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) = 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= 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" = (.877x.877xPi = 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)

  

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. 

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  = 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 = 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 = 0.149587 Gal/Min or in more conventional terms 0.149587 gal/min * 60 min/hour = 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 = 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 = 29.333 Ft^3/ Second

So solving for Area (of opening) = Volume/Length = 2.9333Ft^3/sec/ 175 Ft/Sec = .01676 Ft^2of area which is 0.01676Ft^2  * 144 inch^2/Ft^2 = 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 -----
From: George Lendich
To: Rotary motors in aircraft
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 = 173.6 CFM( Air)/ 14.7 ( Ratio)= 11.8 cfm ( Fuel)
40x3x2,000/1728=138CFM( Air)/14.7( ratio)=9.44cfm( 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)