|
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)
|