Subject: Help neded
From: Rotary Engine
Date: 8/12/2007, 9:56 PM
To: AARotary Engine



    Does anybody know what units are in for air density in this formula
    if V is in feet per second. I'll settle for the metric units.

    I think it is in pounds per cubic inch which
    is .000046676 under standard conditions.


    Paul Lamar ...No rotor no motor.

 Paul,
The small p is density (rho). p = P/RT
Matt


Yes but what are the units? Pounds per cubic inch? Pounds per cubic foot?
etc. etc.


I suspect it is pounds per cubic inch if V is in feet per second.

Here is what I get if it is in pounds per cubic inch and V is in feet per second.
-- 
Paul Lamar ...No rotor no motor.

Paul,
I don't know if this will help any, but figures given to me recently were
that seal level mass/weight of 1 cu ft of Air is approx 0.076lbm/cu ft of
air.
Is that any help ?
George (down under)




I went out to the hangar and dug up my old physics book.

The complete equation is p1 + 1/2 x rho x v1^2 = p2 + 1/2 x rho x v2^2.

I recognized 1/2 X rho X v^2 as dynamic pressure so 1/2 X rho = .0026 if v is in MPH.
Dynamic pressure under standard conditions is 26 pounds per square foot
at sea level.

1/2 x rho is .0012 if v is in feet per second.

p is in pounds per square foot in both cases.

So I fixed the program.

p1 is 2160 pounds per square foot or 15 psi.

p2 = + p1 + .0012 X v1^2 - .0012 X v2^2

p2 in pounds per square foot.

If this is true this square law function is not a good match for
the linear air consumption curve which is a straight line function of RPM
and throttle opening. I am amazed carbs were built like this all those years.
No wonder they have so many extra circuits.

Manifold air pressure is a straight line function of air consumption so if a
mechanical way can be found to control the fuel delivery using manifold air pressure
a vast improvement on the traditional carb will take place.


Paul Lamar ...No rotor no motor.

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