=_NextPart_000_0015_01C7D2AE.9C8EB890--