X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from ms-smtp-05.southeast.rr.com ([24.25.9.104] verified) by logan.com (CommuniGate Pro SMTP 5.1.11) with ESMTP id 2212930 for flyrotary@lancaironline.net; Mon, 30 Jul 2007 07:36:40 -0400 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.104; 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-05.southeast.rr.com (8.13.6/8.13.6) with SMTP id l6UBZrcf001553 for ; Mon, 30 Jul 2007 07:35:53 -0400 (EDT) Message-ID: <001901c7d29d$dfd38bf0$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Fw: [FlyRotary] Air/fuel flow Date: Mon, 30 Jul 2007 07:36:30 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0016_01C7D27C.587CCCA0" 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_0016_01C7D27C.587CCCA0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable George for a two rotor HP=20 My calculations shows a older 13B two rotor at 14.7 A/F ratio at 6000 = rpm=20 RPM A/F CFM Air Lbm/min = Fuel Lbm/Min HP 6000 14.7 272.2 20.86 1.42 153 =20 would produce approx 153 HP using a BSFC of 0.55=20 If you increase the A/F to "best power" of 12.65 then you get 6000 12.65 272.2 20.86 1.65 177 =20 177 HP for the same air flow. =20 Ed ----- Original Message -----=20 From: George Lendich=20 To: Rotary motors in aircraft=20 Sent: Monday, July 30, 2007 3:32 AM Subject: [FlyRotary] Re: Fw: [FlyRotary] Air/fuel flow 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_0016_01C7D27C.587CCCA0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
George for a two rotor HP
 
My calculations shows a older 13B two rotor at = 14.7=20 A/F  ratio at 6000 rpm
     RPM         =           A/F         =        =20 CFM     Air=20 Lbm/min   Fuel=20 Lbm/Min          = ;=20  HP

6000

 14.7

272.2

20.86

1.42

153

 
would produce approx 153 HP using a BSFC of 0.55
 
If you increase the A/F to "best power" of 12.65 then you get
 

6000

12.65

272.2

20.86

1.65

177

 
177 HP for the same air flow. 
 
Ed
----- Original Message -----
From:=20 George=20 Lendich
Sent: Monday, July 30, 2007 = 3:32 AM
Subject: [FlyRotary] Re: Fw: = [FlyRotary]=20 Air/fuel flow

 
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=20 2 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=20 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=20 ratio is based on mass ratio not volume ratio. 

The = reason=20 for this is that a cubic foot of fuel will weigh the same = everywhere but=20 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=20 CF and Ft^2 (foot squared) for Sq ft as it makes it easier to show = where=20 units cancel out.

At sea level the mass/weigth of 1 cubic = foot of=20 air is around 0.076 lbm/Ft^3 (of air).  So with 173.6 = Ft^3/Min =20 * 0.076 lbm/Ft^3  =3D 13.1936 lbm/minute of air(the two CFM = or Ft^3=20 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=20 13.1936 /14.7 =3D 0.8975 lbm/minute of fuel.

Various = figures are used=20 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=20 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=20 up this should be pretty close to the answer.

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

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

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

Hope that=20 helped

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



Lynn,
Following from previous post from = your self,=20 I'm looking at AIR FLOW and FUEL FLOW for my single rotor.
EG = [40 cu"=20 per face 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=20 minute?
2. how does one convert CFM to diameter of = inlet?
George (=20 down under) =
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