X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from mail05.syd.optusnet.com.au ([211.29.132.186] verified) by logan.com (CommuniGate Pro SMTP 5.1.11) with ESMTPS id 2212623 for flyrotary@lancaironline.net; Mon, 30 Jul 2007 03:32:55 -0400 Received-SPF: none receiver=logan.com; client-ip=211.29.132.186; envelope-from=lendich@optusnet.com.au Received: from george (d211-31-248-169.dsl.nsw.optusnet.com.au [211.31.248.169]) by mail05.syd.optusnet.com.au (8.13.1/8.13.1) with SMTP id l6U7W8uS027324 for ; Mon, 30 Jul 2007 17:32:10 +1000 Message-ID: <005301c7d27b$c0c784e0$a9f81fd3@george> From: "George Lendich" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Fw: [FlyRotary] Air/fuel flow Date: Mon, 30 Jul 2007 17:32:13 +1000 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0050_01C7D2CF.9131C610" 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_0050_01C7D2CF.9131C610 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable OK ED, That's 7.134 gallons an hour ( Cruise at 6,000rpm) at 14.7:fuel ratio = at sea level for a single rotor or 14.268 for a 2 rotor, both at 100% = VE. SO 7.134 x 6( lb/gal)/0.6 ( BSFC) =3D 71 HP (single rotor) Wot ( 7,500) will be 8.975 gal an hour and 17.95 gal an hour for a 2 = rotor, at sea level, both at 100 VE. SO 8.975 x6 / 0.6=3D 89hp (single rotor) These numbers are not so good - am I using the wrong BSFC number (0.6 = for 14.7:1)? Should we also be looking for greater VE with the P-port? George (down under) 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_0050_01C7D2CF.9131C610 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
 
OK ED,
That's 7.134 gallons an hour ( Cruise = at=20 6,000rpm) at 14.7:fuel ratio at sea level for a single rotor or 14.268 = for a 2=20 rotor, both at 100% VE.
SO
7.134 x 6( lb/gal)/0.6 ( BSFC) =3D 71 = HP (single=20 rotor)
 
Wot ( 7,500) will be 8.975 gal an = hour and 17.95 gal an hour for a 2 rotor, at sea level, both at = 100=20 VE.
 
SO
8.975 x6 / 0.6=3D 89hp (single = rotor)
 
These numbers are not so good - am I = using the=20 wrong BSFC number (0.6 for 14.7:1)? Should we also be looking for = greater VE=20 with the P-port?
George (down under)
 
 
Well I'll be Ed,
2.4137sq" =3D (.877x.877xPi =3D = 2.416), therefore=20 .877 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=20 Gallons per minute

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

The reason = for this=20 is that a cubic foot of fuel will weigh the same everywhere but = a =20 cubic foot of air will weigh differently at different air densities=20 (altitudes).  I'll use the notation of Ft^3 (foot cube) for CF = and Ft^2=20 (foot squared) for Sq ft as it makes it easier to show where units = cancel=20 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=20 cancel out leaving you with units of pounds mass (lbm) per = minute). =20 Then taking your air/fuel ratio value of 14.7 we have 13.1936 /14.7 = =3D 0.8975=20 lbm/minute of fuel.

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

So 1 minute at your flow = rate we=20 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=20 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=20 which is 0.01676Ft^2  * 144 inch^2/Ft^2 =3D 2.4137 inch^2 of = opening for=20 that assumed velocity.

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

Hope that helped

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



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