Hi! Ken: Maybe I have a different physics book than the one you are
using.
Centripetal force = mass x velocity squared / radius F=mv2/r Now go
back and calculate the water pressure at the exit nozzle based on the
volume of water in the
arm. In a Clem conical drum which is composed of many Phi ratio spiral
tubes
there is considerable reactive thrust from the nozzles which drive the
rotation of
the unit while the centrifugal force is drawing the water into the hub.
Your lawn
sprinkler depends on the city water pressure of 60 PSI or more to
provide the
driving force to turn the head array. In an artificial gravity set up
such as the
Clem engine the radius of the unit is critical as is RPM to achieve the
necessary
pressures needed to drive the assembly beyond the point of frictional
losses
in the distribution tubes. The reason that the distribution tubes are
based on a Phi
ratio is to provide a practically free fall flow through the tubes as
envisioned by
Schauberger in his device. I believe, as does Jerry that there are
thermal factors
and viscosity variations in the oil that are lurking here also. These
are the factors
that we don't have a firm handle on until a unit is built and run. Norm
Ken Carrigan wrote:
> Just had to do this.. it's been bugging me as to how this exit
> forcefrom a centrfugal nozzle head could produce overunity. So here
> issome math (sorry Engineer here) that will explain what a
> Normal"sprinker head" would achieve in power. Forgot you said this..
> butthe raduis squared does come into play.. however.. still do notsee
> where the "kick over" occurs for over unity. OK.. make believe we have
> a sprinkler (twisted L shaped on eachside) which is feed water in the
> center of it. THis sprinker is turningat 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 laminerflow is occuring and no
> resistance is taken in to account (friction orheating). What we want
> to find is the power produced from thissprinkler. So..the moment
> around the z-axis.. will be the summationsof all (r xV)*p*V*A . Opps
> now "r" is radius, and "p" is density ofwater. 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... iszero cause the radius "r" is
> zero. Thus the equation reduces to onlyone sum.. or (r x V)*m. Now
> we have to take into account for therotational velocity and the
> velocity of the zozzle exit jet... in to thenormal atmosphere. Now we
> can subsitute r X V = -r*(Ve - w*r).Thus the moment applied to the
> system by the generator wouldbe -r*(Ve - W*r)*m. BUT the moment
> applied to the generator is thenegative of the moment of the generator
> on the arms. So the equationbecomes positive... no biggy! The power
> now is simple...one we have the moment.. It is the momenttime 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
> hasto be changed to RPS.. etc etc.. and don;t forget 2*pi for
> rotationalvelocity! v/r Ken Carrigan PS.. NOW... WHERE IS THE
> OVERUNITY!!!Help me understand...
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Hi! Ken: Maybe I have a different physics book than the one you areusing.
Ken Carrigan wrote:
Just hadto do this.. it's been bugging me as to how this exit forcefroma centrfugal nozzle head could produce overunity. So here issomemath (sorry Engineer here) that will explain what a Normal"sprinkerhead" would achieve in power. Forgot you said this.. buttheraduis squared does come into play.. however.. still do notseewhere the "kick over" occurs for over unity. OK..make believe we have a sprinkler (twisted L shaped on eachside)which is feed water in the center of it. THis sprinker is turningata 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,andthe area of the exit nozzle is "A". I think that now covers it.Assumptionshave to be made now.. and we concider that laminerflowis occuring and no resistance is taken in to account (friction orheating). What we want to find is the power produced from thissprinkler. So..the moment around the z-axis.. will be the summationsofall (r xV)*p*V*A . Opps now "r" is radius, and "p" is density ofwater. p*V*A is called the mass efflus of the nozzle.. or lets say "m"fornow. The moment going into the hub.. enterance of water... iszerocause the radius "r" is zero. Thus the equation reduces to onlyonesum.. or (r x V)*m. Now we have to take into account for therotationalvelocity and the velocity of the zozzle exit jet... in to thenormalatmosphere. Now we can subsitute r X V = -r*(Ve - w*r).Thusthe moment applied to the system by the generator wouldbe-r*(Ve - W*r)*m. BUT the moment applied to the generator isthenegative of the moment of the generator on thearms. So the equationbecomes positive... nobiggy! The power now is simple...one we havethe moment.. It is the momenttime the rotational velocity..(M = moment) or M*w = P for Power! Forall who would like the in to one equation .. here...P= (w*r*Ve - w*w*r*r)*m ENSURE.. that proper mksunits are used for example ... RPM hasto be changedto RPS.. etc etc.. and don;t forget 2*pi for rotationalvelocity! v/rKen Carrigan PS.. NOW... WHERE IS THE OVERUNITY!!!Helpme understand...
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