Re: Clem... design and equations

Garry Whitman ( (no email) )
Sat, 12 Sep 98 21:39:09 PDT

Ken
For those of us who don't work with engineering equations all day could you
sometime do this problem with a hypothetical engine.
Say 24 one half inch nozzles, light oil, 60" Dia., 1800 RPM, 5000 Gallons
per minute. The hard part for me anyway is not understanding your equation
but keeping the units straight. When you do this all day is probably hard
to understand why people make it so hard but some of us think landing a
hang glider carrying two engines with a total wt. of 105 lb. on our feet at
26 MPH is a piece of cake. If you show the work I can take it from there to
test some ideas I heard thrown out.

Garry Whitman
whitman@kskc.net

----------
> Just had to do this.. it's been bugging me as to how this exit force
> from a centrfugal nozzle head could produce overunity. So here is
> some math (sorry Engineer here) that will explain what a Normal
> "sprinker head" would achieve in power. Forgot you said this.. but
> the raduis squared does come into play.. however.. still do not
> see where the "kick over" occurs for over unity.
>
> OK.. make believe we have a sprinkler (twisted L shaped on each
> side) which is feed water in the center of it. THis sprinker is turning
> at a velocity of "w" and the length of the arms (in the L forms) are
> "L". The exit velocity realitve to the nozzles are "Ve" for exit
velocity,
> and the area of the exit nozzle is "A". I think that now covers it.
> Assumptions have to be made now.. and we concider that laminer
> flow is occuring and no resistance is taken in to account (friction or
> heating). What we want to find is the power produced from this
> sprinkler. So..the moment around the z-axis.. will be the summations
> of all (r xV)*p*V*A . Opps now "r" is radius, and "p" is density of
> water. p*V*A is called the mass efflus of the nozzle.. or lets say "m"
> for now. The moment going into the hub.. enterance of water... is
> zero cause the radius "r" is zero. Thus the equation reduces to only
> one sum.. or (r x V)*m. Now we have to take into account for the
> rotational velocity and the velocity of the zozzle exit jet... in to the
> normal atmosphere. Now we can subsitute r X V = -r*(Ve - w*r).
> Thus the moment applied to the system by the generator would
> be -r*(Ve - W*r)*m. BUT the moment applied to the generator is the
> negative of the moment of the generator on the arms. So the equation
> becomes positive... no biggy!
>
> The power now is simple...one we have the moment.. It is the moment
> time the rotational velocity.. (M = moment) or M*w = P for Power!
>
> For all who would like the in to one equation .. here...
> P = (w*r*Ve - w*w*r*r)*m
>
> ENSURE.. that proper mks units are used for example ... RPM has
> to be changed to RPS.. etc etc.. and don;t forget 2*pi for rotational
> velocity!
>
> v/r Ken Carrigan
>
> PS.. NOW... WHERE IS THE OVERUNITY!!!
> Help me understand...
>