Ed,
I can't remember how the BSFC
numbers are calculated but 153hp/2 =76.5 hp ( single rotor) which is
.56 BSFC ( 7.134x6/.56= 76.43).
SO these figures indicate that .56 =
14.7:1, Close enough to your figure of .55 BSFC.
Doing this for Best Power of
12.65.
177/2= 88.5Hp (single rotor) which is .484 (
7.134x6/.484=88.43)
If .55 BSFC = 14.71, then 12.65 A/F = .48 BSFC
( approx)
George ( down under)
George for a two rotor HP
My calculations shows a older 13B two rotor at 14.7
A/F ratio at 6000 rpm
RPM A/F
CFM Air
Lbm/min Fuel
Lbm/Min
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 -----
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) = 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)
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